Chapter 6 Probability Distribution (Q & A)

A continuous random variable is a variable whose value is obtained by measuring. It can take any value within a given range.

Let's analyze the options:

(A) Number of heads in 5 coin tosses: This is a discrete random variable because it can only take integer values (0, 1, 2, 3, 4, 5).

(B) Number of cars passing a point on a road in an hour: This is a discrete random variable because it represents a count and can only take non-negative integer values.

(C) The height of a randomly selected student: This is a continuous random variable because height can be measured and can take any value within a range (e.g., 1.5 meters, 1.51 meters, 1.512 meters, etc.).

(D) The number of defective items in a batch of 100: This is a discrete random variable as it represents a count of items and can only take integer values from 0 to 100.

Therefore, the correct answer is (C) The height of a randomly selected student.

In probability theory, for any discrete random variable X, the sum of the probabilities of all possible outcomes must equal 1. This is a fundamental property of probability distributions.

The probability mass function (PMF) of a discrete random variable X, denoted by $P(X=x)$, assigns a probability to each possible value $x$ that the random variable can take. The sum of these probabilities over all possible values of $x$ must satisfy the following condition:

This condition ensures that the total probability of all possible outcomes is accounted for, meaning that one of the possible outcomes must occur.

Let's examine the given options:

(A) 0: The sum of probabilities cannot be 0, as there is always a possibility that one of the outcomes will occur.

(B) 1: This aligns with the fundamental axiom of probability that the sum of probabilities of all mutually exclusive and exhaustive events must be 1.

(C) Any positive value less than 1: If the sum were less than 1, it would imply that there are other possible outcomes not accounted for, or that some probabilities are negative, which is not allowed.

(D) Any positive value: This is too broad; probabilities are bounded between 0 and 1, and their sum for a complete distribution must be exactly 1.

Therefore, for a discrete probability distribution, the sum of all probabilities $P(X=x)$ for all possible values of $x$ must be equal to 1.

For any discrete probability distribution, the sum of all probabilities for all possible values of the random variable must be equal to 1. This is a fundamental property of probability distributions.

The given probability distribution is:

$P(X=0) = 0.3$

$P(X=1) = k$

$P(X=2) = 0.4$

According to the property of probability distributions:

Substituting the given probabilities into equation (i):

Plugging in the values:

Combining the constant terms:

To find the value of $k$, subtract 0.7 from both sides of the equation:

Therefore, the value of $k$ is $0.3$.

The mathematical expectation, also known as the expected value or mean, of a discrete random variable $X$ is a measure of the central tendency of its probability distribution. It is calculated by summing the product of each possible value of the random variable and its corresponding probability.

For a discrete random variable $X$ that can take values $x_1, x_2, x_3, \dots, x_n$ with corresponding probabilities $P(X=x_1), P(X=x_2), P(X=x_3), \dots, P(X=x_n)$, the mathematical expectation $E(X)$ is defined as:

This formula essentially represents a weighted average of the possible values of $X$, where the weights are the probabilities of those values occurring.

Let's evaluate the given options based on this definition:

(A) $\sum x P(X=x)$: This formula directly matches the definition of the mathematical expectation of a discrete random variable.

(B) $\sum P(X=x)$: This represents the sum of all probabilities for a discrete random variable, which is always equal to 1. This is a property of probability distributions, not the expected value.

(C) $\sum x^2 P(X=x)$: This formula represents the expected value of $X^2$, denoted as $E(X^2)$, which is used to calculate variance, not the expected value of $X$ itself.

(D) $(\sum x P(X=x))^2$: This is the square of the expected value, $E(X)^2$, which is not the definition of the mathematical expectation.

Therefore, the correct formula for the mathematical expectation $E(X)$ of a discrete random variable $X$ is $\sum x P(X=x)$.

From Question 3, we have the following probability distribution for the random variable $X$:

We found that $k = 0.3$. So, the complete probability distribution is:

The expected value $E(X)$ of a discrete random variable is calculated using the formula:

Applying this formula to the given probability distribution:

$E(X) = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2))$

$E(X) = (0 \times 0.3) + (1 \times 0.3) + (2 \times 0.4)$

Now, let's calculate the terms:

$0 \times 0.3 = 0$

$1 \times 0.3 = 0.3$

$2 \times 0.4 = 0.8$

Summing these values:

Comparing this result with the given options:

(A) $0.3 \times 0 + 0.3 \times 1 + 0.4 \times 2 = 0.3 + 0.8 = 1.1$: This option has the probabilities in the wrong place (multiplied by the values of X instead of values of X multiplied by probabilities) and incorrect calculation for the first term, but the final sum is correct due to the value of k being correctly used in other options. However, the structure of the calculation is important. Let's re-examine the calculation for option A.

Let's re-evaluate option A: $0.3 \times 0$ (Incorrect order). The calculation should be $0 \times 0.3$. The second term $0.3 \times 1$ also has the incorrect order. The third term $0.4 \times 2$ also has the incorrect order. It appears the options might have the terms in a slightly different order, but let's check the structure of the calculation against the formula $E(X) = \sum x P(X=x)$.

(B) $0.3 \times 0 + 0.2 \times 1 + 0.4 \times 2 = 0 + 0.2 + 0.8 = 1.0$: This option uses $0.2$ for $P(X=1)$, which is incorrect. It should be $0.3$.

(C) $0 \times 0.3 + 1 \times 0.2 + 2 \times 0.4 = 0 + 0.2 + 0.8 = 1.0$: This option uses $0.2$ for $P(X=1)$, which is incorrect. It should be $0.3$.

(D) $0 \times 0.3 + 1 \times 0.3 + 2 \times 0.4 = 0 + 0.3 + 0.8 = 1.1$: This option correctly uses the values of $X$ multiplied by their respective probabilities and sums them up correctly.

Therefore, the correct calculation for the expected value $E(X)$ is $0 \times 0.3 + 1 \times 0.3 + 2 \times 0.4 = 1.1$.

The variance of a random variable is a measure of how spread out its distribution is. It quantifies the average squared difference of the random variable from its expected value.

For a discrete random variable $X$, the variance, denoted as Var($X$) or $\sigma^2$, can be calculated in a couple of ways. One of the most common and useful formulas for the variance is:

This definition represents the expected value of the squared deviation of $X$ from its mean $E(X)$.

By expanding the expression in equation (i), we arrive at an alternative and often more practical formula:

$E[(X - E(X))^2] = E[X^2 - 2X E(X) + (E(X))^2]$

Using the linearity of expectation:

$E[X^2] - E[2X E(X)] + E[(E(X))^2]$

Since $E(X)$ is a constant, $E[2X E(X)] = 2 E(X) E(X) = 2 [E(X)]^2$, and $E[(E(X))^2] = [E(X)]^2$.

So, the formula becomes:

Simplifying equation (ii):

Now let's compare this derived formula with the given options:

(A) $E(X^2) - [E(X)]^2$: This matches our derived formula (iii).

(B) $E(X) - [E(X^2)]^2$: This is incorrect.

(C) $E(X^2) + [E(X)]^2$: This is incorrect; the sign should be negative.

(D) $[E(X)]^2 - E(X^2)$: This is the negative of the correct variance formula.

Therefore, the variance Var($X$) of a discrete random variable $X$ is given by $E(X^2) - [E(X)]^2$.

From Question 3, we have the probability distribution for the random variable $X$ with $k=0.3$:

The expected value of $X^2$, denoted as $E(X^2)$, is calculated by summing the product of the square of each possible value of $X$ and its corresponding probability.

The formula is:

Applying this formula to the given probability distribution:

$E(X^2) = (0^2 \times P(X=0)) + (1^2 \times P(X=1)) + (2^2 \times P(X=2))$

$E(X^2) = (0^2 \times 0.3) + (1^2 \times 0.3) + (2^2 \times 0.4)$

Now, let's calculate the terms:

$0^2 \times 0.3 = 0 \times 0.3 = 0$

$1^2 \times 0.3 = 1 \times 0.3 = 0.3$

$2^2 \times 0.4 = 4 \times 0.4 = 1.6$

Summing these values:

Comparing this result with the given options:

(A) $0^2 \times 0.3 + 1^2 \times 0.2 + 2^2 \times 0.4 = 0 + 0.2 + 1.6 = 1.8$: This option uses $P(X=1)=0.2$, which is incorrect.

(B) $(0.3)^2 \times 0 + (0.2)^2 \times 1 + (0.4)^2 \times 2 = 0 + 0.04 + 0.32 = 0.36$: This option squares the probabilities and multiplies by the values of $X$, which is incorrect.

(C) $0 \times 0.3^2 + 1 \times 0.2^2 + 2 \times 0.4^2 = 0 + 0.04 + 0.32 = 0.36$: This option squares the probabilities and multiplies by the values of $X$, which is incorrect.

(D) $0^2 \times 0.3 + 1^2 \times 0.3 + 2^2 \times 0.4 = 0 + 0.3 + 1.6 = 1.9$: This option correctly uses the values of $X$ squared multiplied by their respective probabilities and sums them up correctly.

Therefore, the correct value for $E(X^2)$ is 1.9.

From the previous questions, we have determined the following for the probability distribution in Question 3:

The expected value, $E(X) = 1.1$

The expected value of $X^2$, $E(X^2) = 1.9$

The formula for the variance of a discrete random variable $X$ is:

Substitute the values of $E(X^2)$ and $E(X)$ into the formula:

$\text{Var}(X) = 1.9 - (1.1)^2$

Calculate the square of $E(X)$:

Now, substitute this value back into the variance formula:

Perform the subtraction:

Let's check the provided options. It appears there might be a slight discrepancy in the options or the expected value calculations in previous questions as presented in the options themselves. We will re-evaluate the options based on our calculations of $E(X) = 1.1$ and $E(X^2) = 1.9$.

(A) $1.8 - (1.0)^2 = 0.8$: This option uses an incorrect $E(X^2)$ value of $1.8$ and an incorrect $E(X)$ value of $1.0$.

(B) $1.8 - (1.1)^2 = 1.8 - 1.21 = 0.59$: This option uses an incorrect $E(X^2)$ value of $1.8$.

(C) $1.0 - 1.8 = -0.8$: This option uses incorrect values and results in a negative variance, which is impossible.

(D) $1.0^2 - 1.8 = 1 - 1.8 = -0.8$: This option uses incorrect values and results in a negative variance, which is impossible.

It seems that none of the options perfectly match our calculated variance of $0.69$, given our values of $E(X)=1.1$ and $E(X^2)=1.9$. Let's re-examine the calculations in the options to see if any of them, despite potentially incorrect intermediate values, lead to a correct final form of the variance calculation using *their own* provided intermediate values.

Let's assume one of the options has the correct structure and potentially uses incorrect intermediate values from prior steps, but the calculation shown within the option itself is what we need to evaluate for correctness of the *variance formula application*.

Option (B) shows the calculation as $1.8 - (1.1)^2$. The formula structure $E(X^2) - [E(X)]^2$ is correctly represented here if we consider $E(X^2) = 1.8$ and $E(X) = 1.1$. However, we found $E(X^2)$ to be $1.9$. If we were forced to choose based on the *structure of the formula application*, and assuming there was an error in calculating $E(X^2)$ leading to $1.8$, then option (B) would show the correct formula application.

Let's re-check our calculation for $E(X^2)$. $E(X^2) = (0^2 \times 0.3) + (1^2 \times 0.3) + (2^2 \times 0.4) = 0 + 0.3 + 1.6 = 1.9$. This calculation is correct.

Let's assume there's a typo in the question's options and one of them intended to use the correct values. If we use our correct values: Var(X) = $1.9 - (1.1)^2 = 1.9 - 1.21 = 0.69$.

None of the options yield $0.69$. However, let's look at the intermediate calculations presented within the options. The structure of the variance calculation is $E(X^2) - [E(X)]^2$.

• Option (A) uses $1.8$ for $E(X^2)$ and $1.0$ for $E(X)$.
• Option (B) uses $1.8$ for $E(X^2)$ and $1.1$ for $E(X)$.
• Option (C) uses $1.0$ for $E(X)$ and $1.8$ for $E(X^2)$.
• Option (D) uses $1.0$ for $E(X)$ and $1.8$ for $E(X^2)$.

Our calculated $E(X)$ is $1.1$. Options (B) uses $1.1$ for $E(X)$, which is correct. However, it uses $1.8$ for $E(X^2)$, which is incorrect (it should be $1.9$). The calculation shown in (B) is $1.8 - (1.1)^2 = 1.8 - 1.21 = 0.59$.

Given that the question asks for the variance and provides options with calculations, we need to identify the option that correctly applies the variance formula, even if the input values used within the option might be based on prior errors (as might be the case if this is a multi-part question where an earlier step had a mistake). The structure of the variance formula is $E(X^2) - [E(X)]^2$. Option (B) is the only one that correctly uses $E(X) = 1.1$ and follows the structure of the variance formula, even though it seems to have an incorrect value for $E(X^2)$ within the option's calculation.

If we assume there was a mistake in calculating $E(X^2)$ and it was mistakenly thought to be $1.8$, then option (B) would represent the correct application of the variance formula using that mistaken value. The calculation presented in option (B) is a valid application of the variance formula.

Let's assume that the intended answer relies on the structure of the formula and correct $E(X)$.

The formula is $\text{Var}(X) = E(X^2) - [E(X)]^2$.

We correctly found $E(X) = 1.1$ and $E(X^2) = 1.9$.

Our calculation: $\text{Var}(X) = 1.9 - (1.1)^2 = 1.9 - 1.21 = 0.69$.

Let's re-examine option (B) very carefully: $1.8 - (1.1)^2 = 1.8 - 1.21 = 0.59$. This option correctly uses $(1.1)^2$ and the structure $E(X^2) - [E(X)]^2$. The primary error is the value of $E(X^2)$ being stated as $1.8$ instead of $1.9$. However, the way the calculation is presented follows the correct formula application.

Given the options, it's highly probable that there was an error in providing the options or in the calculation of $E(X^2)$ that led to these options. However, based on the *structure of the calculation and the correct $E(X)$*, option (B) is the most plausible intended answer, assuming an error in the provided $E(X^2)$ value within the option itself.

A binomial distribution describes the number of successes in a fixed number of independent trials, where each trial has only two possible outcomes (success or failure) and the probability of success is constant for each trial.

In the context of tossing a fair coin:

• Trial: Each toss of the coin is an independent trial.
• Two possible outcomes: For each toss, the outcome can be either 'Heads' (success) or 'Tails' (failure).
• Constant probability of success: For a fair coin, the probability of getting a head (success) is $p = 0.5$, and the probability of getting a tail (failure) is $q = 1 - p = 0.5$. These probabilities remain constant for each toss.

The question states that the fair coin is tossed 6 times. This means that the number of times the experiment (coin toss) is performed is fixed.

In a binomial distribution, the "number of trials" is precisely this fixed number of independent experiments.

Therefore, in this case, the number of trials ($n$) is 6.

Let's evaluate the options:

(A) 2: This is the probability of getting a head on a fair coin (if we consider 'Heads' as success), but it's not the number of trials.

(B) 6: This directly corresponds to the number of times the coin is tossed, which is the fixed number of trials.

(C) 1/2: This is the probability of success (getting a head) or failure (getting a tail) in a single toss of a fair coin, not the number of trials.

(D) Varies: The number of trials in a binomial distribution must be fixed, not varying.

Hence, the number of trials in this binomial distribution is 6.

A binomial distribution, denoted as $B(n, p)$, is characterized by two parameters:

• $n$: The number of independent trials.
• $p$: The probability of success on a single trial.

For a binomial distribution, the mean (expected value) and variance have specific formulas derived from its probability mass function. The mean, often denoted as $\mu$ or $E(X)$, represents the average number of successes expected in $n$ trials.

The formula for the mean of a binomial distribution $B(n, p)$ is:

The variance, often denoted as $\sigma^2$ or Var($X$), is given by:

The standard deviation, denoted as $\sigma$, is the square root of the variance:

Now, let's evaluate the given options in relation to the mean:

(A) $np$: This is the correct formula for the mean of a binomial distribution.

(B) $np(1-p)$: This is the formula for the variance of a binomial distribution.

(C) $\sqrt{np(1-p)}$: This is the formula for the standard deviation of a binomial distribution.

(D) $n/p$: This is not a standard formula associated with the mean, variance, or standard deviation of a binomial distribution.

Therefore, in a binomial distribution $B(n, p)$, the mean is given by $np$.

A binomial distribution, denoted as $B(n, p)$, describes the probability of obtaining a certain number of successes in a fixed number of independent Bernoulli trials, where each trial has the same probability of success.

The parameters of a binomial distribution are:

• $n$: The number of trials.
• $p$: The probability of success on a single trial.

The key statistical measures for a binomial distribution are its mean (expected value) and variance.

The mean of a binomial distribution $B(n, p)$ is given by:

The variance of a binomial distribution $B(n, p)$ measures the spread of the distribution around the mean. It is given by:

The standard deviation is the square root of the variance:

Let's examine the provided options:

(A) $np$: This is the formula for the mean of a binomial distribution.

(B) $np(1-p)$: This is the correct formula for the variance of a binomial distribution.

(C) $\sqrt{np(1-p)}$: This is the formula for the standard deviation of a binomial distribution.

(D) $n/p$: This is not a standard formula for the variance of a binomial distribution.

Therefore, in a binomial distribution $B(n, p)$, the variance is given by $np(1-p)$.

The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event.

The parameter of the Poisson distribution is $\lambda$ (lambda), which represents both the mean and the rate parameter.

The probability mass function (PMF) of a Poisson distribution is given by:

For a Poisson distribution, the mean (expected value) and the variance have specific and well-known formulas:

Mean: The mean of a Poisson distribution is equal to its parameter $\lambda$.

Variance: The variance of a Poisson distribution is also equal to its parameter $\lambda$.

Now, let's compare these results with the given options:

(A) Mean = $\lambda$, Variance = $\lambda$: This option correctly states both the mean and the variance of a Poisson distribution.

(B) Mean = $\lambda^2$, Variance = $\lambda$: The mean is incorrect.

(C) Mean = $\lambda$, Variance = $\sqrt{\lambda}$: The variance is incorrect (it is the standard deviation).

(D) Mean = $\sqrt{\lambda}$, Variance = $\lambda$: The mean is incorrect.

Therefore, the correct answer is that the mean and variance of a Poisson distribution with parameter $\lambda$ are both equal to $\lambda$.

The problem states that the number of calls received at a call centre per minute follows a Poisson distribution. The mean number of calls per minute is given as 3.

For a Poisson distribution, the parameter $\lambda$ represents the mean number of events in a given interval. In this case:

• Mean number of calls per minute, $\lambda = 3$.

We are asked to find the probability of receiving exactly 2 calls in a minute. In the context of the Poisson distribution, this means we need to find $P(X=x)$ where $x$ is the number of events we are interested in.

• Number of calls we are interested in, $x = 2$.

The probability mass function (PMF) for a Poisson distribution is:

To find the probability of receiving exactly 2 calls in a minute, we substitute $\lambda = 3$ and $x = 2$ into the PMF:

$P(X=2) = \frac{e^{-3} 3^2}{2!}$

Now, let's compare this result with the given options:

(A) $\frac{e^{-3} 3^2}{2!}$: This option correctly uses $\lambda = 3$ and $x = 2$ in the Poisson PMF formula.

(B) $\frac{e^{-2} 3^2}{2!}$: This option incorrectly uses $e^{-\lambda}$ as $e^{-2}$ instead of $e^{-3}$.

(C) $\frac{e^{-3} 2^3}{3!}$: This option incorrectly uses $\lambda^x$ as $2^3$ and $x!$ as $3!$. It should be $\lambda^x$ as $3^2$ and $x!$ as $2!$.

(D) $\frac{e^{-2} 2^3}{3!}$: This option incorrectly uses $e^{-\lambda}$ as $e^{-2}$, $\lambda^x$ as $2^3$, and $x!$ as $3!$.

Therefore, the probability of receiving exactly 2 calls in a minute is given by $\frac{e^{-3} 3^2}{2!}$.

Let's analyze the Assertion (A) and Reason (R) separately.

Assertion (A): The sum of the probabilities in any probability distribution is 1.

This statement is a fundamental axiom of probability. For any probability distribution (whether discrete or continuous), the sum (or integral, for continuous distributions) of the probabilities over all possible outcomes must equal 1. This ensures that there is a certainty that one of the possible outcomes will occur.

Thus, Assertion (A) is true.

Reason (R): The sample space of a random experiment covers all possible outcomes, and the sum of probabilities of all possible outcomes is always 1.

The sample space ($S$) of a random experiment is the set of all possible outcomes. The reason correctly states that the sample space covers all possible outcomes. Furthermore, it correctly states that the sum of the probabilities of all these mutually exclusive and exhaustive outcomes is always 1. This is the very definition and a foundational rule for constructing probability distributions.

Thus, Reason (R) is true.

Now, we need to determine if Reason (R) correctly explains Assertion (A).

Assertion (A) is a statement about probability distributions. Reason (R) explains *why* this is the case by referring to the definition of the sample space and the rules governing probabilities assigned to its elements. The fact that the sample space encompasses all possibilities, and that probabilities are assigned such that their sum over all these possibilities is 1, is the underlying principle that makes Assertion (A) true.

Therefore, Reason (R) provides the correct theoretical basis for Assertion (A).

Considering both A and R are true, and R correctly explains A, the correct option is (A).

The normal distribution, also known as the Gaussian distribution or bell curve, is a fundamental probability distribution in statistics. It is defined by two parameters: its mean ($\mu$) and its standard deviation ($\sigma$).

Let's examine the characteristics mentioned in the options:

• Skewness: Skewness refers to the asymmetry of a probability distribution. A distribution skewed to the right has a longer tail on the right side, and a distribution skewed to the left has a longer tail on the left side.
• Symmetry: Symmetry means that the distribution can be divided into two mirror-image halves.
• Modality: Modality refers to the number of peaks in the distribution. A unimodal distribution has one peak, a bimodal distribution has two peaks, and so on.

Now, let's evaluate each option in the context of a normal distribution:

(A) It is always skewed to the right: This is incorrect. A normal distribution is symmetric, not skewed.

(B) It is always skewed to the left: This is also incorrect. A normal distribution is symmetric, not skewed.

(C) It is symmetric about its mean: This is a key characteristic of the normal distribution. The probability density function (PDF) of a normal distribution is symmetric around its mean ($\mu$). This means that the mean, median, and mode are all equal in a normal distribution.

(D) It is bimodal: This is incorrect. A normal distribution is unimodal, meaning it has a single peak at its mean.

Therefore, the defining characteristic of a normal distribution among the given options is that it is symmetric about its mean.

A normal distribution is defined by its mean ($\mu$) and standard deviation ($\sigma$). The mean determines the center of the distribution, and the standard deviation determines the spread or width of the distribution.

A standard normal distribution is a special case of the normal distribution where the mean and standard deviation are standardized to specific values. This standardization is useful for comparing different normal distributions and for using standard normal tables (Z-tables) to find probabilities.

By definition, a standard normal distribution has:

• A mean ($\mu$) of 0.
• A standard deviation ($\sigma$) of 1.

The probability density function (PDF) for the standard normal distribution is often denoted by $f(z)$, where $z$ represents a standard normal random variable, and it is given by:

In this formula, the value 0 is implicitly the mean and the value 1 (as part of $z^2/2$) is implicitly related to the standard deviation.

Let's look at the options:

(A) Mean = 0, Standard Deviation = 1: This aligns with the definition of a standard normal distribution.

(B) Mean = 1, Standard Deviation = 0: A standard deviation of 0 would mean all values are the same, which is not a normal distribution.

(C) Mean = 0, Standard Deviation = 0: Similar to (B), a standard deviation of 0 is not characteristic of a normal distribution.

(D) Mean = 1, Standard Deviation = 1: This describes a normal distribution, but not the *standard* normal distribution, which specifically has a mean of 0.

Therefore, for a standard normal distribution, the mean is 0 and the standard deviation is 1.

The normal distribution, also known as the Gaussian distribution, is a continuous probability distribution. Its shape is determined by two parameters: the mean ($\mu$), which indicates the center of the distribution, and the standard deviation ($\sigma$), which indicates the spread or dispersion of the data.

The probability density function (PDF) of a normal random variable $X$ with mean $\mu$ and standard deviation $\sigma$ is given by the formula:

In this formula:

• $x$ is the value of the random variable.
• $\mu$ is the mean of the distribution.
• $\sigma$ is the standard deviation of the distribution.
• $\pi$ is the mathematical constant pi (approximately 3.14159).
• $e$ is the base of the natural logarithm (approximately 2.71828).
• The term $\left(\frac{x-\mu}{\sigma}\right)$ is the z-score, which represents the number of standard deviations $x$ is from the mean.
• The term $\frac{1}{\sigma \sqrt{2\pi}}$ is a normalization constant that ensures the total probability (the area under the curve) is equal to 1.

Now, let's compare this with the given options:

(A) $f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$: This formula exactly matches the probability density function of a normal distribution.

(B) $f(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$: This is the PDF for a standard normal distribution (where $\sigma=1$), but it is not the general form for any normal distribution.

(C) $f(x) = e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$: This is missing both the normalization constant and the $1/\sigma$ term.

(D) $f(x) = \frac{1}{\sigma} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$: This is missing the $\sqrt{2\pi}$ in the normalization constant.

Therefore, the correct probability density function for a normal random variable $X$ with mean $\mu$ and standard deviation $\sigma$ is $f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$.

In probability theory, for a continuous random variable $X$, probabilities are associated with intervals rather than specific points. The probability of $X$ falling within a certain range $[a, b]$ is determined by the area under the curve of its probability density function (PDF), $f(x)$, between the points $a$ and $b$.

The relationship between the probability density function $f(x)$ and the cumulative distribution function (CDF), $F(x)$, is fundamental:

• The CDF, $F(x)$, is the integral of the PDF from $-\infty$ to $x$: $F(x) = \int\limits_{-\infty}^{x} f(t) dt$.
• Conversely, the PDF is the derivative of the CDF: $f(x) = F'(x)$.

The probability that a continuous random variable $X$ falls between two values $a$ and $b$ (inclusive or exclusive, since the probability of a single point is zero for continuous distributions) is given by the definite integral of the PDF from $a$ to $b$.

Therefore, $P(a \leq X \leq b)$ can be calculated as:

Using the Fundamental Theorem of Calculus, this integral can also be expressed in terms of the CDF:

$P(a \leq X \leq b) = F(b) - F(a)$

This is because $F(b) = \int\limits_{-\infty}^{b} f(x) dx$ and $F(a) = \int\limits_{-\infty}^{a} f(x) dx$. Subtracting these gives $\int\limits_{a}^{b} f(x) dx$.

Let's examine the given options:

(A) $F(b) - F(a)$: This is a correct way to calculate the probability using the CDF.

(B) $\int\limits_{a}^{b} f(x) dx$: This is also a correct way to calculate the probability using the PDF.

(C) $\int\limits_{a}^{b} F(x) dx$: This is incorrect. Integrating the CDF does not give the probability of the variable being in the interval $[a, b]$.

(D) Both (A) and (B): Since both (A) and (B) are correct methods for calculating the probability, this option is the most comprehensive answer.

Therefore, the probability $P(a \leq X \leq b)$ for a continuous probability distribution can be given by both $F(b) - F(a)$ and $\int\limits_{a}^{b} f(x) dx$.

In probability theory, the probability of any event, or the probability assigned to any outcome of a random variable, must satisfy certain fundamental properties:

1. Probabilities are non-negative: For any outcome $x$, $P(X=x) \geq 0$.
2. The sum of probabilities over all possible outcomes is 1: $\sum P(X=x) = 1$.

The question asks which of the given values is NOT a valid probability for $P(X=x)$ in a discrete probability distribution.

Let's examine each option based on the properties of probability:

(A) $0.5$: This is a value between 0 and 1, so it is a valid probability.

(B) $1.0$: This is the maximum possible probability, indicating certainty, so it is a valid probability.

(C) $-0.2$: This value is negative. Probabilities cannot be negative. This violates the first fundamental property of probability.

(D) $0.0$: This is the minimum possible probability, indicating an impossible event, so it is a valid probability.

Therefore, the value that is NOT a valid probability for $P(X=x)$ is $-0.2$.

This scenario can be modeled using a binomial distribution. When a fair coin is tossed, there are two possible outcomes for each toss: heads or tails.

Let $X$ be the random variable representing the number of heads obtained.

The parameters for this binomial distribution are:

• Number of trials ($n$): The coin is tossed 100 times, so $n = 100$.
• Probability of success ($p$): For a fair coin, the probability of getting a head (which we define as 'success') in a single toss is $p = 0.5$.

The expected number of successes (in this case, heads) in a binomial distribution is given by the formula:

Substitute the values of $n$ and $p$ into the formula:

$E(X) = 100 \times 0.5$

Calculate the result:

Therefore, the expected number of heads when a fair coin is tossed 100 times is 50.

Let's check the options:

(A) 25

(B) 50

(C) 75

(D) 100

Our calculated expected value matches option (B).

For a binomial random variable $X$ with parameters $n$ (number of trials) and $p$ (probability of success), the mean and variance are given by the following formulas:

Mean: $E(X) = np$

Variance: $\text{Var}(X) = np(1-p)$

We are given that the mean is 4 and the variance is 2.

So, we have the following system of equations:

1. $np = 4$
2. $np(1-p) = 2$

We can use the first equation to substitute $np$ in the second equation:

$4(1-p) = 2$

Now, solve for $p$:

$1-p = \frac{2}{4}$

$1-p = 0.5$

$p = 1 - 0.5$

$p = 0.5$

Now that we have the value of $p$, we can substitute it back into the first equation ($np = 4$) to find $n$:

$n \times 0.5 = 4$

$n = \frac{4}{0.5}$

$n = 8$

So, the parameters are $n=8$ and $p=0.5$.

Let's check the given options:

(A) $n=8, p=0.5$: This matches our calculated values.

(B) $n=10, p=0.4$: Mean = $10 \times 0.4 = 4$. Variance = $10 \times 0.4 \times (1-0.4) = 4 \times 0.6 = 2.4$. This does not match the variance.

(C) $n=16, p=0.25$: Mean = $16 \times 0.25 = 4$. Variance = $16 \times 0.25 \times (1-0.25) = 4 \times 0.75 = 3$. This does not match the variance.

(D) $n=2, p=2$: The probability $p$ cannot be greater than 1, so this option is invalid.

Therefore, the correct parameters are $n=8$ and $p=0.5$.

The problem describes a scenario where an event (a person being affected by a rare disease) occurs with a low probability over a large number of trials. This type of situation can be approximated by a Poisson distribution.

The parameters given are:

• The probability of a person being affected by the disease ($p$) = 1 in 1000, so $p = \frac{1}{1000} = 0.001$.
• The size of the random sample ($n$) = 2000 people.

For a situation that can be approximated by a Poisson distribution, the parameter $\lambda$ (the mean number of events) is calculated as the product of the number of trials and the probability of success in each trial:

Substitute the given values into the formula:

$\lambda = 2000 \times \frac{1}{1000}$

Calculate the value of $\lambda$:

Thus, the value of $\lambda$ for this Poisson distribution is 2.

Let's compare this with the given options:

(A) 1

(B) 2

(C) 20

(D) 0.5

Our calculated value matches option (B).

The normal distribution, characterized by its mean ($\mu$) and standard deviation ($\sigma$), has a well-defined empirical rule (also known as the 68-95-99.7 rule) that describes the proportion of data falling within certain ranges around the mean.

The empirical rule states the following approximate percentages of data within specific intervals:

• Approximately 68% of the data falls within one standard deviation of the mean, i.e., between $\mu - \sigma$ and $\mu + \sigma$.
• Approximately 95% of the data falls within two standard deviations of the mean, i.e., between $\mu - 2\sigma$ and $\mu + 2\sigma$.
• Approximately 99.7% of the data falls within three standard deviations of the mean, i.e., between $\mu - 3\sigma$ and $\mu + 3\sigma$.

The question asks for the interval within which approximately 68% of the data falls.

Based on the empirical rule, this interval is:

Let's compare this with the given options:

(A) $[\mu - \sigma, \mu + \sigma]$: This matches our understanding of the empirical rule for 68% of the data.

(B) $[\mu - 2\sigma, \mu + 2\sigma]$: This interval covers approximately 95% of the data.

(C) $[\mu - 3\sigma, \mu + 3\sigma]$: This interval covers approximately 99.7% of the data.

(D) $[\mu, \mu + \sigma]$: This interval covers approximately half of the 68%, which is about 34%, and represents the data from the mean to one standard deviation above the mean.

Therefore, approximately 68% of the data in a normal distribution falls within the interval $[\mu - \sigma, \mu + \sigma]$.

The variance of a random variable $X$, denoted as Var($X$), measures the spread of the distribution of $X$. It is related to the expected value of $X$ ($E(X)$) and the expected value of $X^2$ ($E(X^2)$) by a fundamental formula.

The formula for the variance is:

We are given the following values:

• $E(X) = 5$
• $E(X^2) = 30$

Substitute these values into the variance formula:

$\text{Var}(X) = 30 - (5)^2$

Now, calculate the square of $E(X)$:

Substitute this result back into the variance calculation:

Perform the subtraction:

Now, let's check the options:

(A) 5: This is our calculated variance.

(B) 25: This is $[E(X)]^2$, not the variance.

(C) 30: This is $E(X^2)$, not the variance.

(D) $30 - 5^2 = 5$: This option correctly shows the application of the variance formula and its result.

Both option (A) and option (D) provide the correct numerical answer for the variance. However, option (D) explicitly shows the calculation steps, which is more informative. Given the format of the questions, it's likely that showing the calculation is preferred.

In probability theory, there are fundamental axioms that govern the assignment of probabilities to events or outcomes of a random variable. For a discrete random variable $X$, these axioms are crucial for defining a valid probability distribution.

The two primary axioms for probabilities are:

1. Non-negativity: The probability of any outcome must be non-negative. That is, for any possible value $x$ that the random variable $X$ can take, $P(X=x) \geq 0$. This is stated in the question.
2. Normalization: The sum of the probabilities of all possible mutually exclusive outcomes must equal 1. This is because one of the possible outcomes is certain to occur.

The question states the first axiom ($P(X=x) \geq 0$ for all $x$) and asks to complete the second axiom regarding the sum of probabilities:

$\sum_{x} P(X=x) = \_\_\_\_$

Here, the summation $\sum_{x}$ implies summing over all possible values that the discrete random variable $X$ can take.

According to the normalization axiom of probability, this sum must always be equal to 1.

Therefore, $\sum_{x} P(X=x) = 1$.

Let's look at the options:

(A) $0$: The sum of probabilities cannot be 0, as it would imply that no outcomes are possible.

(B) $1$: This correctly represents the sum of probabilities over all possible outcomes.

(C) $\infty$: While probabilities are non-negative, their sum is constrained to 1, not infinity.

(D) Depends on $x$: The sum of probabilities over all possible values of $x$ is a constant value (1), and it does not depend on a specific value of $x$.

Hence, the correct completion of the statement is that the sum of probabilities for all possible values of a discrete random variable is 1.

A binomial distribution models a sequence of independent experiments that meet specific criteria. Let's review the necessary conditions for a binomial distribution:

1. Fixed Number of Trials ($n$): The experiment consists of a predetermined, fixed number of trials.
2. Independent Trials: Each trial must be independent of the others. The outcome of one trial does not affect the outcome of any other trial.
3. Two Outcomes per Trial: Each trial must have only two possible outcomes, typically referred to as "success" and "failure".
4. Constant Probability of Success ($p$): The probability of success must be the same for every trial.

Now let's evaluate each option based on these conditions:

(A) The trials are independent: This is a necessary condition for a binomial distribution.

(B) There are exactly two outcomes for each trial: This is also a necessary condition, often called a Bernoulli trial.

(C) The probability of success changes from trial to trial: This statement contradicts the fourth necessary condition, which requires the probability of success ($p$) to be constant for all trials. If the probability of success changes, it is not a binomial distribution.

(D) The number of trials is fixed: This is the first necessary condition for a binomial distribution.

Therefore, the condition that is NOT necessary for a binomial distribution is that the probability of success changes from trial to trial.

In a binomial experiment, there are only two possible outcomes for each trial: success or failure.

Let $p$ be the probability of success in a single trial, and $q$ be the probability of failure in a single trial.

We need to evaluate each statement to determine which one is false:

(A) $p + q = 1$: Since success and failure are the only two possible outcomes for each trial, their probabilities must add up to 1. This is a fundamental property of probabilities for exhaustive and mutually exclusive events. So, this statement is true.

(B) $p \ge 0$: Probability values cannot be negative. The smallest possible probability is 0, representing an impossible event. So, this statement is true.

(C) $q \ge 0$: Similarly, the probability of failure cannot be negative. So, this statement is true.

(D) $p \times q = 1$: Let's consider a common scenario, like a fair coin toss, where $p = 0.5$ (for heads) and $q = 0.5$ (for tails). In this case, $p \times q = 0.5 \times 0.5 = 0.25$, which is not equal to 1. For any probability $p$ between 0 and 1 (exclusive), $q = 1-p$. The product $p(1-p)$ is maximized when $p=0.5$, giving $0.25$. If $p=0$ or $p=1$, the product is 0. Thus, $p \times q$ can never be 1 unless $p$ or $q$ is not a valid probability (e.g., if $p=1$ and $q=1$, which is impossible as they must sum to 1). So, this statement is false.

Therefore, the false statement is $p \times q = 1$.

The notation $X \sim B(n, p)$ means that $X$ is a binomial random variable with $n$ trials and a probability of success $p$ on each trial.

In this question, $X \sim B(10, 0.2)$, which means:

• Number of trials, $n = 10$.
• Probability of success, $p = 0.2$.

The probability of failure in a single trial is $q = 1 - p = 1 - 0.2 = 0.8$.

The probability mass function (PMF) for a binomial distribution is given by:

We need to find $P(X=0)$. This means we are looking for the probability of getting exactly 0 successes in 10 trials.

Using the PMF with $k=0$, $n=10$, $p=0.2$, and $q=0.8$:

$P(X=0) = \binom{10}{0} (0.2)^0 (0.8)^{10-0}$

Let's evaluate the terms:

• $\binom{10}{0} = 1$ (The number of ways to choose 0 items from 10 is 1).
• $(0.2)^0 = 1$ (Any non-zero number raised to the power of 0 is 1).
• $(0.8)^{10-0} = (0.8)^{10}$.

Substituting these values back into the equation:

$P(X=0) = 1 \times 1 \times (0.8)^{10}$

$P(X=0) = (0.8)^{10}$

Now, let's check the options:

(A) $(0.8)^{10}$: This matches our calculated probability.

(B) $(0.2)^{10}$: This would be the probability of getting 10 successes ($P(X=10)$).

(C) $1 - (0.8)^{10}$: This would represent the probability of getting at least one success ($P(X \geq 1)$).

(D) $1 - (0.2)^{10}$: This is not directly interpretable in terms of the basic binomial probabilities.

Therefore, $P(X=0)$ is $(0.8)^{10}$.

The Poisson distribution is a discrete probability distribution that models the number of events occurring in a fixed interval of time or space, given a constant average rate ($\lambda$). As the mean $\lambda$ increases, the shape of the Poisson distribution changes.

Let's consider the behavior of the Poisson distribution for different values of $\lambda$:

• Small $\lambda$ (e.g., $\lambda < 5$): The distribution is highly skewed to the right, with the probability concentrated near 0.
• Moderate $\lambda$: The distribution becomes less skewed and starts to look more symmetric.
• Large $\lambda$ (e.g., $\lambda > 10$ or $\lambda > 20$): The distribution becomes very symmetric and bell-shaped. It closely approximates a normal distribution.

This approximation is a result of the Central Limit Theorem, which states that the sum (or average) of a large number of independent and identically distributed random variables, regardless of their original distribution, tends to be normally distributed. Since a Poisson random variable can be seen as the sum of many independent Bernoulli random variables, its distribution approaches normal as the number of underlying events (and thus $\lambda$) increases.

Now let's look at the options:

(A) Binomial distribution: While there are relationships between Poisson and Binomial (e.g., Poisson can approximate Binomial for large $n$ and small $p$), the Poisson distribution itself doesn't *resemble* Binomial as $\lambda$ increases; rather, it approximates another distribution.

(B) Uniform distribution: A uniform distribution has equal probabilities over a range, which is not the shape the Poisson distribution approaches.

(C) Exponential distribution: The exponential distribution models the time between events in a Poisson process, not the count of events itself. The Poisson and exponential distributions are related, but the shape of the Poisson distribution for large $\lambda$ does not resemble an exponential distribution.

(D) Normal distribution: As explained above, for large values of $\lambda$, the Poisson distribution becomes approximately symmetric and bell-shaped, closely resembling a normal distribution.

Therefore, as the mean $\lambda$ increases, the Poisson distribution tends to resemble a normal distribution.

A Poisson distribution is used to model the number of events occurring in a fixed interval of time or space, given a constant average rate of occurrence, and assuming that these events are independent. Key characteristics suitable for a Poisson model are:

• The events occur randomly and independently.
• The average rate of events ($\lambda$) is constant over the interval.
• The probability of an event occurring is proportional to the length of the interval.
• The probability of two or more events occurring in a very short interval is negligible.

Let's analyze each scenario:

(A) The number of defective items in a sample of 100 from a large production line: This is typically modelled by a binomial distribution if the sample size is fixed and the probability of defect is constant. If the population is very large and the defect rate is very low, it can be approximated by a Poisson distribution.

(B) The height of adults in a country: Height is a continuous variable that is typically measured, not counted. Distributions like the normal distribution are used to model continuous measurements like height.

(C) The number of road accidents on a specific stretch of highway in a month: This scenario fits the criteria for a Poisson distribution well.

• The events (accidents) occur over a fixed interval (a month).
• Accidents are generally considered to occur randomly and independently.
• There's an average rate of accidents per month ($\lambda$).

(D) The outcome of rolling a die 50 times: Each roll of a die has multiple outcomes (1, 2, 3, 4, 5, 6). If we are interested in the number of times a specific outcome occurs (e.g., rolling a '6'), it would be a binomial distribution if we consider 'rolling a 6' as success and 'not rolling a 6' as failure. If we are interested in the sequence of outcomes, other models might apply.

Comparing the scenarios, the number of road accidents on a specific stretch of highway in a month (C) is the most direct and appropriate fit for a Poisson distribution because it involves counting the number of random, independent events occurring over a fixed interval.

A standard normal variable, denoted by $Z$, is a normal random variable with a mean ($\mu$) of 0 and a standard deviation ($\sigma$) of 1. Its probability density function (PDF) is symmetric about its mean.

The probability $P(Z \le 0)$ represents the probability that a standard normal random variable takes a value less than or equal to 0. Graphically, this corresponds to the area under the standard normal curve to the left of 0.

Since the standard normal distribution is symmetric about its mean, which is 0, the mean divides the distribution exactly in half. This means that the area under the curve to the left of the mean is equal to the area under the curve to the right of the mean. The total area under the curve is 1.

Therefore:

• The area to the left of the mean (0) is 0.5.
• The area to the right of the mean (0) is 0.5.

So, $P(Z \le 0) = 0.5$.

We can also express this using the cumulative distribution function (CDF) of the standard normal distribution, often denoted as $\Phi(z)$. The CDF gives $P(Z \le z)$. For the standard normal distribution, $\Phi(0)$ is exactly 0.5.

Looking at the options:

(A) 0: This would imply that the probability is concentrated entirely at values less than 0, which is incorrect for a continuous distribution centered at 0.

(B) 0.5: This is the correct probability, representing half of the total probability, due to the symmetry of the standard normal distribution about its mean.

(C) 1: This would imply that all values are less than or equal to 0, which is incorrect for a distribution that extends to positive values.

(D) Cannot be determined: The probability can be determined from the properties of the standard normal distribution.

Therefore, $P(Z \le 0)$ is 0.5.

The Z-score is a statistical measure that describes a value's relationship to the mean of a group of values, measured in terms of standard deviation from the mean. For a normal random variable $X$ with mean $\mu$ and standard deviation $\sigma$, the Z-score of a particular value $x$ is calculated using the formula:

In this problem, we are given:

• The normal random variable is $X$.
• The mean, $\mu = 10$.
• The standard deviation, $\sigma = 2$.
• The specific value of $X$ for which we need to find the Z-score is $x = 12$.

Substitute these values into the Z-score formula:

$Z = \frac{12 - 10}{2}$

Now, perform the calculations:

$Z = \frac{2}{2}$

The Z-score for $X=12$ is 1. This means that the value 12 is exactly one standard deviation above the mean.

Let's check the options:

(A) 1: This matches our calculated Z-score.

(B) 2: This is the standard deviation itself, not the Z-score for $X=12$.

(C) 0: This would be the Z-score for the mean value itself ($X=10$).

(D) -1: This would be the Z-score for a value one standard deviation below the mean ($X=10-2=8$).

Therefore, the Z-score for $X=12$ is 1.

Let's match each distribution with its typical application:

(i) Binomial Distribution: This distribution is used to model the number of successes in a fixed number of independent trials, where each trial has only two possible outcomes (success or failure) with a constant probability of success.
Typical application: (b) Probability of getting a specific number of successes in fixed independent trials.

(ii) Poisson Distribution: This distribution models the number of events occurring in a fixed interval of time or space, given a constant average rate of occurrence.
Typical application: (c) Probability of a specific number of events occurring in a fixed interval of time/space.

(iii) Normal Distribution: This distribution is a continuous probability distribution that is symmetric about its mean. It is often used to model natural phenomena where values cluster around a central mean.
Typical application: (a) Modelling heights or weights.

(iv) Discrete Probability Distribution: This is a general term for distributions where the random variable can only take a finite number of values or a countably infinite number of values. It encompasses distributions like Binomial, Poisson, Geometric, etc.
Typical application: (d) Representing probabilities for a finite or countably infinite set of outcomes.

Based on these matches, we have:

• (i) matches with (b)
• (ii) matches with (c)
• (iii) matches with (a)
• (iv) matches with (d)

This corresponds to the pairing (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d).

Let's check the options:

(A) (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d): This option matches our derived pairings.

(B) (i)-(c), (ii)-(b), (iii)-(a), (iv)-(d): Incorrect for (i) and (ii).

(C) (i)-(b), (ii)-(a), (iii)-(c), (iv)-(d): Incorrect for (ii) and (iii).

(D) (i)-(a), (ii)-(c), (iii)-(b), (iv)-(d): Incorrect for (i) and (iii).

Therefore, the correct matching is provided in option (A).

The expected value of a random variable is a linear operator. This means that for any constants $a$ and $b$, and any random variable $X$, the expected value of a linear transformation of $X$ has specific properties.

The linearity property of expectation states that:

1. The expected value of a constant times a random variable is the constant times the expected value of the random variable: $E(aX) = a E(X)$.
2. The expected value of a random variable plus a constant is the expected value of the random variable plus the constant: $E(X + b) = E(X) + b$.

Combining these two properties, we can find the expected value of $aX + b$:

$E(aX + b) = E(aX) + E(b)$ (using the additive property)

$E(aX + b) = a E(X) + b$ (since $E(b) = b$ for a constant $b$, and using the multiplicative property)

So, $E(aX + b) = a E(X) + b$.

Let's examine the options:

(A) $a E(X)$: This is only part of the expression, missing the constant $b$.

(B) $b E(X)$: This is incorrect; it swaps the roles of $a$ and $b$ and misses the additive constant.

(C) $a E(X) + b$: This correctly applies the linearity property of expectation.

(D) $a + b E(X)$: This incorrectly applies the linearity property, reversing the roles of $a$ and $b$ and adding $a$ instead of multiplying it by $E(X)$.

Therefore, $E(aX + b)$ is equal to $a E(X) + b$.

The variance of a random variable measures its dispersion or spread. When a random variable undergoes a linear transformation, its variance changes in a specific way.

Let $X$ be a random variable with variance Var($X$). We are interested in finding the variance of $aX + b$, where $a$ and $b$ are constants.

The variance is defined as Var($Y$) = $E[(Y - E(Y))^2]$.

Let $Y = aX + b$.

First, we find the expected value of $Y$: $E(Y) = E(aX + b) = aE(X) + b$ (using the linearity of expectation).

Now, we can find the variance of $Y$:

Var($Y$) = Var($aX + b$) = $E[( (aX + b) - E(aX + b) )^2]$

Substitute $E(aX + b) = aE(X) + b$:

Var($aX + b$) = $E[( (aX + b) - (aE(X) + b) )^2]$

Simplify the expression inside the square:

$(aX + b) - (aE(X) + b) = aX + b - aE(X) - b = aX - aE(X) = a(X - E(X))$

Now, substitute this back into the variance expression:

Var($aX + b$) = $E[ (a(X - E(X)))^2 ]$

Var($aX + b$) = $E[ a^2 (X - E(X))^2 ]$

Since $a^2$ is a constant, we can pull it out of the expectation operator:

Var($aX + b$) = $a^2 E[ (X - E(X))^2 ]$

We know that $E[ (X - E(X))^2 ]$ is the definition of Var($X$).

Therefore,

Notice that the constant $b$ (the additive term) does not affect the variance, as variance measures spread, and adding a constant shifts the distribution without changing its spread.

Let's examine the options:

(A) $a$ Var($X$): This is incorrect; the constant $a$ should be squared.

(B) $a^2$ Var($X$): This matches our derived formula.

(C) $a^2$ Var($X$) + $b^2$: The $b^2$ term is incorrect; the additive constant $b$ does not affect the variance.

(D) $a$ Var($X$) + $b$: This is incorrect; it has the wrong coefficient for Var($X$) and includes an incorrect additive term.

Thus, the variance of $aX + b$ is $a^2$ Var($X$).

For a binomial distribution, we need to identify the number of trials ($n$) and the probability of success ($p$) on each trial.

The problem states:

• A ball is drawn with replacement 4 times. This means the experiment is repeated a fixed number of times, and each draw is independent of the others because the ball is replaced. Therefore, the number of trials, $n$, is 4.
• We are interested in the number of red balls drawn, so drawing a red ball is considered a "success".
• The bag contains 3 red balls and 2 blue balls, making a total of $3 + 2 = 5$ balls.
• Since the ball is drawn with replacement, the probability of drawing a red ball remains constant for each draw. The probability of success ($p$) is the ratio of the number of red balls to the total number of balls.

$p = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{3}{5}$.

So, the parameters for this binomial distribution are $n=4$ and $p=3/5$.

Let's check the options:

(A) $n=4, p=3/5$: This matches our determined parameters.

(B) $n=4, p=2/5$: Here $p=2/5$ would represent the probability of drawing a blue ball, not a red ball, and we are counting red balls.

(C) $n=5, p=3/5$: The number of trials is incorrect; it should be 4, not 5.

(D) $n=5, p=4/5$: Both $n$ and $p$ are incorrect.

Therefore, the binomial distribution has parameters $n=4$ and $p=3/5$.

From Question 36, we established that the random variable $X$ (number of red balls drawn) follows a binomial distribution with parameters:

• Number of trials, $n = 4$.
• Probability of success (drawing a red ball), $p = 3/5$.
• Probability of failure (drawing a blue ball), $q = 1 - p = 1 - 3/5 = 2/5$.

We need to find the probability of getting exactly 2 red balls, which means finding $P(X=2)$.

The probability mass function (PMF) for a binomial distribution is:

Here, $k$ is the number of successes we are interested in. In this case, we want exactly 2 red balls, so $k=2$.

Substituting the values $n=4$, $k=2$, $p=3/5$, and $q=2/5$ into the PMF:

$P(X=2) = \binom{4}{2} (3/5)^2 (2/5)^{4-2}$

$P(X=2) = \binom{4}{2} (3/5)^2 (2/5)^2$

Now, let's compare this with the given options:

(A) $\binom{4}{2} (3/5)^2 (2/5)^2$: This exactly matches our derived probability formula.

(B) $\binom{4}{2} (2/5)^2 (3/5)^2$: This option swaps the probability of success and failure in the base of the exponents. While the values $(3/5)^2$ and $(2/5)^2$ are correct, their positions are swapped, which might be confusing, but mathematically $(a^2 b^2) = (b^2 a^2)$. However, the formula specifically requires $p^k q^{n-k}$, so $p$ corresponds to the success probability and $q$ to the failure probability.

(C) $\binom{5}{2} (3/5)^2 (2/5)^3$: This has incorrect $n$ and the exponent for $q$.

(D) $\binom{4}{2} (3/5)^2 (2/5)^0$: This would be the probability of getting exactly 2 red balls and 0 blue balls, which is not what's asked.

Option (A) is the most direct and correct representation of the binomial probability formula for this scenario.

The notation $X \sim P(\lambda)$ means that $X$ is a Poisson random variable with parameter $\lambda$. The parameter $\lambda$ represents the average number of events in a given interval.

The probability mass function (PMF) for a Poisson distribution is given by:

We need to find $P(X > 0)$. This is the probability that the random variable $X$ takes any value greater than 0. The possible values for $X$ are $0, 1, 2, 3, \dots$.

It is often easier to calculate the probability of the complementary event. The complement of "$X > 0$" is "$X \le 0$". Since $X$ can only take non-negative integer values, "$X \le 0$" is equivalent to "$X = 0$".

So, $P(X > 0) = 1 - P(X \le 0) = 1 - P(X = 0)$.

Now, let's calculate $P(X = 0)$ using the Poisson PMF (equation i) with $x=0$:

$P(X=0) = \frac{e^{-\lambda} \lambda^0}{0!}$

We know that $\lambda^0 = 1$ (for any $\lambda \ne 0$) and $0! = 1$.

$P(X=0) = \frac{e^{-\lambda} \times 1}{1}$

$P(X=0) = e^{-\lambda}$

Now, substitute this back into the expression for $P(X > 0)$:

$P(X > 0) = 1 - P(X = 0)$

$P(X > 0) = 1 - e^{-\lambda}$

Let's check the options:

(A) $e^{-\lambda}$: This is $P(X=0)$, not $P(X>0)$.

(B) $1 - e^{-\lambda}$: This matches our calculated probability.

(C) $\lambda e^{-\lambda}$: This is the probability $P(X=1)$.

(D) $1 - \lambda e^{-\lambda}$: This is $1 - P(X=1)$, which is $P(X \ne 1)$.

Therefore, $P(X > 0)$ for a Poisson distribution is $1 - e^{-\lambda}$.

The heights of adult Indian males are approximately normally distributed. We are given:

• Mean height, $\mu = 165$ cm.
• Standard deviation, $\sigma = 7$ cm.

We want to find the probability that a randomly selected adult Indian male is taller than 172 cm. Let $X$ be the height of a randomly selected adult Indian male. We want to find $P(X > 172)$.

To find this probability using the standard normal distribution, we need to convert the value $x=172$ to a Z-score. The formula for the Z-score is:

Substitute the given values:

$Z = \frac{172 - 165}{7}$

Calculate the Z-score:

$Z = \frac{7}{7}$

The probability $P(X > 172)$ is equivalent to the probability that the standard normal variable $Z$ is greater than its corresponding Z-score. Therefore:

$P(X > 172) = P(Z > 1)$

Now, let's examine the options:

(A) $P(Z > 1)$: This matches our result.

(B) $P(Z < 1)$: This would be the probability of being shorter than 172 cm.

(C) $P(Z > 7/172)$: This uses the values in the Z-score formula incorrectly.

(D) $P(Z < 7/172)$: This also uses the values incorrectly and is for the wrong tail of the distribution.

Thus, the probability that a randomly selected adult Indian male is taller than 172 cm is $P(Z > 1)$.

In probability theory, for a continuous random variable, the probability density function (PDF), denoted by $f(x)$, describes the relative likelihood for the random variable to take on a given value. The PDF itself is not a probability; rather, probabilities are obtained by integrating the PDF over a specific range.

The fundamental property of a probability density function is that the total area under its curve over the entire range of possible values of the random variable must be equal to 1.

Mathematically, if $X$ is a continuous random variable with PDF $f(x)$ and its range is from $-\infty$ to $\infty$ (or over its defined domain), then:

This means that the total probability of all possible outcomes must sum to 1, which signifies certainty that one of the possible outcomes will occur.

Let's examine the options:

(A) 0: An area of 0 would imply that no outcomes are possible, which is not the case for a valid probability distribution.

(B) 1: This correctly represents the total probability of all possible outcomes occurring.

(C) $\infty$: While the range might theoretically be infinite, the area under the PDF is normalized to 1.

(D) A value between 0 and 1: Probabilities of specific intervals are between 0 and 1, but the total area over the entire range must be exactly 1.

Therefore, the area under the probability density function curve for a continuous random variable over its entire range is always 1.

The variance of a discrete random variable $X$, denoted as Var($X$) or $\sigma^2$, is a measure of the spread or dispersion of the probability distribution. It quantifies how far, on average, the possible values of $X$ are from its mean.

There are two common formulas to calculate the variance:

1. Using the definition: The variance is the expected value of the squared difference between the random variable and its mean ($\mu$).
2. Using the computational formula: Variance is the expected value of the square of the random variable minus the square of the expected value.

Let's look at the first definition. If $\mu = E(X)$ is the mean of the random variable $X$, then the variance is given by:

For a discrete random variable, this definition translates to summing the product of the squared deviation from the mean and the probability of that value:

The second, computational formula is:

Where $E(X^2) = \sum_{x} x^2 P(X=x)$.

Now, let's evaluate the given options based on these formulas:

(A) $\sum (x - \mu)^2 P(X=x)$: This directly matches the definition of variance (equation ii).

(B) $\sum (x - \mu) P(X=x)$: This is the formula for the expected value of $(X - \mu)$, which is $E(X - \mu) = E(X) - \mu = \mu - \mu = 0$. This is not the variance.

(C) $\sum x^2 P(X=x) - \mu$: This is $E(X^2) - \mu$. It is close to the computational formula but should be $E(X^2) - \mu^2$ (or $E(X^2) - [E(X)]^2$).

(D) $\sum x P(X=x^2) - \mu^2$: This involves $P(X=x^2)$, which is not a standard term in variance calculation, and $P(X=x^2)$ is generally not well-defined or useful in this context.

Therefore, the variance of a discrete random variable $X$ can also be calculated as $\sum (x - \mu)^2 P(X=x)$.

The binomial distribution, $B(n, p)$, describes the probability of $k$ successes in $n$ independent Bernoulli trials, each with a probability of success $p$.

There are two common approximations for the binomial distribution based on the values of its parameters $n$ and $p$:

1. Poisson Approximation: When $n$ is large and $p$ is small, the binomial distribution can be approximated by a Poisson distribution with parameter $\lambda = np$. This approximation is particularly good when $n \ge 20$ and $p \le 0.05$, or when $n \ge 100$ and $np \le 10$.
2. Normal Approximation: When $n$ is large and $p$ is not too close to 0 or 1 (typically, when $np \ge 5$ and $n(1-p) \ge 5$), the binomial distribution can be approximated by a normal distribution with mean $\mu = np$ and variance $\sigma^2 = np(1-p)$.

The question specifically states that "$n$ is large while $p$ is small". This is the condition under which the Poisson approximation is most appropriate.

Let's examine the options:

(A) Normal distribution: This approximation is used when $n$ is large and $p$ is close to 0.5 (i.e., not small). The condition "p is small" makes this approximation less suitable than the Poisson approximation.

(B) Poisson distribution: This is the correct approximation when $n$ is large and $p$ is small. The parameter $\lambda$ for the Poisson distribution is set to $np$.

(C) Exponential distribution: The exponential distribution models the time between events in a Poisson process, not the number of successes in trials.

(D) Uniform distribution: A uniform distribution implies that all outcomes in a range are equally likely, which is not characteristic of the binomial distribution, even for large $n$ and small $p$.

Therefore, when $n$ is large and $p$ is small, the binomial distribution can be approximated by the Poisson distribution.

Let's analyze the scenario to determine the appropriate probability distribution:

• Number of trials ($n$): The quality control manager inspects a batch of 500 bulbs. This is a fixed number of trials, so $n=500$.
• Two possible outcomes: For each bulb inspected, there are two outcomes: it is either defective (success) or not defective (failure).
• Probability of success ($p$): The probability of a bulb being defective is given as $0.01$. This is the probability of "success" in this context.
• Independence: We assume that the defectiveness of one bulb is independent of the defectiveness of other bulbs.

These characteristics (fixed number of trials, two outcomes, constant probability of success, and independence of trials) are the defining features of a binomial distribution.

We should also consider if a Poisson approximation is suitable. The conditions for Poisson approximation to the binomial distribution are a large $n$ and a small $p$. Here, $n=500$ is large, and $p=0.01$ is small. Thus, a Poisson distribution could be used as an approximation.

The parameter $\lambda$ for the Poisson approximation would be $np = 500 \times 0.01 = 5$.

However, the question asks for the *appropriate* distribution, and the binomial distribution is the exact model for this scenario, whereas Poisson is an approximation. The normal distribution is used for continuous data or when $np$ and $n(1-p)$ are both sufficiently large, which might not be the case here ($np=5$, $n(1-p)=495$, $n(1-p)$ is large but $np=5$ is borderline for normal approximation). A uniform distribution is for equally likely outcomes over a range.

Given the options, the binomial distribution is the most direct and exact model for counting the number of defective items in a fixed sample size with a constant probability of defect.

Let's review the options:

(A) Normal distribution: Generally used for continuous data or when $np$ and $n(1-p)$ are large.

(B) Binomial distribution: Fits all the criteria for this scenario (fixed trials, two outcomes, constant probability, independence).

(C) Poisson distribution: A good approximation when $n$ is large and $p$ is small, but binomial is the exact model.

(D) Uniform distribution: Not applicable here.

Therefore, the most appropriate distribution to model the number of defective bulbs in the batch is the binomial distribution.

From Question 43, we determined that the number of defective bulbs in the batch follows a binomial distribution.

The parameters of this binomial distribution are:

• Number of trials, $n = 500$ (the number of bulbs inspected).
• Probability of success (a bulb being defective), $p = 0.01$.

The expected number of successes (in this case, defective bulbs) in a binomial distribution is given by the formula:

Substitute the values of $n$ and $p$ into the formula:

$E(X) = 500 \times 0.01$

Calculate the result:

Now, let's check the options:

(A) 500: This is the number of trials ($n$), not the expected number of defective bulbs.

(B) 0.01: This is the probability of a single bulb being defective ($p$), not the expected number of defective bulbs in the batch.

(C) $500 \times 0.01 = 5$: This correctly shows the calculation of the expected number of defective bulbs ($np$) and the result.

(D) $500 \times (1 - 0.01) = 495$: This calculates $n \times q$, which would be the expected number of non-defective bulbs.

Therefore, the expected number of defective bulbs is $500 \times 0.01 = 5$.

From Question 43, we have a scenario modeled by a binomial distribution with parameters $n=500$ (number of trials) and $p=0.01$ (probability of success, i.e., a bulb being defective).

The conditions for approximating a binomial distribution with a Poisson distribution are that $n$ is large and $p$ is small. In this case:

• $n = 500$, which is considered large.
• $p = 0.01$, which is considered small.

Therefore, the binomial distribution can be reasonably approximated by a Poisson distribution.

When a binomial distribution $B(n, p)$ is approximated by a Poisson distribution, the parameter $\lambda$ of the Poisson distribution is equal to the mean of the binomial distribution, which is given by $np$.

So, $\lambda = n \times p$.

Substitute the values:

$\lambda = 500 \times 0.01$

$\lambda = 5$

Now let's check the options:

(A) Yes, $\lambda = 500$: This uses $n$ as $\lambda$, which is incorrect.

(B) Yes, $\lambda = 0.01$: This uses $p$ as $\lambda$, which is incorrect.

(C) Yes, $\lambda = 5$: This correctly states that the scenario can be approximated by a Poisson distribution and correctly calculates $\lambda = np = 5$.

(D) No, it cannot be approximated by a Poisson distribution: This is incorrect because the conditions ($n$ large, $p$ small) are met.

Therefore, the scenario can be reasonably approximated by a Poisson distribution with $\lambda = 5$.

The expected value (or mean) of a random variable is the weighted average of all possible values that the random variable can take. The weights used are the probabilities associated with each value.

For a discrete random variable $X$ with probability mass function $P(X=x)$ (or $f(x)$ in discrete context), the expected value is calculated as:

For a continuous random variable $X$ with probability density function (PDF) $f(x)$, the expected value is calculated by integrating the product of each possible value $x$ and its PDF $f(x)$ over the entire range of possible values of $X$. Assuming the range is from $-\infty$ to $\infty$ (or its defined domain):

Let's evaluate the options based on these definitions:

(A) $\sum x f(x)$: This uses summation ($\sum$), which is appropriate for discrete random variables, not continuous ones. While $f(x)$ is used here as a notation, the summation makes it incorrect for a continuous PDF.

(B) $\int\limits_{-\infty}^{\infty} x f(x) dx$: This is the correct formula for the expected value of a continuous random variable using its PDF.

(C) $\int\limits_{-\infty}^{\infty} f(x) dx$: This represents the total area under the PDF curve over its entire range, which by definition is always 1 (the total probability), not the expected value.

(D) $\sum f(x)$: This uses summation and does not multiply by $x$, so it's incorrect for both discrete and continuous expected values.

Therefore, the expected value $E(X)$ for a continuous random variable $X$ with PDF $f(x)$ is given by $\int\limits_{-\infty}^{\infty} x f(x) dx$.

For any probability density function (PDF), $f(x)$, the total area under the curve over its entire range must be equal to 1. This is a fundamental property of PDFs.

The PDF is given as $f(x) = cx$ for $0 \le x \le 2$, and $f(x) = 0$ otherwise. We need to find the value of the constant $c$ such that the total probability is 1.

We use the property that the integral of the PDF over its entire domain must be 1:

Since $f(x) = 0$ outside the interval $[0, 2]$, the integral becomes:

Now, we evaluate the integral:

$\int_{0}^{2} cx dx = c \int_{0}^{2} x dx$

The integral of $x$ with respect to $x$ is $\frac{x^2}{2}$.

$c \left[ \frac{x^2}{2} \right]_{0}^{2} = 1$

Evaluate the expression at the limits:

$c \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = 1$

$c \left( \frac{4}{2} - 0 \right) = 1$

$c (2) = 1$

Solve for $c$:

So, the value of $c$ is 1/2.

Let's check the options:

(A) 1/2: This matches our calculated value.

(B) 1: If $c=1$, the integral would be 2, not 1.

(C) 1/4: If $c=1/4$, the integral would be $1/4 \times 2 = 1/2$, not 1.

(D) 2: If $c=2$, the integral would be $2 \times 2 = 4$, not 1.

Therefore, the value of $c$ is 1/2.

From Question 47, we found that the probability density function (PDF) is $f(x) = cx$ for $0 \le x \le 2$, and we determined that $c = 1/2$. So, the PDF is $f(x) = \frac{1}{2}x$ for $0 \le x \le 2$, and $0$ otherwise.

The expected value $E(X)$ of a continuous random variable is given by the integral of $x f(x)$ over the entire range of the variable:

Using the specific PDF and its domain:

$E(X) = \int_{0}^{2} x \left(\frac{1}{2}x\right) dx$

$E(X) = \int_{0}^{2} \frac{1}{2} x^2 dx$

Now, we evaluate this integral:

$E(X) = \frac{1}{2} \int_{0}^{2} x^2 dx$

The integral of $x^2$ with respect to $x$ is $\frac{x^3}{3}$.

$E(X) = \frac{1}{2} \left[ \frac{x^3}{3} \right]_{0}^{2}$

Evaluate at the limits:

$E(X) = \frac{1}{2} \left( \frac{2^3}{3} - \frac{0^3}{3} \right)$

$E(X) = \frac{1}{2} \left( \frac{8}{3} - 0 \right)$

$E(X) = \frac{1}{2} \times \frac{8}{3}$

Let's examine the options:

(A) $\int\limits_{0}^{2} x (cx) dx = c \int\limits_{0}^{2} x^2 dx = c \left[\frac{x^3}{3}\right]\limits_{0}^{2} = c \frac{8}{3} = \frac{1}{2} \times \frac{8}{3} = \frac{4}{3}$: This option correctly sets up the integral for $E(X)$, substitutes $c=1/2$, performs the integration, and arrives at the correct answer. It shows the full calculation process.

(B) $\int\limits_{0}^{2} cx dx = c \left[\frac{x^2}{2}\right]\limits_{0}^{2} = 2c = 1$: This calculates $\int c x dx$, which is not the expected value $E(X)$, but rather related to $P(X \le 2)$. It also incorrectly concludes $2c=1$ which refers to the condition for the PDF to integrate to 1, not for $E(X)$.

(C) $\int\limits_{0}^{2} x^2 dx = \frac{8}{3}$: This calculates the integral of $x^2$, but misses the $x$ and $f(x)$ components of $E(X)$ and the constant $c$. It also doesn't include $c$ or the final value of $E(X)$.

(D) $\int\limits_{0}^{2} \frac{1}{2} x dx = \frac{1}{2} \left[\frac{x^2}{2}\right]\limits_{0}^{2} = \frac{1}{2} \times 2 = 1$: This calculates the integral of $\frac{1}{2}x$, which is $E(X)$ if the PDF was $f(x)=\frac{1}{2}$ (uniform distribution) or part of $E(X)$ if $f(x)$ was different. It uses the correct value of $c$ but the wrong function ($x$ instead of $x^2$) in the integral for $E(X)$.

Option (A) correctly sets up the integral for $E(X)$, substitutes the value of $c$, performs the integration, and arrives at the correct expected value.

The variance of a random variable is the square of its standard deviation. Conversely, the standard deviation is the square root of the variance.

Let $\sigma$ be the standard deviation and $\sigma^2$ be the variance.

The relationship between them is:

Squaring both sides of equation (i) gives:

In this problem, we are given that the standard deviation of a normal distribution is 5.

So, $\sigma = 5$.

Using the relationship from equation (ii), the variance is the square of the standard deviation:

Var($X$) = $\sigma^2 = 5^2$

Var($X$) = $25$

Let's check the options:

(A) 5: This is the standard deviation itself.

(B) $\sqrt{5}$: This is the square root of the standard deviation, not related.

(C) 25: This is the square of the standard deviation (5), which is the variance.

(D) 10: This is twice the standard deviation, which is not the variance.

Therefore, if the standard deviation is 5, the variance is 25.

A random variable is a variable whose value is a numerical outcome of a random phenomenon. Random variables are broadly classified into two types: discrete and continuous.

Let's define the characteristics of a discrete random variable and compare them with the given options:

• Definition of Discrete Random Variable: A random variable is discrete if its set of possible values is finite or countably infinite. This means that the values can be listed out, even if the list is infinitely long (like the set of non-negative integers).

Now, let's analyze each option:

(A) It can take any value within a given range: This describes a continuous random variable. For example, height can take any value between 150 cm and 180 cm.

(B) Its possible values can be listed (finite or countably infinite): This is the definition of a discrete random variable. For example, the number of heads in coin tosses (0, 1, 2, 3,...) or the number of defective items (0, 1, 2, ..., 100) are discrete.

(C) The probability of any specific value is always 0: This is a characteristic of a continuous random variable. For continuous variables, $P(X=x) = 0$ for any specific value $x$. For discrete variables, $P(X=x)$ can be greater than 0 for specific values.

(D) Its distribution is represented by a probability density function: The probability of a continuous random variable is represented by a probability density function (PDF). A discrete random variable's distribution is represented by a probability mass function (PMF).

Therefore, the characteristic of a discrete random variable among the given options is that its possible values can be listed (finite or countably infinite).

When a fair six-sided die is rolled, the possible outcomes are the integers from 1 to 6. Since the die is fair, each outcome has an equal probability of occurring.

The possible values of the random variable $X$ are $\{1, 2, 3, 4, 5, 6\}$.

The probability of each outcome is $P(X=x) = \frac{1}{6}$ for $x \in \{1, 2, 3, 4, 5, 6\}$.

The expected value $E(X)$ of a discrete random variable is calculated as the sum of each possible value multiplied by its probability:

Substituting the values:

$E(X) = (1 \times \frac{1}{6}) + (2 \times \frac{1}{6}) + (3 \times \frac{1}{6}) + (4 \times \frac{1}{6}) + (5 \times \frac{1}{6}) + (6 \times \frac{1}{6})$

We can factor out the common probability $\frac{1}{6}$:

$E(X) = \frac{1}{6} (1 + 2 + 3 + 4 + 5 + 6)$

Now, sum the numbers from 1 to 6:

$1 + 2 + 3 + 4 + 5 + 6 = 21$

Alternatively, using the formula for the sum of the first $n$ integers, $S_n = \frac{n(n+1)}{2}$, with $n=6$: $S_6 = \frac{6(6+1)}{2} = \frac{6 \times 7}{2} = \frac{42}{2} = 21$.

Substitute the sum back into the expected value calculation:

$E(X) = \frac{1}{6} \times 21$

$E(X) = \frac{21}{6}$

$E(X) = \frac{7}{2}$

$E(X) = 3.5$

Let's check the options:

(A) 3

(B) 3.5

(C) 4

(D) 7

Our calculated expected value matches option (B).

Assuming "Question 51" refers to a scenario where a fair six-sided die is rolled, and $X$ is the random variable representing the outcome of the roll, the possible values of $X$ are $1, 2, 3, 4, 5, 6$, each with a probability of $\frac{1}{6}$.

The expected value of $X^2$, denoted as $E(X^2)$, is calculated as the sum of the squares of each possible value of $X$ multiplied by its corresponding probability.

The formula for $E(X^2)$ is:

In this case, $n=6$, and $P(x_i) = \frac{1}{6}$ for all $i=1, 2, 3, 4, 5, 6$.

So, $E(X^2)$ can be calculated as:

We can factor out $\frac{1}{6}$:

Now, let's calculate the sum of the squares:

Substituting this back into the equation for $E(X^2)$:

As a decimal, $\frac{91}{6} \approx 15.17$.

Comparing this with the given options:

(A) $(3.5)^2 = 12.25$ (This is $E(X)^2$, not $E(X^2)$)

(B) $\frac{1}{6}(1^2+2^2+3^2+4^2+5^2+6^2) = \frac{1}{6}(1+4+9+16+25+36) = \frac{91}{6} \approx 15.17$ (This matches our calculation.)

(C) $\frac{1}{6}(1+2+3+4+5+6)^2 = \frac{21^2}{6} = \frac{441}{6} = 73.5$ (This is incorrect as it squares the sum of values instead of summing the squares.)

(D) $\frac{91}{36}$ (This is incorrect.)

Thus, the correct option is (B).

Assuming $X$ is the outcome of a fair six-sided die roll (from Question 51), we have $E(X) = \frac{21}{6} = 3.5$ and $E(X^2) = \frac{91}{6}$.

The variance is calculated as Var($X$) = $E(X^2) - [E(X)]^2$.

Var($X$) = $\frac{91}{6} - \left(\frac{21}{6}\right)^2$

Finding a common denominator (36):

Var($X$) = $\frac{105}{36}$

Simplifying the fraction by dividing by 3:

This matches option (D) and is the result shown in option (C) after calculation. Option (C) illustrates the process correctly.

The correct option is (C) as it shows the full calculation, or (D) for the final value.

The notation $X \sim B(n, p)$ indicates that the random variable $X$ follows a Binomial distribution with parameters $n$ (the number of trials) and $p$ (the probability of success in a single trial).

The probability mass function (PMF) for a Binomial distribution, which gives the probability of obtaining exactly $k$ successes in $n$ independent Bernoulli trials, is:

where:

• $\binom{n}{k}$ represents the binomial coefficient, which is the number of ways to choose $k$ successes from $n$ trials. It is calculated as $\frac{n!}{k!(n-k)!}$.
• $p^k$ is the probability of getting $k$ successes.
• $(1-p)^{n-k}$ is the probability of getting $n-k$ failures (where the probability of failure is $1-p$).

Let's examine the given options:

• (A) $\binom{n}{k} p^k (1-p)^{n-k}$ - This is the standard formula for the binomial probability.
• (B) $\binom{n}{k} p^{n-k} (1-p)^k$ - This swaps the probabilities of success and failure, which is incorrect.
• (C) $p^k (1-p)^{n-k}$ - This is missing the binomial coefficient, which accounts for the different combinations of successes and failures.
• (D) $\frac{p^k e^{-p}}{k!}$ - This is the probability mass function for a Poisson distribution, not a Binomial distribution.

Therefore, the correct formula for the probability of getting exactly $k$ successes in a Binomial distribution $B(n, p)$ is given by option (A).

This problem describes a scenario that fits the Binomial distribution. We are given:

• The number of trials, $n = 20$ (the sample size).
• The probability of success (a person being left-handed), $p = 0.10$ (or $10\%$).
• The number of successes we are interested in, $k = 3$ (exactly 3 people are left-handed).

From the previous question, the probability of getting exactly $k$ successes in a Binomial distribution $B(n, p)$ is given by the formula:

Here, the probability of failure (a person NOT being left-handed) is $1-p = 1 - 0.10 = 0.90$.

Substituting the given values into the formula:

• $n = 20$
• $k = 3$
• $p = 0.1$
• $1-p = 0.9$
• $n-k = 20 - 3 = 17$

So, the probability is:

Now, let's check the options:

• (A) $\binom{20}{3} (0.1)^3 (0.9)^{17}$ - This matches our derived formula.
• (B) $\binom{20}{3} (0.9)^3 (0.1)^{17}$ - This incorrectly assigns the probabilities.
• (C) $\frac{e^{-2} 2^3}{3!}$ - This is the probability from a Poisson distribution with $\lambda=2$ and $k=3$. It's not related to the binomial scenario described.
• (D) $(0.1)^3 (0.9)^{17}$ - This is missing the binomial coefficient $\binom{20}{3}$, which accounts for the different ways to select 3 left-handed people from 20.

Therefore, the correct probability is given by option (A).

For a Poisson distribution with parameter $\lambda$ (lambda), the mean and variance are both equal to $\lambda$.

The properties of a Poisson distribution are:

• Mean, $E(X) = \lambda$
• Variance, Var($X$) = $\lambda$

The standard deviation (SD) is the square root of the variance.

In this question, we are given that $\lambda = 4$.

Therefore, the variance is Var($X$) = $4$.

The standard deviation is:

Comparing this with the given options:

• (A) 4 - This is the value of $\lambda$, which is the mean and variance, not the standard deviation.
• (B) 2 - This matches our calculated standard deviation.
• (C) 16 - This is $\lambda^2$, which is not a relevant property here.
• (D) $\sqrt{4} = 2$ - This option correctly shows the calculation of the standard deviation from $\lambda$.

Both options (B) and (D) provide the correct numerical value of the standard deviation. However, option (D) also shows the step of taking the square root of $\lambda$, which is the correct method to find the standard deviation. Therefore, option (D) is the most complete and explanatory correct answer.

The probability density function (PDF) of a normal distribution is peaked at its mean ($\mu$). The value of the PDF at the peak is given by $\frac{1}{\sigma \sqrt{2\pi}}$, where $\sigma$ is the standard deviation.

Given: Mean $\mu = 50$, Standard Deviation $\sigma = 10$.

The value at the peak is $\frac{1}{\sigma \sqrt{2\pi}}$.

Option (B) states $\frac{1}{10 \sqrt{2\pi}} e^0 = \frac{1}{10 \sqrt{2\pi}}$. This option correctly identifies the peak value and shows the intermediate step with $e^0$ (since the exponent in the PDF formula becomes zero at the mean). Therefore, option (B) is the correct answer.

For a continuous random variable $X$, the probability of it taking any single, specific value $a$ is always zero. This is a fundamental property derived from the definition of probability for continuous distributions, where probabilities are associated with intervals rather than discrete points.

Mathematically, $P(X=a) = \int_{a}^{a} f(x) dx = 0$, where $f(x)$ is the probability density function.

Therefore, the correct option is (A).

The number of customer arrivals follows a Poisson distribution. The probability mass function of a Poisson distribution is given by:

$P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$

where:

$\lambda$ is the average number of events in the given interval.

$k$ is the actual number of events.

$e$ is the base of the natural logarithm (approximately 2.71828).

In this problem:

The average number of arrivals per interval ($\lambda$) = 5.

The number of arrivals we are interested in ($k$) = 3.

Substituting these values into the Poisson probability formula:

$P(X=3) = \frac{e^{-5} 5^3}{3!}$

Comparing this with the given options, the correct option is (B).

The probability of at least one arrival is the complement of the probability of no arrivals.

Using the Poisson probability formula $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$:

Here, $\lambda = 5$ (average arrivals per interval).

The probability of no arrivals ($k=0$) is:

$P(X=0) = \frac{e^{-5} 5^0}{0!}$

Since $5^0 = 1$ and $0! = 1$, this simplifies to:

$P(X=0) = \frac{e^{-5} \times 1}{1} = e^{-5}$

The probability of at least one arrival is $P(X \geq 1) = 1 - P(X=0)$.

$P(X \geq 1) = 1 - e^{-5}$

Option (B) is $1 - e^{-5}$.

Option (C) is $1 - \frac{e^{-5} 5^0}{0!}$, which is also $1 - e^{-5}$.

Therefore, both (B) and (C) are correct.

The Z-score is calculated using the formula:

$Z = \frac{X - \mu}{\sigma}$

where:

$X$ is the individual score.

$\mu$ is the mean of the distribution.

$\sigma$ is the standard deviation of the distribution.

Given:

Individual score ($X$) = 75

Mean ($\mu$) = 60

Standard deviation ($\sigma$) = 15

Substitute the values into the formula:

$Z = \frac{75 - 60}{15}$

$Z = \frac{15}{15}$

$Z = 1$

Therefore, the Z-score of a student scoring 75 is 1.

The correct option is (A).

Let's analyze each property in relation to the normal distribution and other common distributions:

(A) The mean is equal to the variance. This property is unique to the Poisson distribution, not the normal distribution. For a normal distribution, the mean and variance are independent parameters ($\mu$ and $\sigma^2$ respectively).

(B) It is symmetric. While the normal distribution is indeed symmetric about its mean, other distributions also exhibit symmetry. For example, the t-distribution and the Cauchy distribution are also symmetric.

(C) It is used for discrete data. The normal distribution is a continuous probability distribution and is used for continuous data. Discrete data is typically modeled by distributions like the Binomial, Poisson, or Geometric distributions.

(D) The sum of probabilities is 1. This is a fundamental property of *all* probability distributions, whether discrete or continuous. The total probability of all possible outcomes must always sum to 1.

Revisiting option (B): Although symmetry is not *strictly* unique to the normal distribution, in the context of common distributions encountered in introductory statistics, the bell-shaped, unimodal symmetry is a defining characteristic that is often emphasized as distinguishing it. However, the question asks for a unique property *among the ones listed*. Let's re-evaluate if any property is *more* unique or a more defining characteristic that sets it apart if considered in contrast to *all* other distributions, or perhaps it's asking which property *only* the normal distribution has from the given options.

Let's consider the options again:

A. Mean = Variance: Unique to Poisson.

B. Symmetric: Many distributions are symmetric (t-distribution, Cauchy, uniform on a symmetric interval).

C. Used for discrete data: Incorrect, normal is for continuous data.

D. Sum of probabilities = 1: True for all probability distributions.

There seems to be a misunderstanding or a poorly phrased question if the intent was to find a property *exclusive* to the normal distribution.

However, if the question implies a property that is *most characteristic* or *uniquely defining in its typical presentation and application*, then symmetry (B) is often highlighted.

Let's assume the question intends to ask which property *from the given options* is a defining feature of the normal distribution that might not be as universally true for *all other* distributions being implicitly contrasted (even if other symmetric distributions exist).

In many contexts, the specific shape and the fact that the mean, median, and mode are all equal due to perfect symmetry are considered hallmarks of the normal distribution. While symmetry itself isn't exclusive, the *degree* and *nature* of symmetry (perfectly bell-shaped) are key.

Let's reconsider the prompt's options and typical textbook emphasis:

• A is definitively not unique to normal.
• C is factually incorrect about normal distribution usage.
• D is true for all probability distributions.

This leaves (B) as the only plausible answer, implying that the *type* or *degree* of symmetry is what is meant to be unique or characteristic in this context.

The most distinguishing characteristic among the provided options for the normal distribution is its symmetry about its mean, where the mean, median, and mode all coincide. While other distributions can be symmetric, the specific bell-shaped curve of the normal distribution is uniquely defined by its symmetry and the role of its standard deviation in shaping this curve.

Therefore, the property unique to the normal distribution among the ones listed is that it is symmetric.

The correct option is (B).

The expected value $E(X)$ of a discrete random variable $X$ is calculated by summing the product of each possible value of $X$ and its corresponding probability $P(X=x)$.

The formula is: $E(X) = \sum [x \cdot P(X=x)]$

Given the probability distribution:

$X$ values are 1, 2, 3.

$P(X=x)$ values are 0.1, 0.2, 0.7.

Now, we calculate the expected value:

$E(X) = (1 \times P(X=1)) + (2 \times P(X=2)) + (3 \times P(X=3))$

$E(X) = (1 \times 0.1) + (2 \times 0.2) + (3 \times 0.7)$

$E(X) = 0.1 + 0.4 + 2.1$

$E(X) = 2.6$

This calculation matches option (A).

Let's look at why other options are incorrect:

(B) $0.1+0.2+0.7 = 1.0$: This is the sum of probabilities, which should always be 1 for a valid probability distribution, but it's not the expected value.

(C) $\frac{1+2+3}{3} = 2$: This is the arithmetic mean of the possible values of $X$, not the expected value, as it doesn't consider the probabilities.

(D) $1^2 \times 0.1 + 2^2 \times 0.2 + 3^2 \times 0.7$: This calculation represents the expected value of $X^2$, i.e., $E(X^2)$, not $E(X)$.

The correct option is (A).

The expected value of $X^2$, denoted as $E(X^2)$, is calculated by summing the product of each possible value of $X^2$ and its corresponding probability $P(X=x)$.

The formula is: $E(X^2) = \sum [x^2 \cdot P(X=x)]$

From Question 63, the probability distribution is:

$X$ values are 1, 2, 3.

$P(X=x)$ values are 0.1, 0.2, 0.7.

First, we find the values of $X^2$:

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

Now, we calculate $E(X^2)$:

$E(X^2) = (1^2 \times P(X=1)) + (2^2 \times P(X=2)) + (3^2 \times P(X=3))$

$E(X^2) = (1 \times 0.1) + (4 \times 0.2) + (9 \times 0.7)$

$E(X^2) = 0.1 + 0.8 + 6.3$

$E(X^2) = 7.2$

This calculation matches option (B).

Let's look at why other options are incorrect:

(A) $2.6^2 = 6.76$: This is the square of the expected value $E(X)$, which is $E(X)^2$, not $E(X^2)$.

(C) $1 \times 0.1^2 + 2 \times 0.2^2 + 3 \times 0.7^2 = 0.01 + 0.08 + 1.47 = 1.56$: This calculation incorrectly squares the probabilities ($P(X=x)^2$) instead of the $X$ values.

(D) $1+4+9=14$: This is the sum of the squared $X$ values, not weighted by their probabilities.

The correct option is (B).

The variance of a random variable $X$, denoted as Var($X$), can be calculated using the formula:

Var($X$) = $E(X^2) - [E(X)]^2$

From the previous questions:

The expected value $E(X) = 2.6$ (from Question 63).

The expected value of $X^2$, $E(X^2) = 7.2$ (from Question 64).

Now, we can calculate the variance:

Var($X$) = $E(X^2) - [E(X)]^2$

Var($X$) = $7.2 - (2.6)^2$

Calculate $(2.6)^2$:

$2.6 \times 2.6 = 6.76$

Substitute this value back into the variance formula:

Var($X$) = $7.2 - 6.76$

Var($X$) = $0.44$

This calculation matches option (A).

Let's look at why other options are incorrect:

(B) $2.6$: This is the expected value $E(X)$, not the variance.

(C) $7.2$: This is the expected value of $X^2$, $E(X^2)$, not the variance.

(D) $1.56 - 2.6^2$: The value 1.56 is incorrect for $E(X^2)$ (it was calculated incorrectly in option (C) of the previous question).

The correct option is (A).

For a binomial distribution $X \sim B(n, p)$, the formulas for the mean and variance are:

Mean ($E(X)$) = $np$

Variance (Var($X$)) = $np(1-p)$

Given the parameters:

Number of trials ($n$) = 10

Probability of success ($p$) = 0.3

Calculate the mean:

Mean = $n \times p$

Mean = $10 \times 0.3$

Mean = $3$

Calculate the variance:

Variance = $n \times p \times (1-p)$

First, calculate $(1-p)$: $1 - 0.3 = 0.7$

Variance = $10 \times 0.3 \times 0.7$

Variance = $3 \times 0.7$

Variance = $2.1$

So, the mean is 3 and the variance is 2.1.

Comparing this with the given options:

Option (A) states Mean = 3, Variance = 2.1. This matches our calculations.

The correct option is (A).

The number of typos per page follows a Poisson distribution. The probability mass function of a Poisson distribution is given by:

$P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$

where:

$\lambda$ is the average number of events (typos) in the given interval (page).

$k$ is the actual number of events (typos).

In this problem:

The average number of typos per page ($\lambda$) = 0.5.

We want to find the probability of no typos, which means the number of typos ($k$) = 0.

Substitute these values into the Poisson probability formula:

$P(X=0) = \frac{e^{-0.5} (0.5)^0}{0!}$

We know that any non-zero number raised to the power of 0 is 1, so $(0.5)^0 = 1$. Also, $0! = 1$.

$P(X=0) = \frac{e^{-0.5} \times 1}{1}$

$P(X=0) = e^{-0.5}$

This matches option (A).

Let's analyze why other options are incorrect:

(B) $\frac{e^{-0.5} 0^0}{0!} = e^{-0.5}$: While the result is correct, the expression uses $0^0$, which is indeterminate or often defined as 1 in contexts like this. However, the formula for Poisson requires $\lambda^k$, so it should be $(0.5)^0$, not $0^0$. The base should be the mean ($\lambda$), not the number of occurrences ($k$).

(C) $\frac{e^{-0} 0.5^0}{0!} = 1$: This incorrectly uses $e^{-0}$ (which is 1) instead of $e^{-0.5}$.

(D) $1 - e^{-0.5}$: This would be the probability of having at least one typo, not the probability of having no typos.

The correct option is (A).

Let's analyze the Assertion (A) and the Reason (R) separately.

Assertion (A): For a binomial distribution $B(n, p)$, the variance is always less than the mean.

The mean of a binomial distribution is $E(X) = np$.

The variance of a binomial distribution is Var($X$) = $np(1-p)$.

To determine if the variance is always less than the mean, we compare $np(1-p)$ with $np$.

For a binomial distribution, $n \ge 0$ and $0 < p < 1$.

If $n > 0$ and $0 < p < 1$, then $1-p$ will be between 0 and 1 (i.e., $0 < 1-p < 1$).

Multiplying $np$ by a number between 0 and 1 ($1-p$) will result in a value less than $np$.

So, $np(1-p) < np$ as long as $n>0$ and $0 < p < 1$.

Therefore, the variance is always less than the mean for a binomial distribution under these conditions. The assertion (A) is true.

Reason (R): Variance is $np(1-p)$ and Mean is $np$. Since $0 < p < 1$, $1-p < 1$, so $np(1-p) < np$.

The formulas for mean ($np$) and variance ($np(1-p)$) are correct for a binomial distribution.

The condition $0 < p < 1$ is also correct for a binomial distribution (if $p=0$ or $p=1$, the distribution becomes degenerate with zero variance). If $0 < p < 1$, then $1-p$ will be between 0 and 1, which means $1-p < 1$.

Multiplying $np$ by a number less than 1 ($1-p$) will indeed result in a value less than $np$. Thus, $np(1-p) < np$.

The reasoning provided is logically sound and correctly explains why the variance is less than the mean.

Therefore, the reason (R) is true.

Since both A and R are true, and R correctly explains why A is true, option (A) is the correct choice.

The correct option is (A).

The standard normal distribution (Z-distribution) is a bell-shaped curve that is symmetric about its mean, which is 0. The total area under the curve represents a probability of 1.

The question asks for the area to the left of $Z=1.96$, which is $P(Z \le 1.96)$.

This value corresponds to a specific entry in the standard normal distribution table (Z-table).

Looking up the value $1.96$ in a standard normal distribution table:

Find the row corresponding to 1.9.

Find the column corresponding to 0.06.

The intersection of this row and column gives the cumulative probability $P(Z \le 1.96)$.

The value found in most Z-tables for $Z=1.96$ is approximately 0.9750.

This means that approximately 97.5% of the data falls below a Z-score of 1.96.

Let's consider the options:

(A) 0.95: This corresponds to a Z-score of approximately $\pm 1.645$.

(B) 0.975: This corresponds to a Z-score of approximately $1.96$.

(C) 0.025: This is the area to the *right* of $Z=1.96$ ($P(Z > 1.96)$), or the area to the *left* of $Z=-1.96$ ($P(Z \le -1.96)$).

(D) 0.5: This represents the area to the left of the mean ($Z=0$).

Therefore, the approximate value of $P(Z \le 1.96)$ is 0.975.

The correct option is (B).

Let's examine each statement about the Normal distribution:

(A) It is bell-shaped.

This statement is TRUE. The characteristic shape of the Normal distribution is a symmetric bell curve.

(B) The mean, median, and mode are equal.

This statement is TRUE. Due to its perfect symmetry, the mean, median, and mode of a Normal distribution all coincide at the center of the distribution.

(C) The total area under the curve is 1.

This statement is TRUE. Like all probability distributions, the total area under the Normal distribution curve represents the total probability, which must be equal to 1.

(D) It is a discrete probability distribution.

This statement is FALSE. The Normal distribution is a continuous probability distribution. Discrete probability distributions deal with countable outcomes (like the number of heads in coin flips), whereas continuous probability distributions deal with outcomes that can take any value within a range (like height, weight, or temperature).

Since the question asks for the FALSE statement, option (D) is the correct answer.

The correct option is (D).

The problem describes a scenario that fits a binomial distribution. The parameters are:

Number of trials ($n$) = 4 (the number of children).

Probability of success ($p$) = 0.5 (the probability of having a male child).

We are looking for the probability of having exactly 2 male children, so $k = 2$.

The probability mass function for a binomial distribution is given by:

$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$

Substituting the given values ($n=4$, $p=0.5$, $k=2$):

$P(X=2) = \binom{4}{2} (0.5)^2 (1-0.5)^{4-2}$

$P(X=2) = \binom{4}{2} (0.5)^2 (0.5)^2$

Let's evaluate the components:

The binomial coefficient $\binom{4}{2}$ is calculated as $\frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6$.

$(0.5)^2 = 0.25$.

$(0.5)^2 = 0.25$.

So, $P(X=2) = 6 \times 0.25 \times 0.25$.

$P(X=2) = 6 \times 0.0625$

$P(X=2) = 0.375$

Now let's compare this with the options:

(A) $\binom{4}{2} (0.5)^2 (0.5)^2 = 6 \times 0.25 \times 0.25 = 6 \times 0.0625 = 0.375$. This option provides the correct formula and the correct calculated value.

(B) $\binom{4}{2} (0.5)^4$. This is equivalent to $\binom{4}{2} (0.5)^2 (0.5)^2$ because $(0.5)^2 \times (0.5)^2 = (0.5)^{2+2} = (0.5)^4$. So, this option is also correct in terms of the expression, although option (A) provides the full calculation and result.

(C) $0.5^4$. This is just $(0.5)^k(0.5)^{n-k}$ and misses the binomial coefficient $\binom{n}{k}$.

(D) $2 \times 0.5$. This is $k \times p$, which is related to the mean but not the probability of a specific outcome.

Given that option (A) provides the complete calculation and the final answer, and correctly applies the binomial probability formula, it is the most comprehensive correct answer. Option (B) is also mathematically correct as an expression, but option (A) shows the steps and the final numerical answer.

The correct option is (A).

The probability mass function (PMF) of a Poisson distribution is given by:

$P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$

where:

$\lambda$ is the average number of events in a given interval (the mean).

$x$ is the actual number of occurrences.

$e$ is the base of the natural logarithm (approximately 2.71828).

We need to find the probability $P(X=x)$ when $x=0$.

Substitute $x=0$ into the PMF:

$P(X=0) = \frac{e^{-\lambda} \lambda^0}{0!}$

Recall that any non-zero number raised to the power of 0 is 1 ($\lambda^0 = 1$, assuming $\lambda \neq 0$).

Also, by definition, $0! = 1$.

So, the expression becomes:

$P(X=0) = \frac{e^{-\lambda} \times 1}{1}$

$P(X=0) = e^{-\lambda}$

This matches option (B).

Let's look at the other options:

(A) $\lambda$: This is the mean, not the probability of zero events.

(C) $1 - e^{-\lambda}$: This represents the probability of having at least one event ($P(X \ge 1)$).

(D) 0: This would be the probability of an impossible event, but having zero events in a Poisson distribution is possible.

The correct option is (B).

We are given the expected value of a random variable $X$, $E(X) = 10$.

We need to find the expected value of a linear transformation of $X$, specifically $E(2X + 5)$.

We use the linearity property of expectation, which states that for any random variable $X$ and constants $a$ and $b$:

$E(aX + b) = aE(X) + b$

In this case, $a=2$ and $b=5$.

Substitute the given value of $E(X)$ into the formula:

$E(2X + 5) = 2 \times E(X) + 5$

$E(2X + 5) = 2 \times 10 + 5$

$E(2X + 5) = 20 + 5$

$E(2X + 5) = 25$

This matches option (C).

Let's look at the other options:

(A) 10: This is $E(X)$, not $E(2X+5)$.

(B) 20: This is $2 \times E(X)$, which is the expected value of $2X$ but does not include the $+5$ term.

(D) 15: This would be $E(X) + 5$, neglecting the multiplication by 2.

The correct option is (C).

We are given the variance of a random variable $X$, Var($X$) = 4.

We need to find the variance of a linear transformation of $X$, specifically Var($3X - 2$).

We use the properties of variance:

1. Var($aX$) = $a^2$ Var($X$), where $a$ is a constant.

2. Var($X + b$) = Var($X$), where $b$ is a constant (adding or subtracting a constant does not change the variance).

Combining these properties for Var($aX + b$):

Var($aX + b$) = Var($aX$) = $a^2$ Var($X$).

In this problem, $a=3$ and $b=-2$ (or we consider the form $3X - 2$).

Substitute the given value of Var($X$) into the formula:

Var($3X - 2$) = $3^2 \times$ Var($X$)

Var($3X - 2$) = $9 \times 4$

Var($3X - 2$) = 36

This matches option (C).

Let's look at the other options:

(A) $3 \times 4 - 2 = 10$: This incorrectly applies the linearity of expectation to variance and subtracts the constant.

(B) $9 \times 4 - 4 = 32$: This correctly squares the coefficient of X ($3^2=9$) and multiplies by the variance, but then incorrectly subtracts 4.

(D) $3 \times 4 = 12$: This incorrectly uses the coefficient (3) instead of its square ($3^2=9$) when multiplying by the variance.

The correct option is (C).

For a continuous random variable, the probability that the variable falls within a certain range is found by integrating its probability density function (PDF) over that range.

The given PDF is $f(x) = 2x$ for $0 \le x \le 1$, and $f(x) = 0$ otherwise.

We need to find the probability $P(0.5 \le X \le 1)$.

This is calculated by integrating the PDF from the lower bound (0.5) to the upper bound (1):

$P(0.5 \le X \le 1) = \int\limits_{0.5}^{1} f(x) dx$

$P(0.5 \le X \le 1) = \int\limits_{0.5}^{1} 2x dx$

Now, we perform the integration:

The integral of $2x$ with respect to $x$ is $x^2$.

$ \int 2x dx = x^2 + C $

We evaluate this from $0.5$ to $1$:

$ [x^2] \Big|_{0.5}^{1} = (1)^2 - (0.5)^2 $

$ = 1 - 0.25 $

$ = 0.75 $

This calculation matches option (B).

Let's examine why the other options are incorrect:

(A) $\int\limits_{0}^{1} 2x dx = [x^2]\limits_{0}^{1} = 1$: This calculates the total probability from 0 to 1, which is indeed 1 (as it should be for a valid PDF over its entire range), but it does not answer the specific question of $P(0.5 \le X \le 1)$.

(C) $f(1) - f(0.5) = 2(1) - 2(0.5) = 2 - 1 = 1$: This is a subtraction of function values, not an integration. Integration is required to find probabilities for continuous distributions.

(D) $f(0.75)$: This is evaluating the PDF at a single point, which for a continuous distribution has a probability of 0. Also, it's not the correct way to find the probability over an interval.

The correct option is (B).

The mode of a binomial distribution $B(n, p)$ is determined by the value of $(n+1)p$.

The rules are:

1. If $(n+1)p$ is not an integer, the mode is $\lfloor (n+1)p \rfloor$ (the greatest integer less than or equal to $(n+1)p$).

2. If $(n+1)p$ is an integer, then there are two modes: $(n+1)p - 1$ and $(n+1)p$.

In this problem, we have a binomial distribution $B(10, 0.4)$.

Here, $n=10$ and $p=0.4$.

First, calculate $(n+1)p$:

$(n+1)p = (10+1) \times 0.4$

$(n+1)p = 11 \times 0.4$

$(n+1)p = 4.4$

Since $4.4$ is not an integer, we apply the first rule. The mode is the floor of $4.4$.

Mode = $\lfloor 4.4 \rfloor = 4$.

This result matches option (A).

Let's examine the other options:

(B) 4 and 5: This would be the case if $(n+1)p$ were an integer, like 5.

(C) 5: This would be the mode if $(n+1)p = 5$, or if $(n+1)p = 5.x$ and the floor was 5.

(D) $(10+1) \times 0.4 = 4.4$: This is the value of $(n+1)p$, but not the mode itself.

The correct option is (A).

For a Poisson random variable $X$ with mean $\lambda$, the probability mass function is:

$P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$

Given that the mean $\lambda = 2$. We want to find $P(X \le 1)$.

$P(X \le 1)$ means the probability of having 0 occurrences or 1 occurrence. So, $P(X \le 1) = P(X=0) + P(X=1)$.

First, calculate $P(X=0)$:

$P(X=0) = \frac{e^{-2} 2^0}{0!}$

Since $2^0 = 1$ and $0! = 1$, this simplifies to:

$P(X=0) = \frac{e^{-2} \times 1}{1} = e^{-2}$

Next, calculate $P(X=1)$:

$P(X=1) = \frac{e^{-2} 2^1}{1!}$

Since $2^1 = 2$ and $1! = 1$, this simplifies to:

$P(X=1) = \frac{e^{-2} \times 2}{1} = 2e^{-2}$

Now, add these probabilities together:

$P(X \le 1) = P(X=0) + P(X=1)$

$P(X \le 1) = e^{-2} + 2e^{-2}$

$P(X \le 1) = (1 + 2)e^{-2}$

$P(X \le 1) = 3e^{-2}$

This matches option (A).

Let's look at the other options:

(B) $e^{-2}$: This is $P(X=0)$ only.

(C) $2e^{-2}$: This is $P(X=1)$ only.

(D) $1 - 3e^{-2}$: This would represent the probability of having 2 or more occurrences, $P(X \ge 2)$.

The correct option is (A).

Let's analyze the nature of the exponential distribution and the given options.

The exponential distribution is used to model the time until an event occurs in a Poisson process. For example, the time between arrivals of customers, or the time it takes to serve a customer.

The key characteristic of the exponential distribution is that it describes a quantity that can take any non-negative real value. Time, in this context, is a continuous variable; it can be measured with any degree of precision and can take any value within a certain range (e.g., 1.5 minutes, 1.55 minutes, 1.557 minutes, etc.).

Now let's look at the options:

(A) Discrete probability distribution: Discrete distributions deal with countable outcomes (like the number of customers, number of heads in coin flips). Since time is continuous, this is incorrect.

(B) Continuous probability distribution: Continuous distributions deal with outcomes that can take any value within a range. Since time is continuous, this is correct.

(C) Binomial distribution: The binomial distribution models the number of successes in a fixed number of independent Bernoulli trials (e.g., number of male children in a family). It deals with discrete counts, not continuous time.

(D) Poisson distribution: The Poisson distribution models the number of events occurring in a fixed interval of time or space, given a constant average rate. It also deals with discrete counts, not continuous time intervals.

Therefore, the exponential distribution is an example of a continuous probability distribution.

The correct option is (B).

A parameter is a numerical characteristic of a population or a probability distribution that is fixed but usually unknown. Parameters are used to define a distribution. Let's examine each option:

(A) Mean of a normal distribution ($\mu$): The mean ($\mu$) and standard deviation ($\sigma$) are the two defining parameters of a normal distribution. So, this is a parameter.

(B) Variance of a normal distribution ($\sigma^2$): The variance ($\sigma^2$) is directly related to the standard deviation ($\sigma$), which is a parameter of the normal distribution. So, this is also considered a parameter defining the spread.

(C) Number of trials in a binomial distribution ($n$): In a binomial distribution $B(n, p)$, both $n$ (the number of trials) and $p$ (the probability of success) are parameters that define the distribution.

(D) Expected value of a random variable ($E(X)$): The expected value $E(X)$ is a calculated property or statistic of a random variable based on its distribution and its possible values. While it's a numerical characteristic, it is typically a *result* of the distribution's parameters and definition, rather than a parameter itself that *defines* the distribution's shape or location in the same way as $\mu$, $\sigma^2$, $n$, or $p$. For example, $E(X)$ can be derived from the parameters of the distribution (e.g., $E(X)=np$ for binomial, $E(X)=\mu$ for normal).

Therefore, the expected value $E(X)$ is not considered a parameter of a distribution; rather, it is a statistic or characteristic that is derived from the parameters.

The correct option is (D).

A discrete random variable $X$ is a variable that takes on a finite number of values or a countably infinite number of values. It assigns a numerical value to each outcome of a random experiment.

In this problem, the random variable $X$ represents the number of heads obtained when tossing a coin twice.

Let's list all possible outcomes when tossing a coin twice:

Outcome 1: HH (Head on the first toss, Head on the second toss)

Outcome 2: HT (Head on the first toss, Tail on the second toss)

Outcome 3: TH (Tail on the first toss, Head on the second toss)

Outcome 4: TT (Tail on the first toss, Tail on the second toss)

Now, let's determine the value of the random variable $X$ (the number of heads) for each outcome:

For HH: Number of heads = 2. So, $X=2$.

For HT: Number of heads = 1. So, $X=1$.

For TH: Number of heads = 1. So, $X=1$.

For TT: Number of heads = 0. So, $X=0$.

The possible numerical values that the random variable $X$ can take are 0, 1, and 2.

Therefore, the set of possible values of $X$ is {0, 1, 2}.

Let's look at the options:

(A) {H, T}: These are the possible outcomes of a single coin toss, not the numerical values of the number of heads in two tosses.

(B) {0, 1, 2}: These are the possible numerical values for the number of heads when tossing a coin twice.

(C) {HH, HT, TH, TT}: These are the sample space (all possible outcomes), not the numerical values of the random variable X.

(D) {0, 1}: This would be the possible values if the coin was tossed only once, or if the variable represented something else.

The correct option is (B).

The expected value $E(X)$ of a discrete random variable $X$ is calculated by summing the products of each possible value of $X$ and its corresponding probability $P(X=x)$.

The formula is: $E(X) = \sum [x \cdot P(X=x)]$

From the given probability distribution table:

Possible values of $X$ are 0, 1, and 2.

The corresponding probabilities $P(X=x)$ are $1/4$, $1/2$, and $1/4$.

Now, we calculate the expected value:

$E(X) = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2))$

$E(X) = (0 \times \frac{1}{4}) + (1 \times \frac{1}{2}) + (2 \times \frac{1}{4})$

$E(X) = 0 + \frac{1}{2} + \frac{2}{4}$

$E(X) = \frac{1}{2} + \frac{1}{2}$

$E(X) = 1$

This calculation matches option (A).

Let's examine why the other options are incorrect:

(B) $(0+1+2)/3 = 1$: This is the average of the $X$ values, but it does not take into account the probabilities. It would only be correct if all probabilities were equal.

(C) $(1/4 + 1/2 + 1/4) = 1$: This is the sum of probabilities, which must be 1 for any valid probability distribution, but it is not the expected value.

(D) $0 \times 1/4 + 1 \times 1/4 + 2 \times 1/4 = 3/4$: This calculation incorrectly uses $1/4$ as the probability for $X=1$, instead of the correct probability of $1/2$.

The correct option is (A).

The expected value of $X^2$, denoted as $E(X^2)$, is calculated by summing the products of each possible value of $X^2$ and its corresponding probability $P(X=x)$.

The formula is: $E(X^2) = \sum [x^2 \cdot P(X=x)]$

From Question 80 and 81, the probability distribution is:

First, we find the values of $X^2$:

If $X=0$, then $X^2 = 0^2 = 0$.

If $X=1$, then $X^2 = 1^2 = 1$.

If $X=2$, then $X^2 = 2^2 = 4$.

Now, we calculate $E(X^2)$:

$E(X^2) = (0^2 \times P(X=0)) + (1^2 \times P(X=1)) + (2^2 \times P(X=2))$

$E(X^2) = (0 \times \frac{1}{4}) + (1 \times \frac{1}{2}) + (4 \times \frac{1}{4})$

$E(X^2) = 0 + \frac{1}{2} + \frac{4}{4}$

$E(X^2) = \frac{1}{2} + 1$

$E(X^2) = 1.5$

This calculation matches option (B).

Let's examine why the other options are incorrect:

(A) $1^2 = 1$: This only calculates $E(X)^2$ (the square of the mean from the previous question) and not $E(X^2)$, or it incorrectly assumes $E(X^2)$ is just the square of one of the $X$ values.

(C) $(0+1+4)/3 = 5/3$: This is the average of the squared $X$ values, not weighted by their probabilities.

(D) $(1/4 + 1/2 + 1) = 1.75$: This calculation seems to involve probabilities and squared values incorrectly.

The correct option is (B).

The variance of a random variable $X$ can be calculated using the formula:

Var($X$) = $E(X^2) - [E(X)]^2$

From the previous questions (Question 81 and 82):

Expected value $E(X) = 1$.

Expected value of $X^2$, $E(X^2) = 1.5$.

Now, we substitute these values into the variance formula:

Var($X$) = $E(X^2) - [E(X)]^2$

Var($X$) = $1.5 - (1)^2$

Var($X$) = $1.5 - 1$

Var($X$) = $0.5$

This calculation matches option (A).

Let's examine why the other options are incorrect:

(B) $1 - 1.5 = -0.5$: This incorrectly subtracts $E(X^2)$ from $[E(X)]^2$. Variance cannot be negative.

(C) $1.5$: This is the value of $E(X^2)$, not the variance.

(D) $1$: This is the value of $E(X)$, not the variance.

The correct option is (A).

A standard normal variable, denoted by $Z$, is a normal random variable with a mean of 0 and a variance of 1 (or standard deviation of 1). This means $Z \sim N(0, 1)$.

To transform any normally distributed random variable $X$ with mean $\mu$ and standard deviation $\sigma$ into a standard normal variable $Z$, we use the standardization formula.

The formula for standardization is:

$Z = \frac{X - \mu}{\sigma}$

Let's verify this transformation:

If $X \sim N(\mu, \sigma^2)$, then $E(X) = \mu$ and Var($X$) = $\sigma^2$ (so the standard deviation is $\sigma$).

Applying the transformation $Z = \frac{X - \mu}{\sigma}$:

Expected value of Z: $E(Z) = E\left(\frac{X - \mu}{\sigma}\right) = \frac{1}{\sigma} E(X - \mu) = \frac{1}{\sigma} (E(X) - \mu) = \frac{1}{\sigma} (\mu - \mu) = \frac{0}{\sigma} = 0$.

Variance of Z: Var($Z$) = Var$\left(\frac{X - \mu}{\sigma}\right) = \left(\frac{1}{\sigma}\right)^2$ Var($X - \mu$) = $\frac{1}{\sigma^2}$ Var($X$) = $\frac{1}{\sigma^2} \times \sigma^2 = 1$.

Since $E(Z)=0$ and Var($Z$)=1, the transformation $Z = \frac{X - \mu}{\sigma}$ indeed results in a standard normal variable.

Comparing this with the given options:

(A) $Z = \frac{X + \mu}{\sigma}$: This transformation would shift the mean by $\mu$, not center it at 0.

(B) $Z = \frac{X - \mu}{\sigma}$: This matches the correct standardization formula.

(C) $Z = \frac{\mu - X}{\sigma}$: This transformation would invert the variable ($X-\mu$ becomes $\mu-X$), resulting in a mean of 0 but a potentially different interpretation depending on the original distribution's skewness if it were not symmetric.

(D) $Z = \frac{X - \sigma}{\mu}$: This transformation incorrectly uses the standard deviation ($\sigma$) in the numerator and the mean ($\mu$) in the denominator, and it's not the standard way to standardize.

The correct option is (B).

A random variable can be classified as either discrete or continuous based on the nature of the values it can take.

A discrete random variable can only take on a finite number of values or a countably infinite number of values. These values are typically counts or whole numbers.

A continuous random variable can take on any value within a given range. These values are typically measurements.

In this question, the random variable represents the "number of siblings" a person has.

The number of siblings is a count. A person can have 0 siblings, 1 sibling, 2 siblings, 3 siblings, and so on. It is not possible to have 1.5 siblings or $\pi$ siblings.

Since the possible values are whole numbers (counts), the random variable is discrete.

Let's look at the options:

(A) Continuous: This is incorrect because the number of siblings is a count, not a measurement that can take any value in a range.

(B) Discrete: This is correct because the number of siblings can only take on specific, countable values (0, 1, 2, ...).

(C) Normal: The normal distribution is a type of probability distribution that can be used to model discrete variables under certain conditions (like approximating a binomial distribution with a large n), but the variable itself is fundamentally discrete, not inherently normal. Normal is a distribution type, not a classification of variable type.

(D) Both discrete and continuous: A random variable is either discrete or continuous, not both.

The correct option is (B).

The standard deviation is the square root of the variance.

The formula relating standard deviation ($\sigma$) and variance (Var($X$)) is:

$\sigma = \sqrt{\text{Var}(X)}$

We are given:

Expected value $E(X) = \textsf{₹}500$ (This information is not needed to calculate the standard deviation).

Variance Var($X$) = $\textsf{₹}100$.

Now, we calculate the standard deviation:

$\sigma = \sqrt{\textsf{₹}100}$

$\sigma = \textsf{₹}10$

This matches option (C).

Let's look at the other options:

(A) $\textsf{₹}100$: This is the variance itself.

(B) $\textsf{₹}250000$: This is the square of the expected value ($500^2$), which is incorrect.

(D) $\textsf{₹}500$: This is the expected value.

The correct option is (C).

A random variable is a function that maps the outcomes of a random experiment to numerical values. In simpler terms, it's a variable whose possible values are numerical outcomes of a random phenomenon.

Random variables can be broadly classified into two types: discrete random variables and continuous random variables.

Discrete Random Variable:

A discrete random variable is a variable whose possible values are countable and finite or countably infinite. The values can often be listed. There are gaps between the possible values.

Example of a Discrete Random Variable:

Consider the experiment of tossing a fair coin three times. Let $X$ be the random variable representing the number of heads obtained.

The possible outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.

The values $X$ can take are the number of heads in each outcome:

• HHH $\to$ $X=3$
• HHT $\to$ $X=2$
• HTH $\to$ $X=2$
• THH $\to$ $X=2$
• HTT $\to$ $X=1$
• THT $\to$ $X=1$
• TTH $\to$ $X=1$
• TTT $\to$ $X=0$

Thus, the possible values for $X$ are $\{0, 1, 2, 3\}$. These values are countable and finite, making $X$ a discrete random variable.

Continuous Random Variable:

A continuous random variable is a variable that can take any value within a given interval or range. Its possible values are uncountable. There are no gaps between the possible values within its range.

Example of a Continuous Random Variable:

Consider measuring the height of students in a class. Let $Y$ be the random variable representing the height of a randomly selected student.

The height can take any value within a certain range, for example, between $1.50$ meters and $1.80$ meters. A student's height could be $1.65$ m, $1.655$ m, $1.6552$ m, and so on. It's not limited to specific, distinct values.

Thus, $Y$ can take any value in the interval, say, $(1.50, 1.80)$ (assuming heights are measured in meters). Since there are infinitely many possible values within this interval, $Y$ is a continuous random variable.

Probability Distribution

A probability distribution is a mathematical function that describes the likelihood of obtaining the possible values that a random variable can assume. In other words, it provides the probabilities of all possible outcomes of a random experiment.

A probability distribution allows us to quantify the uncertainty associated with a random variable. It can be represented in various forms, such as a table, graph, or formula. There are two main types of probability distributions, based on whether the random variable is discrete or continuous:

1. Probability Mass Function (PMF): Used for discrete random variables (variables that can take on a finite or countably infinite number of distinct values).

2. Probability Density Function (PDF): Used for continuous random variables (variables that can take on any value within a given range).

Probability Mass Function (PMF)

A Probability Mass Function (PMF) is a function that gives the probability that a discrete random variable is exactly equal to some specific value. It is denoted as $P(X=x)$, where $X$ is the random variable and $x$ is a particular value that $X$ can take.

For instance, if we consider the experiment of rolling a fair six-sided die, the random variable $X$ representing the outcome can take values from the set $\{1, 2, 3, 4, 5, 6\}$. The PMF for this random variable is $P(X=x) = \frac{1}{6}$ for each $x$ in this set, and $P(X=x) = 0$ for any other value of $x$.

Conditions for a function $P(x)$ to be a PMF

For any function $P(x)$ to be a valid Probability Mass Function for a discrete random variable $X$, it must satisfy the following two fundamental conditions:

1. Non-negativity Condition:

The probability of any possible outcome $x$ must be non-negative. It is impossible to have a negative probability for an event.

2. Summation Condition:

The sum of the probabilities for all possible values that the random variable $X$ can take must be exactly equal to 1. This ensures that the total probability covers all possible outcomes and that we are 100% certain that one of the outcomes will occur.

If a function $P(x)$ fulfills both of these conditions, it is a valid PMF.

Given:

A fair coin is tossed twice. The random variable X is defined as the number of heads obtained.

Sample Space

When a coin is tossed twice, the possible outcomes can be represented by pairs of letters, where H stands for Heads and T stands for Tails.

The sample space (S), which is the set of all possible outcomes, is:

The total number of possible outcomes is 4.

Possible Values of the Random Variable X

The random variable X represents the number of heads in each outcome.

For the outcome TT, the number of heads is 0. Thus, $X=0$.

For the outcomes HT and TH, the number of heads is 1. Thus, $X=1$.

For the outcome HH, the number of heads is 2. Thus, $X=2$.

So, the possible values that the random variable X can take are 0, 1, and 2.

Calculating the Probabilities

We now calculate the probability for each value of X.

1. Probability of getting 0 heads, $P(X=0)$

There is only one outcome with 0 heads: {TT}.

$P(X=0) = \frac{\text{Number of outcomes with 0 heads}}{\text{Total number of outcomes}} = \frac{1}{4}$

2. Probability of getting 1 head, $P(X=1)$

There are two outcomes with 1 head: {HT, TH}.

$P(X=1) = \frac{\text{Number of outcomes with 1 head}}{\text{Total number of outcomes}} = \frac{2}{4} = \frac{1}{2}$

3. Probability of getting 2 heads, $P(X=2)$

There is one outcome with 2 heads: {HH}.

$P(X=2) = \frac{\text{Number of outcomes with 2 heads}}{\text{Total number of outcomes}} = \frac{1}{4}$

Probability Distribution Table for X

The probability distribution of the random variable X is given by the following table:

We can check that the sum of probabilities is 1, as required for a probability distribution:

$\sum P(X=x) = P(X=0) + P(X=1) + P(X=2) = \frac{1}{4} + \frac{1}{2} + \frac{1}{4} = \frac{1+2+1}{4} = \frac{4}{4} = 1$.

To Check:

Whether the given distribution is a valid probability distribution for a random variable X.

Justification:

For a function to be a valid probability distribution (specifically, a Probability Mass Function or PMF), it must satisfy two essential conditions:

1. The probability of each event, $P(X=x)$, must be non-negative. That is, $P(X=x) \ge 0$ for all values of $x$.

2. The sum of all probabilities must be equal to 1. That is, $\sum P(X=x) = 1$.

Let's check these two conditions for the given distribution.

Step 1: Checking the Non-negativity Condition

We examine the probabilities for each value of X:

$P(X=0) = 0.3$

$P(X=1) = 0.4$

$P(X=2) = 0.3$

All these probabilities (0.3, 0.4, and 0.3) are greater than 0. Therefore, the first condition, $P(X=x) \ge 0$, is satisfied.

Step 2: Checking the Summation Condition

We sum the probabilities for all possible values of X:

Substituting the given values:

The sum of the probabilities is exactly 1. Therefore, the second condition, $\sum P(X=x) = 1$, is also satisfied.

Conclusion

Since both conditions for a valid probability distribution are met, we can conclude that the given table represents a valid probability distribution for the random variable X.

Mathematical Expectation (Mean) of a Discrete Random Variable

The Mathematical Expectation, also known as the Expected Value or the Mean, of a discrete random variable is the long-run average value of the random variable. It represents the weighted average of all possible values that the random variable can take, where the weights are the probabilities of those values occurring.

In simpler terms, if you were to repeat a random experiment an infinite number of times, the average of all the outcomes would be the expected value. It is a fundamental concept in probability theory that helps to summarize the central tendency of a probability distribution.

The expected value is often denoted by $E(X)$ or by the Greek letter $\mu$ (mu).

Formula for the Mathematical Expectation, E(X)

Let X be a discrete random variable that can take on a finite number of values $x_1, x_2, x_3, \ldots, x_n$ with corresponding probabilities $P(X=x_1), P(X=x_2), P(X=x_3), \ldots, P(X=x_n)$.

The formula for the mathematical expectation (or mean) of X is given by the sum of the product of each possible value of X and its corresponding probability:

Where:

• $E(X)$ or $\mu$ is the expected value or mean of the random variable X.
• $x_i$ is the $i$-th possible value of the random variable X.
• $P(X=x_i)$ is the probability that the random variable X takes the value $x_i$.
• $\sum$ is the summation symbol, indicating that we sum the products for all possible values of X (from $i=1$ to $n$).

Essentially, to find the expected value, you multiply each possible outcome by its probability and then add all these products together.

Given:

The probability distribution for a random variable X is given as follows:

To Find:

The expected value of the random variable X, which is denoted by $E(X)$.

Formula Used:

The expected value (or mean) of a discrete random variable X is calculated by summing the products of each possible value of X and its corresponding probability. The formula is:

Solution:

To find the expected value, we multiply each value of $x$ by its probability $P(X=x)$ and then add up these products. We can organize this calculation in a table.

Now, we sum the values in the last column to get the expected value:

Thus, the expected value of the random variable X is 0.1.

Variance of a Discrete Random Variable

The Variance of a discrete random variable is a measure of the spread or dispersion of its probability distribution. It quantifies how much the values of a random variable differ from its expected value (mean) on average.

Specifically, the variance is defined as the expected value of the squared deviation of the random variable from its mean. A small variance indicates that the values of the random variable are clustered closely around the mean, while a large variance indicates that the values are more spread out.

The variance is always non-negative, since it is an average of squared values. It is denoted by $\text{Var}(X)$ or $\sigma^2$ (sigma-squared).

Formula for Variance, Var(X)

Let X be a discrete random variable with a set of possible values $\{x_1, x_2, \ldots, x_n\}$, corresponding probabilities $P(X=x_i)$, and a mean (expected value) of $\mu = E(X)$.

There are two common formulas to calculate the variance:

1. The Definitional Formula

This formula directly follows from the definition of variance as the expected value of the squared difference from the mean:

For a discrete random variable, this is calculated using the summation:

This means you calculate the squared difference between each value $x_i$ and the mean $\mu$, multiply it by its probability, and sum up all the results.

2. The Computational Formula

A more convenient formula for manual calculations is derived from the definitional formula. It states that the variance is the mean of the squares minus the square of the mean.

Where:

• $E[X]$ is the expected value (mean) of X.
• $E[X^2]$ is the expected value of $X^2$, which is calculated as:

Both formulas will yield the same result, but the computational formula is often easier to apply as it avoids calculating the deviation from the mean for each value.

Given:

For a random variable X, we are given:

1. The expected value (mean) of X, $E(X) = 3$.

2. The expected value of $X^2$, $E(X^2) = 15$.

To Find:

The variance of the random variable X, denoted by $\text{Var}(X)$.

Formula Used:

The variance of a random variable X can be calculated using the computational formula, which relates the variance to the mean of the squares and the square of the mean:

Solution:

We will substitute the given values of $E(X)$ and $E(X^2)$ into the formula for variance.

Starting with the formula:

Substitute the given values, $E(X) = 3$ and $E(X^2) = 15$:

Now, we calculate the square of the mean:

Finally, we perform the subtraction:

Therefore, the variance of the random variable X is 6.

The properties of expectation are fundamental in probability theory and are collectively known as the Linearity of Expectation. Here are the specific properties for the given cases.

Property 1: Expectation of a Linear Transformation E(aX + b)

This property describes how the expectation behaves when a random variable X is scaled by a constant 'a' and shifted by a constant 'b'.

If X is a random variable and 'a' and 'b' are constants, then the expectation of the linear transformation $aX + b$ is given by:

This property can be broken down into two simpler rules:

• Scaling by a constant: The expectation of a random variable multiplied by a constant is the constant multiplied by the expectation of the random variable.

  $E(aX) = aE(X)$

• Shifting by a constant: The expectation of a random variable plus a constant is the expectation of the random variable plus that constant.

  $E(X + b) = E(X) + b$

A special case of this is the expectation of a constant 'b', which is just the constant itself: $E(b) = b$.

Property 2: Expectation of a Sum of Random Variables E(X + Y)

This property, also known as the Additivity of Expectation, describes the expectation of the sum of two or more random variables.

If X and Y are any two random variables, the expectation of their sum is the sum of their individual expectations:

A very important aspect of this property is that it holds true for any random variables X and Y, regardless of whether they are independent or dependent.

This rule can be generalized to the sum of any finite number of random variables:

Here are the fundamental properties of variance for linear transformations and sums of random variables.

Property 1: Variance of a Linear Transformation Var(aX + b)

This property describes how the variance is affected when a random variable X is scaled by a constant 'a' and shifted by a constant 'b'.

If X is a random variable and 'a' and 'b' are constants, then the variance of the linear transformation $aX + b$ is given by:

This property can be understood by considering the effects of scaling and shifting separately:

• Effect of Scaling (multiplying by 'a'): When a random variable is multiplied by a constant 'a', its variance is multiplied by the square of that constant, $a^2$. This is because variance measures the spread in squared units.

  $\text{Var}(aX) = a^2 \text{Var}(X)$

• Effect of Shifting (adding 'b'): Adding a constant 'b' to a random variable shifts the entire distribution, including its mean, by 'b'. However, the spread or dispersion of the values around the mean does not change. Therefore, the variance remains unaffected by an additive constant.

  $\text{Var}(X + b) = \text{Var}(X)$

Combining these two effects gives the general rule.

Property 2: Variance of a Sum of Independent Random Variables Var(X + Y)

This property describes the variance of the sum of two random variables.

If X and Y are two independent random variables, the variance of their sum is the sum of their individual variances:

Important Note: This property crucially relies on the condition that X and Y are independent. If the variables are not independent, the formula is more complex and involves their covariance: $\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) + 2\text{Cov}(X, Y)$.

Similarly, for the difference of two independent random variables, the variances still add:

This is because $\text{Var}(X - Y) = \text{Var}(X + (-1)Y)$. Using the properties, if X and Y are independent, then $\text{Var}(X + (-1)Y) = \text{Var}(X) + \text{Var}(-1 \cdot Y) = \text{Var}(X) + (-1)^2\text{Var}(Y) = \text{Var}(X) + \text{Var}(Y)$.

Given:

The expected value of a random variable X is given as:

To Find:

The value of the expression $E(2X + 3)$.

Formula Used:

We will use the property of linearity of expectation. For a random variable X and any constants 'a' and 'b', the property is given by:

Solution:

We are asked to find the value of $E(2X + 3)$.

By applying the property of linearity of expectation with $a=2$ and $b=3$, we can write:

Now, we substitute the given value, $E(X) = 5$, into this equation:

Next, we perform the arithmetic operations:

Thus, the expected value of $2X + 3$ is 13.

Given:

The variance of a random variable X is given as:

To Find:

The value of the expression $\text{Var}(3X - 2)$.

Formula Used:

We will use the property of variance for a linear transformation. For a random variable X and any constants 'a' and 'b', the property is given by:

This property states that the variance is affected by scaling (multiplication) but not by shifting (addition or subtraction).

Solution:

We are asked to find the value of $\text{Var}(3X - 2)$.

By applying the property of variance with $a=3$ and $b=-2$, we can write:

Now, we substitute the given value, $\text{Var}(X) = 4$, into this equation:

Next, we calculate the square of the constant 'a':

Finally, we perform the multiplication:

Thus, the variance of $3X - 2$ is 36.

Binomial Distribution

The Binomial distribution is a fundamental discrete probability distribution that models the number of successes in a fixed number of independent trials. It is used when an experiment, known as a Bernoulli trial, is repeated multiple times, and each trial has only two possible outcomes (e.g., success/failure, yes/no, heads/tails).

An experiment can be described by a Binomial distribution if it satisfies the following four conditions:

• Fixed Number of Trials (n): The experiment consists of a fixed number of identical trials, denoted by $n$.
• Independent Trials: The outcome of each trial is independent of the outcomes of all other trials.
• Two Possible Outcomes: Each trial results in one of only two possible outcomes, which are conventionally labeled as "success" and "failure".
• Constant Probability of Success (p): The probability of a "success," denoted by $p$, remains the same for every trial. The probability of a "failure" is therefore $(1-p)$, often denoted by $q$.

A random variable X that counts the number of successes in such an experiment is said to follow a Binomial distribution, denoted as $X \sim B(n, p)$.

Probability Mass Function (PMF) of a Binomial Distribution

The Probability Mass Function (PMF) of a Binomial distribution gives the probability of obtaining exactly $k$ successes in $n$ independent trials. The formula for the PMF is:

This formula is valid for $k = 0, 1, 2, \ldots, n$.

The components of the formula are:

• $n$ is the total number of trials.
• $k$ is the specific number of successes for which we are calculating the probability.
• $p$ is the probability of success on a single trial.
• $(1-p)$ is the probability of failure on a single trial.
• $\binom{n}{k}$ is the binomial coefficient, which represents the number of different ways to choose $k$ successful trials from a total of $n$ trials. It is calculated as:

  $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Given:

A random variable X follows a Binomial distribution, which can be written as $X \sim B(n, p)$.

The parameters for this distribution are:

• Number of trials, $n = 4$
• Probability of success, $p = 0.5$

To Find:

The probability of obtaining exactly 2 successes, which is denoted as $P(X=2)$.

Formula Used:

The Probability Mass Function (PMF) for a Binomial distribution is given by the formula:

where $k$ is the number of successes, and the binomial coefficient $\binom{n}{k}$ is calculated as $\frac{n!}{k!(n-k)!}$.

Solution:

We need to calculate $P(X=2)$ using the given values: $n=4$, $p=0.5$, and the desired number of successes $k=2$.

First, substitute these values into the PMF formula:

Simplify the expression inside the parentheses:

Now, let's calculate each component of the formula:

1. Calculate the binomial coefficient $\binom{4}{2}$:

2. Calculate the probability terms:

3. Substitute these results back into the main equation:

Alternate Solution (Using Fractions)

We can also solve this using fractions, where $p=0.5 = \frac{1}{2}$.

Simplifying the fraction:

Thus, the probability of getting exactly 2 successes is 0.375 or $\frac{3}{8}$.

Given:

A random variable X follows a Binomial distribution, which is denoted as $X \sim B(n, p)$.

The parameters for this distribution are:

• Number of trials, $n = 5$
• Probability of success, $p = 0.4$

To Find:

The mean ($E(X)$) and the variance ($\text{Var}(X)$) of the random variable X.

Formulas Used:

For a Binomial distribution with parameters $n$ and $p$:

1. The formula for the mean (expected value) is:

2. The formula for the variance is:

Solution:

1. Calculation of the Mean, E(X)

We use the formula for the mean and substitute the given values, $n=5$ and $p=0.4$.

2. Calculation of the Variance, Var(X)

We use the formula for the variance. First, we calculate the probability of failure, $(1-p)$:

Now, we substitute the values into the variance formula:

Since we already know that $np = 2$, we can simplify the calculation:

Conclusion:

For the given Binomial distribution $X \sim B(5, 0.4)$, the results are:

• The mean is $E(X) = 2$.
• The variance is $\text{Var}(X) = 1.2$.

Poisson Distribution

The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space. It is used to model situations where events happen independently and at a constant average rate.

A random process follows a Poisson distribution if it meets the following criteria:

• Events occur independently. The occurrence of one event does not affect the probability of another event occurring.
• The average rate at which events occur is constant. This rate is denoted by the Greek letter $\lambda$ (lambda).
• The probability of an event occurring is proportional to the length of the interval.
• Two events cannot occur at exactly the same instant.

Common examples of phenomena that can be modeled by a Poisson distribution include:

• The number of phone calls a call center receives per hour.
• The number of typographical errors on a single page of a book.
• The number of customers arriving at a store in a 10-minute interval.

A random variable X that follows a Poisson distribution is denoted as $X \sim \text{Poisson}(\lambda)$.

Probability Mass Function (PMF) of a Poisson Distribution

The Probability Mass Function (PMF) of a Poisson distribution gives the probability of observing exactly $k$ events in a given interval, when the average number of events is $\lambda$.

The formula for the PMF is:

This formula is valid for $k = 0, 1, 2, \ldots$.

The components of the formula are:

• $k$ is the actual number of occurrences for which the probability is being calculated (a non-negative integer).
• $\lambda$ (lambda) is the average rate of occurrences (mean) for the given interval. It must be a positive number ($\lambda > 0$).
• $e$ is Euler's number, the base of the natural logarithm, approximately equal to 2.71828.
• $k!$ is the factorial of $k$.

For a Poisson distribution, the probability mass function is given by:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the mean and $k$ is the number of events.

In this case, $\lambda = 2$ and we want to find $P(X = 0)$.

So, $P(X = 0) = \frac{e^{-2} 2^0}{0!} = \frac{e^{-2} \cdot 1}{1} = e^{-2}$

Given that $e^{-2} \approx 0.1353$, therefore,

$P(X = 0) \approx 0.1353$

If X follows a Poisson distribution with parameter $\lambda$, denoted as X ~ P($\lambda$), then:

Mean of X, $E(X) = \lambda$

Variance of X, $Var(X) = \lambda$

For a Poisson distribution with mean $\lambda$, the probability mass function is given by:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$

Given that $P(X = 1) = P(X = 2)$, we have:

$\frac{e^{-\lambda} \lambda^1}{1!} = \frac{e^{-\lambda} \lambda^2}{2!}$

We can cancel out $e^{-\lambda}$ from both sides:

$\frac{\lambda}{1} = \frac{\lambda^2}{2}$

Multiplying both sides by 2, we get:

$2\lambda = \lambda^2$

Rearranging the terms:

$\lambda^2 - 2\lambda = 0$

$\lambda(\lambda - 2) = 0$

This gives us two possible values for $\lambda$: $\lambda = 0$ or $\lambda = 2$.

Since $\lambda$ represents the mean of the Poisson distribution, it must be a positive value. Therefore, $\lambda = 0$ is not a valid solution in this context.

Thus, the mean of the distribution is $\lambda = 2$.

A Normal distribution, also known as the Gaussian distribution, is a continuous probability distribution that is symmetric around its mean. It is characterized by its bell-shaped curve and is defined by two parameters: the mean ($\mu$) and the standard deviation ($\sigma$). The probability density function (pdf) of a normal distribution is given by:

$f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2}(\frac{x - \mu}{\sigma})^2}$

Properties of a Normal Distribution Curve:

1. Symmetry: The normal distribution curve is perfectly symmetric around its mean ($\mu$). This means that the left and right halves of the curve are mirror images of each other.

2. Mean, Median, and Mode: In a normal distribution, the mean, median, and mode are all equal and located at the center of the distribution.

3. Bell-Shaped: The curve has a bell shape with a single peak at the mean. The curve tapers off symmetrically away from the mean.

4. Asymptotic to the x-axis: The curve approaches the x-axis (horizontal axis) but never actually touches or crosses it. It extends infinitely in both directions.

5. Total Area Under the Curve: The total area under the normal distribution curve is equal to 1, representing the total probability of all possible outcomes.

6. Empirical Rule (68-95-99.7 Rule): This rule states that:

   • Approximately 68% of the data falls within one standard deviation of the mean ($\mu \pm \sigma$).
   • Approximately 95% of the data falls within two standard deviations of the mean ($\mu \pm 2\sigma$).
   • Approximately 99.7% of the data falls within three standard deviations of the mean ($\mu \pm 3\sigma$).

7. Unimodal: The normal distribution has only one mode (peak), which occurs at the mean.

8. Defined by $\mu$ and $\sigma$: The shape and position of the normal distribution are completely determined by its mean ($\mu$) and standard deviation ($\sigma$). The mean determines the center of the curve, and the standard deviation determines the spread or dispersion of the data.

For a Normal distribution with mean $\mu$ and standard deviation $\sigma$, the probability density function (PDF) is given by:

$f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2}(\frac{x - \mu}{\sigma})^2}$

Where:

• $x$ is the random variable.
• $\mu$ is the mean of the distribution.
• $\sigma$ is the standard deviation of the distribution.
• $\pi \approx 3.14159$
• $e \approx 2.71828$

The mean ($\mu$) of a Normal distribution determines the center of the distribution. Changing the mean shifts the entire distribution curve along the x-axis (horizontal axis) without changing its shape or spread.

Effect of increasing $\mu$ (keeping $\sigma$ constant):

If we increase the value of $\mu$, the entire normal distribution curve shifts to the right along the x-axis. The peak of the curve, which represents the mean, moves to a higher value on the x-axis. The shape and spread (standard deviation) of the curve remain the same.

Effect of decreasing $\mu$ (keeping $\sigma$ constant):

If we decrease the value of $\mu$, the entire normal distribution curve shifts to the left along the x-axis. The peak of the curve moves to a lower value on the x-axis. Again, the shape and spread of the curve remain unchanged.

In summary, changing the mean $\mu$ of a Normal distribution results in a horizontal translation (shift) of the curve. It does not affect the shape or spread of the distribution.

The standard deviation ($\sigma$) of a Normal distribution determines the spread or dispersion of the distribution. Changing the standard deviation affects the shape of the curve, making it either wider and flatter or narrower and taller, while the mean remains at the same location.

Effect of increasing $\sigma$ (keeping $\mu$ constant):

If we increase the value of $\sigma$, the normal distribution curve becomes wider and flatter. This means that the data is more spread out, and the peak of the curve is lower. The probability of observing values far from the mean increases.

Effect of decreasing $\sigma$ (keeping $\mu$ constant):

If we decrease the value of $\sigma$, the normal distribution curve becomes narrower and taller. This means that the data is more concentrated around the mean, and the peak of the curve is higher. The probability of observing values far from the mean decreases.

In summary, changing the standard deviation $\sigma$ of a Normal distribution affects the spread (dispersion) of the curve. A larger $\sigma$ results in a wider, flatter curve, while a smaller $\sigma$ results in a narrower, taller curve. The mean of the distribution remains unchanged.

For a Normal distribution, approximately 68% of the data lies within one standard deviation of the mean ($\mu \pm \sigma$). This is a part of the empirical rule (68-95-99.7 rule) for normal distributions.

Specifically, about 68% of the values fall between $\mu - \sigma$ and $\mu + \sigma$.

The standard Normal distribution is a special case of the Normal distribution. It is a Normal distribution with a mean of 0 and a standard deviation of 1.

Mean: $\mu = 0$

Standard Deviation: $\sigma = 1$

The probability density function (PDF) of the standard Normal distribution is given by:

$f(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}z^2}$

Where $z$ is the standard score (z-score), representing the number of standard deviations a data point is from the mean.

This is a binomial probability problem. We can use the binomial probability formula:

$P(X = k) = {n \choose k} p^k (1-p)^{n-k}$

Where:

• $n$ is the number of trials (coin tosses).
• $k$ is the number of successful trials we want (number of heads).
• $p$ is the probability of success on a single trial (probability of getting a head on one toss).
• ${n \choose k}$ is the binomial coefficient, which represents the number of ways to choose $k$ successes from $n$ trials.

In this case:

• $n = 5$
• $k = 3$
• $p = 0.5$ (since the coin is fair)

So, we have:

$P(X = 3) = {5 \choose 3} (0.5)^3 (1-0.5)^{5-3}$

$P(X = 3) = {5 \choose 3} (0.5)^3 (0.5)^2$

Calculate the binomial coefficient:

${5 \choose 3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(2 \times 1)} = \frac{5 \times 4}{2} = 10$

Now, plug the values back into the formula:

$P(X = 3) = 10 \times (0.5)^3 \times (0.5)^2 = 10 \times (0.125) \times (0.25) = 10 \times 0.03125 = 0.3125$

Therefore, the probability of getting exactly 3 heads in 5 coin tosses is 0.3125.

For a Poisson distribution, the probability mass function is given by:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the mean and $k$ is the number of events.

In this case, $\lambda = 0.5$ and we want to find $P(X = 1)$.

So, $P(X = 1) = \frac{e^{-0.5} (0.5)^1}{1!} = e^{-0.5} \times 0.5$

Given that $e^{-0.5} \approx 0.6065$, therefore,

$P(X = 1) \approx 0.6065 \times 0.5 = 0.30325$

Thus, the probability of finding exactly 1 defect in an item is approximately 0.30325.

We can use the linearity property of expectation, which states that for any random variable X and constants a and b:

$E(aX + b) = aE(X) + b$

In this case, we want to find $E(X - 10)$. We can rewrite this as $E(1 \cdot X + (-10))$.

So, $a = 1$ and $b = -10$.

Using the linearity property, we have:

$E(X - 10) = E(X) - 10$

Given that $E(X) = 10$, we can substitute this value:

$E(X - 10) = 10 - 10 = 0$

Therefore, $E(X - 10) = 0$.

If X and Y are independent random variables, then:

$Var(X - Y) = Var(X) + Var(-Y)$

Also, $Var(aY) = a^2 Var(Y)$, where a is a constant. In this case, $a = -1$, so $Var(-Y) = (-1)^2 Var(Y) = Var(Y)$.

Therefore, $Var(X - Y) = Var(X) + Var(Y)$

Given that $Var(X) = 5$ and $Var(Y) = 7$, we have:

$Var(X - Y) = 5 + 7 = 12$

Thus, $Var(X - Y) = 12$.

For a distribution to be a valid probability distribution, the following two conditions must be met:

1. The probability of each outcome must be between 0 and 1, inclusive. That is, $0 \le P(Y=y) \le 1$ for all y.
2. The sum of the probabilities of all possible outcomes must equal 1. That is, $\sum P(Y=y) = 1$.

Let's check the given distribution:

• $P(Y=1) = 0.2$
• $P(Y=2) = 0.3$
• $P(Y=3) = 0.6$

All probabilities are between 0 and 1, so the first condition is met.

Now, let's check if the sum of the probabilities equals 1:

$P(Y=1) + P(Y=2) + P(Y=3) = 0.2 + 0.3 + 0.6 = 1.1$

The sum of the probabilities is 1.1, which is not equal to 1. Therefore, the second condition is not met.

Conclusion: The given distribution is not a valid probability distribution because the sum of the probabilities is not equal to 1.

X can take values 0, 1, or 2, representing the number of red balls drawn.

Total number of ways to draw 2 balls from 10: ${10 \choose 2} = \frac{10!}{2!8!} = \frac{10 \times 9}{2} = 45$

1. P(X = 0): This means both balls drawn are blue.

Number of ways to draw 2 blue balls from 7: ${7 \choose 2} = \frac{7!}{2!5!} = \frac{7 \times 6}{2} = 21$

$P(X = 0) = \frac{21}{45} = \frac{7}{15}$

2. P(X = 1): This means one red ball and one blue ball are drawn.

Number of ways to draw 1 red ball from 3: ${3 \choose 1} = 3$

Number of ways to draw 1 blue ball from 7: ${7 \choose 1} = 7$

Number of ways to draw 1 red and 1 blue ball: $3 \times 7 = 21$

$P(X = 1) = \frac{21}{45} = \frac{7}{15}$

3. P(X = 2): This means both balls drawn are red.

Number of ways to draw 2 red balls from 3: ${3 \choose 2} = \frac{3!}{2!1!} = \frac{3 \times 2}{2} = 3$

$P(X = 2) = \frac{3}{45} = \frac{1}{15}$

Probability Distribution of X:

For a Binomial distribution with parameters $n$ (number of trials) and $p$ (probability of success), the mean and variance are given by:

• Mean: $\mu = np$
• Variance: $\sigma^2 = np(1-p)$

Given that the mean is 4 and the variance is 2, we have:

Now we can substitute (i) into (ii):

$4(1-p) = 2$

Divide both sides by 4:

$1-p = \frac{2}{4} = 0.5$

Solve for $p$:

$p = 1 - 0.5 = 0.5$

Now substitute the value of $p$ back into equation (i):

$n(0.5) = 4$

Solve for $n$:

$n = \frac{4}{0.5} = 8$

Therefore, the parameters are $n = 8$ and $p = 0.5$.

A Binomial distribution can be approximated by a Poisson distribution under the following conditions:

1. Large number of trials (n): The number of trials, $n$, should be sufficiently large. Generally, $n \ge 30$ is considered a good rule of thumb.

2. Small probability of success (p): The probability of success on a single trial, $p$, should be small. Generally, $p \le 0.1$ is considered a good guideline.

3. The product np is moderate: The product of $n$ and $p$, which represents the mean ($\lambda = np$), should be a moderate value. Typically, it should be less than 10. If $np$ is too large, the Poisson approximation may not be accurate.

In summary, when $n$ is large, $p$ is small, and $np$ is a moderate value (usually less than 10), the Poisson distribution provides a good approximation to the Binomial distribution. This approximation is useful because the Poisson distribution is often easier to work with than the Binomial distribution, especially when dealing with large values of $n$.

For a Normal distribution:

Mode: The mode is the value that appears most frequently in the distribution. In a Normal distribution, the mode is equal to the mean ($\mu$).

Median: The median is the middle value when the data is arranged in ascending order. In a Normal distribution, the median is also equal to the mean ($\mu$).

Relationship to the Mean:

In a Normal distribution, the mean, median, and mode are all equal. This is because the distribution is symmetric around the mean. Therefore:

Mean = Median = Mode = $\mu$

The Standard Deviation of a discrete random variable is a measure of the spread or dispersion of the variable's values around its mean. It quantifies the average distance of the values from the mean.

The formula for the standard deviation ($\sigma$) of a discrete random variable X is:

$\sigma = \sqrt{\sum_{i=1}^{n} (x_i - \mu)^2 P(x_i)}$

Where:

• $x_i$ are the possible values of the random variable.
• $\mu$ is the mean of the random variable.
• $P(x_i)$ is the probability of the random variable taking the value $x_i$.
• $n$ is the number of possible values of the random variable.

Relationship to the Variance:

The standard deviation is the square root of the variance. In other words, the variance is the square of the standard deviation.

If $\sigma^2$ represents the variance, then:

$\sigma = \sqrt{\sigma^2}$

And conversely:

$\sigma^2 = \sigma^2$

The variance is calculated as:

$\sigma^2 = \sum_{i=1}^{n} (x_i - \mu)^2 P(x_i)$

So, the standard deviation is simply the square root of this value. The standard deviation is often preferred over the variance because it is in the same units as the random variable, making it easier to interpret.

Since a ticket is drawn at random from tickets numbered 1 to 10, each ticket has an equal probability of being drawn.

The possible values for the random variable X are the numbers 1 to 10. Since there are 10 tickets in total, and each ticket has an equal chance of being drawn, the probability of drawing any particular ticket is $\frac{1}{10}$.

Therefore, the probability distribution of X is:

$P(X = x) = \frac{1}{10}$ for $x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$

We can represent the probability distribution in a table:

The expected value (or mean) of a discrete random variable X is given by:

$E(X) = \sum_{i=1}^{n} x_i P(x_i)$

Where:

• $x_i$ are the possible values of the random variable.
• $P(x_i)$ is the probability of the random variable taking the value $x_i$.
• $n$ is the number of possible values of the random variable.

From Question 36, we have the following probability distribution:

$P(X = x) = \frac{1}{10}$ for $x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$

Therefore, we can calculate E(X) as follows:

$E(X) = (1 \times \frac{1}{10}) + (2 \times \frac{1}{10}) + (3 \times \frac{1}{10}) + (4 \times \frac{1}{10}) + (5 \times \frac{1}{10}) + (6 \times \frac{1}{10}) + (7 \times \frac{1}{10}) + (8 \times \frac{1}{10}) + (9 \times \frac{1}{10}) + (10 \times \frac{1}{10})$

We can factor out $\frac{1}{10}$:

$E(X) = \frac{1}{10} (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)$

The sum of the integers from 1 to 10 is $\frac{10 \times (10 + 1)}{2} = \frac{10 \times 11}{2} = 55$

So, $E(X) = \frac{1}{10} \times 55 = 5.5$

Therefore, the expected value E(X) is 5.5.

If X follows a Binomial distribution, denoted as X ~ B(n, p), then by definition:

$n$ represents the number of trials.

$p$ represents the probability of success in a single trial.

Therefore, the probability of success in a single trial is simply $p$.

For a Poisson distribution, the probability mass function is given by:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the mean and $k$ is the number of events.

In this case, $\lambda = 1$ and we want to find the probability of no arrivals, which means $P(X = 0)$.

So, $P(X = 0) = \frac{e^{-1} (1)^0}{0!} = \frac{e^{-1} \cdot 1}{1} = e^{-1}$

Given that $e^{-1} \approx 0.3679$, therefore,

$P(X = 0) \approx 0.3679$

Thus, the probability of no arrivals in a minute is approximately 0.3679.

If the height of students in a class is normally distributed with a mean of 160 cm and a standard deviation of 5 cm, it means the following:

1. Central Tendency: The average height of students in the class is 160 cm. This is the center of the distribution.

2. Symmetry: The heights are symmetrically distributed around the mean of 160 cm. This means that there are approximately as many students taller than 160 cm as there are students shorter than 160 cm.

3. Spread: The standard deviation of 5 cm indicates the spread or variability of the heights. A smaller standard deviation means that the heights are more clustered around the mean, while a larger standard deviation would indicate a wider spread.

4. Empirical Rule:

   • Approximately 68% of the students have heights between 155 cm (160 - 5) and 165 cm (160 + 5).
   • Approximately 95% of the students have heights between 150 cm (160 - 2*5) and 170 cm (160 + 2*5).
   • Approximately 99.7% of the students have heights between 145 cm (160 - 3*5) and 175 cm (160 + 3*5).

5. Shape: The distribution of heights follows a bell-shaped curve (Gaussian curve), with the highest point at the mean (160 cm) and tapering off symmetrically on either side.

In essence, this description provides a good understanding of how the heights are distributed within the class, with the majority of students having heights close to the average, and fewer students having heights that are significantly taller or shorter than the average.

For a probability distribution to be valid, the sum of the probabilities for all possible values of the random variable must equal 1.

In this case, we have P(X=x) = $cx$ for $x = 1, 2, 3$. So we need to find $c$ such that:

$P(X=1) + P(X=2) + P(X=3) = 1$

Substituting the given formula, we get:

$c(1) + c(2) + c(3) = 1$

$c + 2c + 3c = 1$

$6c = 1$

Solving for $c$:

$c = \frac{1}{6}$

Therefore, the value of the constant $c$ is $\frac{1}{6}$.

The expected value (or mean) of a discrete random variable X is given by:

$E(X) = \sum_{i=1}^{n} x_i P(x_i)$

Where:

• $x_i$ are the possible values of the random variable.
• $P(x_i)$ is the probability of the random variable taking the value $x_i$.
• $n$ is the number of possible values of the random variable.

From Question 41, we have P(X=x) = $cx$ for $x = 1, 2, 3$, and we found that $c = \frac{1}{6}$. So, $P(X=x) = \frac{x}{6}$.

Therefore, we can calculate E(X) as follows:

$E(X) = (1 \times P(X=1)) + (2 \times P(X=2)) + (3 \times P(X=3))$

$E(X) = (1 \times \frac{1}{6}) + (2 \times \frac{2}{6}) + (3 \times \frac{3}{6})$

$E(X) = \frac{1}{6} + \frac{4}{6} + \frac{9}{6}$

$E(X) = \frac{1 + 4 + 9}{6} = \frac{14}{6} = \frac{7}{3}$

Therefore, the expected value E(X) is $\frac{7}{3}$.

For a Binomial distribution with parameters $n$ (number of trials) and $p$ (probability of success), the mean and variance are given by:

• Mean: $\mu = np$
• Variance: $\sigma^2 = np(1-p)$

Given that the mean is 6 and the variance is 4, we have:

Now we can substitute (i) into (ii):

$6(1-p) = 4$

Divide both sides by 6:

$1-p = \frac{4}{6} = \frac{2}{3}$

Solve for $p$:

$p = 1 - \frac{2}{3} = \frac{1}{3}$

Now substitute the value of $p$ back into equation (i):

$n(\frac{1}{3}) = 6$

Solve for $n$:

$n = 6 \times 3 = 18$

So we have $n = 18$ and $p = \frac{1}{3}$.

We want to find $P(X = 0)$. The formula for the probability mass function of a Binomial distribution is:

$P(X = k) = {n \choose k} p^k (1-p)^{n-k}$

In this case, $k = 0$, so:

$P(X = 0) = {18 \choose 0} (\frac{1}{3})^0 (1 - \frac{1}{3})^{18-0}$

$P(X = 0) = {18 \choose 0} (\frac{1}{3})^0 (\frac{2}{3})^{18}$

Since ${18 \choose 0} = 1$ and $(\frac{1}{3})^0 = 1$, we have:

$P(X = 0) = (\frac{2}{3})^{18}$

Therefore, the probability $P(X=0)$ is $(\frac{2}{3})^{18}$.

The Poisson distribution is useful for modeling the number of events that occur in a fixed interval of time or space, given that these events occur with a known average rate and independently of the time since the last event.

Here are two real-life examples where the Poisson distribution can be applied:

1. Number of customer arrivals at a store in an hour: The Poisson distribution can be used to model the number of customers who arrive at a store's checkout counter during a specific hour. If the average arrival rate is known (e.g., 20 customers per hour), the Poisson distribution can help determine the probability of observing a certain number of arrivals in that hour (e.g., the probability of exactly 25 customers arriving).

2. Number of emails received per day: The number of emails an individual receives in a day can often be modeled using a Poisson distribution. If the average number of emails received per day is known, the Poisson distribution can be used to calculate the probability of receiving a particular number of emails on any given day (e.g., the probability of receiving exactly 10 emails).

The empirical rule, also known as the 68-95-99.7 rule, is a guideline for understanding the spread of data in a Normal distribution. It states the percentage of values that lie within a certain number of standard deviations from the mean.

1. Within $\pm 1\sigma$ (one standard deviation) of the mean: Approximately 68% of the data values fall within one standard deviation of the mean. This means that about 68% of the data lies between $\mu - \sigma$ and $\mu + \sigma$, where $\mu$ is the mean and $\sigma$ is the standard deviation.

2. Within $\pm 2\sigma$ (two standard deviations) of the mean: Approximately 95% of the data values fall within two standard deviations of the mean. This means that about 95% of the data lies between $\mu - 2\sigma$ and $\mu + 2\sigma$.

3. Within $\pm 3\sigma$ (three standard deviations) of the mean: Approximately 99.7% of the data values fall within three standard deviations of the mean. This means that about 99.7% of the data lies between $\mu - 3\sigma$ and $\mu + 3\sigma$.

In summary, the empirical rule provides a quick way to estimate the proportion of data that falls within certain ranges in a Normal distribution. It highlights that most of the data is clustered around the mean, and very few values are far from the mean.

In this scenario, the distribution of X, the number of correct answers, follows a Binomial distribution.

Here's why:

• Fixed number of trials: There are a fixed number of questions, which is 4. Each question represents a trial.

• Independent trials: The outcome of guessing on one question does not affect the outcome of guessing on any other question. Therefore, the trials are independent.

• Two possible outcomes: For each question, there are two possible outcomes: either the student guesses the correct answer (success) or the student guesses an incorrect answer (failure).

• Constant probability of success: The probability of guessing the correct answer is the same for each question. Since there are 3 options and only one is correct, the probability of guessing correctly is $\frac{1}{3}$.

Therefore, X follows a Binomial distribution, denoted as X ~ B(n, p), where:

• $n = 4$ (number of trials, i.e., number of questions)
• $p = \frac{1}{3}$ (probability of success in a single trial, i.e., probability of guessing correctly on one question)

Thus, the distribution of X is Binomial with parameters n = 4 and p = 1/3, which can be written as X ~ B(4, $\frac{1}{3}$).

For a Poisson distribution, the probability mass function is given by:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the mean and $k$ is the number of events.

In this case, $\lambda = 3$ and we want to find $P(X = 2)$.

So, $P(X = 2) = \frac{e^{-3} (3)^2}{2!} = \frac{e^{-3} \cdot 9}{2}$

Given that $e^{-3} \approx 0.0498$, therefore,

$P(X = 2) \approx \frac{0.0498 \times 9}{2} = \frac{0.4482}{2} = 0.2241$

Thus, the probability of exactly 2 accidents on a particular day is approximately 0.2241.

The standard deviation is the square root of the variance.

Given that the variance is 100, we have:

$\sigma^2 = 100$

Therefore, the standard deviation is:

$\sigma = \sqrt{100} = 10$

Thus, the standard deviation of X is 10.

We can use the linearity property of expectation, which states that for any random variable X and constant b:

$E(X + b) = E(X) + b$

In this case, we want to find $E(X + 5)$.

So, $b = 5$.

Using the linearity property, we have:

$E(X + 5) = E(X) + 5$

Given that $E(X) = 6$, we can substitute this value:

$E(X + 5) = 6 + 5 = 11$

Therefore, $E(X + 5) = 11$.

X can take values 0, 1, or 2, representing the number of faulty pens drawn.

Total number of ways to draw 2 pens from 10: ${10 \choose 2} = \frac{10!}{2!8!} = \frac{10 \times 9}{2} = 45$

1. P(X = 0): This means both pens drawn are good.

Number of ways to draw 2 good pens from 8: ${8 \choose 2} = \frac{8!}{2!6!} = \frac{8 \times 7}{2} = 28$

$P(X = 0) = \frac{28}{45}$

2. P(X = 1): This means one faulty pen and one good pen are drawn.

Number of ways to draw 1 faulty pen from 2: ${2 \choose 1} = 2$

Number of ways to draw 1 good pen from 8: ${8 \choose 1} = 8$

Number of ways to draw 1 faulty and 1 good pen: $2 \times 8 = 16$

$P(X = 1) = \frac{16}{45}$

3. P(X = 2): This means both pens drawn are faulty.

Number of ways to draw 2 faulty pens from 2: ${2 \choose 2} = \frac{2!}{2!0!} = 1$

$P(X = 2) = \frac{1}{45}$

Probability Distribution of X:

We can use the binomial probability formula:

$P(X = k) = {n \choose k} p^k (1-p)^{n-k}$

Where:

• $n$ is the number of trials.
• $k$ is the number of successful trials we want.
• $p$ is the probability of success on a single trial.
• ${n \choose k}$ is the binomial coefficient.

In this case:

• $n = 10$
• $k = 1$
• $p = 0.1$

So, we have:

$P(X = 1) = {10 \choose 1} (0.1)^1 (1-0.1)^{10-1}$

$P(X = 1) = {10 \choose 1} (0.1)^1 (0.9)^9$

Calculate the binomial coefficient:

${10 \choose 1} = \frac{10!}{1!(10-1)!} = \frac{10!}{1!9!} = \frac{10}{1} = 10$

Now, plug the values back into the formula:

$P(X = 1) = 10 \times (0.1) \times (0.9)^9$

$P(X = 1) = 10 \times 0.1 \times 0.387420489$

$P(X = 1) = 0.387420489$

Therefore, the probability $P(X=1)$ is approximately 0.3874.

For a Poisson distribution, the probability mass function is given by:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the mean and $k$ is the number of events.

Given that $2P(X=2) = P(X=1) + 2P(X=0)$, we can substitute the Poisson probabilities:

$2 \cdot \frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-\lambda} \lambda^1}{1!} + 2 \cdot \frac{e^{-\lambda} \lambda^0}{0!}$

We can cancel out $e^{-\lambda}$ from all terms:

$2 \cdot \frac{\lambda^2}{2} = \lambda + 2 \cdot 1$

Simplifying, we get:

$\lambda^2 = \lambda + 2$

Rearrange to form a quadratic equation:

$\lambda^2 - \lambda - 2 = 0$

Factor the quadratic equation:

$(\lambda - 2)(\lambda + 1) = 0$

This gives us two possible values for $\lambda$: $\lambda = 2$ or $\lambda = -1$.

Since $\lambda$ represents the mean of the Poisson distribution, it must be a positive value. Therefore, $\lambda = -1$ is not a valid solution.

Thus, the mean of X is $\lambda = 2$.

Given that the mean weight of apples is 200 grams and the standard deviation is 20 grams, one standard deviation above the mean is calculated as:

$\mu + \sigma = 200 + 20 = 220$

Therefore, a weight of 220 grams corresponds to one standard deviation above the mean.

The expected value of $X^2$, denoted as $E(X^2)$, is calculated as follows:

$E(X^2) = \sum_{i=1}^{n} x_i^2 P(x_i)$

Where:

• $x_i$ are the possible values of the random variable.
• $P(x_i)$ is the probability of the random variable taking the value $x_i$.
• $n$ is the number of possible values of the random variable.

Using the given probability distribution, we have:

$E(X^2) = (1^2 \times 0.1) + (2^2 \times 0.2) + (3^2 \times 0.7)$

$E(X^2) = (1 \times 0.1) + (4 \times 0.2) + (9 \times 0.7)$

$E(X^2) = 0.1 + 0.8 + 6.3$

$E(X^2) = 7.2$

Therefore, $E(X^2) = 7.2$.

For a Binomial distribution with parameters $n$ (number of trials) and $p$ (probability of success), the mean and variance are given by:

• Mean: $\mu = np$
• Variance: $\sigma^2 = np(1-p)$

Given that the mean is 5 and the variance is 3, we have:

Now we can substitute (i) into (ii):

$5(1-p) = 3$

Divide both sides by 5:

$1-p = \frac{3}{5} = 0.6$

Solve for $p$:

$p = 1 - 0.6 = 0.4$

Now substitute the value of $p$ back into equation (i):

$n(0.4) = 5$

Solve for $n$:

$n = \frac{5}{0.4} = 12.5$

However, $n$ must be an integer since it represents the number of trials. There might be an error in the problem statement because it's unusual to have a non-integer value for $n$. Assuming the mean and variance are correct, there's no Binomial distribution that perfectly fits those parameters. However, if we are forced to round, we would round to the nearest integer. Therefore, we would say that $n \approx 13$.

If it is necessary to provide a possible $n$ assuming possible errors on the mean and variance values, one would use algebra to determine possible integer values of $n$

1. Find the value of k:

For a probability distribution to be valid, the sum of the probabilities for all possible values of the random variable must equal 1. Therefore:

$P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1$

$k + 2k + 3k + 2k + k = 1$

$9k = 1$

$k = \frac{1}{9}$

2. Calculate E(X):

The expected value (or mean) of a discrete random variable X is given by:

$E(X) = \sum_{i=1}^{n} x_i P(x_i)$

$E(X) = (0 \times k) + (1 \times 2k) + (2 \times 3k) + (3 \times 2k) + (4 \times k)$

$E(X) = 0 + 2k + 6k + 6k + 4k$

$E(X) = 18k$

Since $k = \frac{1}{9}$, we have:

$E(X) = 18 \times \frac{1}{9} = 2$

3. Calculate Var(X):

The variance of a discrete random variable X is given by:

$Var(X) = E(X^2) - [E(X)]^2$

First, we need to calculate $E(X^2)$:

$E(X^2) = \sum_{i=1}^{n} x_i^2 P(x_i)$

$E(X^2) = (0^2 \times k) + (1^2 \times 2k) + (2^2 \times 3k) + (3^2 \times 2k) + (4^2 \times k)$

$E(X^2) = 0 + 2k + 12k + 18k + 16k$

$E(X^2) = 48k$

Since $k = \frac{1}{9}$, we have:

$E(X^2) = 48 \times \frac{1}{9} = \frac{16}{3}$

Now, we can calculate Var(X):

$Var(X) = E(X^2) - [E(X)]^2 = \frac{16}{3} - (2)^2 = \frac{16}{3} - 4 = \frac{16 - 12}{3} = \frac{4}{3}$

4. Calculate P(X $\geq$ 2):

$P(X \geq 2) = P(X=2) + P(X=3) + P(X=4)$

$P(X \geq 2) = 3k + 2k + k = 6k$

Since $k = \frac{1}{9}$, we have:

$P(X \geq 2) = 6 \times \frac{1}{9} = \frac{2}{3}$

Summary:

• $k = \frac{1}{9}$
• $E(X) = 2$
• $Var(X) = \frac{4}{3}$
• $P(X \geq 2) = \frac{2}{3}$

Given a Binomial distribution with $n=10$ and $p=0.1$, we have the probability mass function:

$P(X = k) = {n \choose k} p^k (1-p)^{n-k}$

$P(X = k) = {10 \choose k} (0.1)^k (0.9)^{10-k}$

i) Exactly 2 faulty items:

We want to find $P(X = 2)$:

$P(X = 2) = {10 \choose 2} (0.1)^2 (0.9)^{10-2}$

$P(X = 2) = {10 \choose 2} (0.1)^2 (0.9)^8$

${10 \choose 2} = \frac{10!}{2!8!} = \frac{10 \times 9}{2} = 45$

$P(X = 2) = 45 \times (0.01) \times (0.43046721)$

$P(X = 2) = 45 \times 0.0043046721 = 0.1937102445 \approx 0.1937$

ii) At most 1 faulty item:

We want to find $P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = 0) = {10 \choose 0} (0.1)^0 (0.9)^{10-0} = 1 \times 1 \times (0.9)^{10} = 0.3486784401 \approx 0.3487$

$P(X = 1) = {10 \choose 1} (0.1)^1 (0.9)^{10-1} = 10 \times (0.1) \times (0.9)^9 = 10 \times 0.1 \times 0.387420489 \approx 0.3874$

$P(X \leq 1) = 0.3487 + 0.3874 = 0.7361$

iii) At least 3 faulty items:

We want to find $P(X \geq 3) = 1 - P(X < 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]$

$P(X \geq 3) = 1 - [0.3487 + 0.3874 + 0.1937]$

$P(X \geq 3) = 1 - 0.9298 = 0.0702$

Summary:

• $P(X = 2) \approx 0.1937$
• $P(X \leq 1) \approx 0.7361$
• $P(X \geq 3) \approx 0.0702$

Given a Poisson distribution with $\lambda = 4$, the probability mass function is:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$

$P(X = k) = \frac{e^{-4} (4)^k}{k!}$

i) Exactly 5 calls are received in a minute:

We want to find $P(X = 5)$:

$P(X = 5) = \frac{e^{-4} (4)^5}{5!} = \frac{0.0183 \times 1024}{120} = \frac{18.7392}{120} = 0.15616 \approx 0.1562$

ii) At most 2 calls are received in a minute:

We want to find $P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 0) = \frac{e^{-4} (4)^0}{0!} = e^{-4} = 0.0183$

$P(X = 1) = \frac{e^{-4} (4)^1}{1!} = 4e^{-4} = 4 \times 0.0183 = 0.0732$

$P(X = 2) = \frac{e^{-4} (4)^2}{2!} = \frac{16e^{-4}}{2} = 8e^{-4} = 8 \times 0.0183 = 0.1464$

$P(X \leq 2) = 0.0183 + 0.0732 + 0.1464 = 0.2379$

Summary:

• $P(X = 5) \approx 0.1562$
• $P(X \leq 2) \approx 0.2379$

Given a Normal distribution with $\mu = 60$ and $\sigma = 10$, we first convert the marks to z-scores using the formula:

$z = \frac{x - \mu}{\sigma}$

i) More than 75 marks:

$x = 75$, so $z = \frac{75 - 60}{10} = \frac{15}{10} = 1.5$

We want to find $P(X > 75) = P(Z > 1.5)$. From the standard normal table, the area between $Z=0$ and $Z=1.5$ is 0.4332. Since the total area to the right of $Z=0$ is 0.5, we have:

$P(Z > 1.5) = 0.5 - 0.4332 = 0.0668$

ii) Less than 50 marks:

$x = 50$, so $z = \frac{50 - 60}{10} = \frac{-10}{10} = -1$

We want to find $P(X < 50) = P(Z < -1)$. From the standard normal table, the area between $Z=0$ and $Z=-1$ is 0.3413. Since the total area to the left of $Z=0$ is 0.5, we have:

$P(Z < -1) = 0.5 - 0.3413 = 0.1587$

iii) Between 55 and 70 marks:

$x_1 = 55$, so $z_1 = \frac{55 - 60}{10} = \frac{-5}{10} = -0.5$

$x_2 = 70$, so $z_2 = \frac{70 - 60}{10} = \frac{10}{10} = 1$

We want to find $P(55 < X < 70) = P(-0.5 < Z < 1)$. From the standard normal table:

• The area between $Z=0$ and $Z=-0.5$ is 0.1915.
• The area between $Z=0$ and $Z=1$ is 0.3413.

$P(-0.5 < Z < 1) = 0.1915 + 0.3413 = 0.5328$

Summary:

• $P(X > 75) = 0.0668$
• $P(X < 50) = 0.1587$
• $P(55 < X < 70) = 0.5328$

1. Find the value of k:

The sum of the probabilities must equal 1:

$0.1 + 0.2 + k + 0.3 + 0.2 = 1$

$0.8 + k = 1$

$k = 1 - 0.8 = 0.2$

2. Calculate E(X):

$E(X) = \sum x P(X=x)$

$E(X) = (-2 \times 0.1) + (-1 \times 0.2) + (0 \times 0.2) + (1 \times 0.3) + (2 \times 0.2)$

$E(X) = -0.2 - 0.2 + 0 + 0.3 + 0.4 = 0.3$

3. Calculate E($X^2$):

$E(X^2) = \sum x^2 P(X=x)$

$E(X^2) = ((-2)^2 \times 0.1) + ((-1)^2 \times 0.2) + (0^2 \times 0.2) + (1^2 \times 0.3) + (2^2 \times 0.2)$

$E(X^2) = (4 \times 0.1) + (1 \times 0.2) + (0 \times 0.2) + (1 \times 0.3) + (4 \times 0.2)$

$E(X^2) = 0.4 + 0.2 + 0 + 0.3 + 0.8 = 1.7$

4. Calculate Var(X):

$Var(X) = E(X^2) - [E(X)]^2$

$Var(X) = 1.7 - (0.3)^2 = 1.7 - 0.09 = 1.61$

5. Calculate Standard Deviation of X:

Standard Deviation = $\sqrt{Var(X)}$

Standard Deviation = $\sqrt{1.61} \approx 1.2689$

Summary:

• $k = 0.2$
• $E(X) = 0.3$
• $E(X^2) = 1.7$
• $Var(X) = 1.61$
• Standard Deviation of X $\approx 1.2689$

We have a binomial distribution with $n = 6$ and $p = \frac{1}{3}$. The probability mass function is given by:

$P(X = k) = {n \choose k} p^k (1-p)^{n-k}$

$P(X = k) = {6 \choose k} (\frac{1}{3})^k (\frac{2}{3})^{6-k}$

i) Exactly 4 heads:

$P(X = 4) = {6 \choose 4} (\frac{1}{3})^4 (\frac{2}{3})^{6-4}$

${6 \choose 4} = \frac{6!}{4!2!} = \frac{6 \times 5}{2} = 15$

$P(X = 4) = 15 \times (\frac{1}{81}) \times (\frac{4}{9}) = 15 \times \frac{4}{729} = \frac{60}{729} = \frac{20}{243} \approx 0.0823$

ii) At least 3 heads:

$P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

$P(X = 3) = {6 \choose 3} (\frac{1}{3})^3 (\frac{2}{3})^3 = 20 \times \frac{1}{27} \times \frac{8}{27} = \frac{160}{729} \approx 0.2195$

$P(X = 5) = {6 \choose 5} (\frac{1}{3})^5 (\frac{2}{3})^1 = 6 \times \frac{1}{243} \times \frac{2}{3} = \frac{12}{729} \approx 0.0165$

$P(X = 6) = {6 \choose 6} (\frac{1}{3})^6 (\frac{2}{3})^0 = 1 \times \frac{1}{729} \times 1 = \frac{1}{729} \approx 0.0014$

$P(X \geq 3) = 0.2195 + 0.0823 + 0.0165 + 0.0014 = 0.3197$

iii) At most 2 heads:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 0) = {6 \choose 0} (\frac{1}{3})^0 (\frac{2}{3})^6 = 1 \times 1 \times \frac{64}{729} = \frac{64}{729} \approx 0.0878$

$P(X = 1) = {6 \choose 1} (\frac{1}{3})^1 (\frac{2}{3})^5 = 6 \times \frac{1}{3} \times \frac{32}{243} = \frac{192}{729} \approx 0.2634$

$P(X = 2) = {6 \choose 2} (\frac{1}{3})^2 (\frac{2}{3})^4 = 15 \times \frac{1}{9} \times \frac{16}{81} = \frac{240}{729} \approx 0.3292$

$P(X \leq 2) = 0.0878 + 0.2634 + 0.3292 = 0.6804$

Summary:

• $P(X = 4) \approx 0.0823$
• $P(X \geq 3) \approx 0.3197$
• $P(X \leq 2) \approx 0.6804$

Binomial Distribution:

• Parameters: A Binomial distribution is characterized by two parameters:

  • $n$: The number of trials.
  • $p$: The probability of success in a single trial.

• Mean: The mean ($\mu$) of a Binomial distribution is given by:

  $\mu = np$

  The mean represents the average number of successes expected in $n$ trials.

• Variance: The variance ($\sigma^2$) of a Binomial distribution is given by:

  $\sigma^2 = np(1-p)$

  The variance measures the spread or dispersion of the distribution. A larger variance indicates a greater variability in the number of successes.

Poisson Distribution:

• Parameters: A Poisson distribution is characterized by one parameter:

  • $\lambda$: The average rate of events occurring in a fixed interval of time or space.

• Mean: The mean ($\mu$) of a Poisson distribution is given by:

  $\mu = \lambda$

  The mean represents the average number of events expected in the given interval.

• Variance: The variance ($\sigma^2$) of a Poisson distribution is given by:

  $\sigma^2 = \lambda$

  In a Poisson distribution, the mean and variance are equal. This is a key characteristic of the distribution.

Relationship to Parameters:

• For the Binomial distribution, both the mean and variance are directly related to the number of trials ($n$) and the probability of success ($p$). As $n$ increases, the mean and variance tend to increase. Similarly, the value of $p$ influences both the mean and variance.
• For the Poisson distribution, both the mean and variance are solely determined by the average rate ($\lambda$). A larger average rate results in a higher mean and variance, indicating a greater frequency of events.

Given a Normal distribution with $\mu = 800$ and $\sigma = 50$, we first convert the lifetimes to z-scores using the formula:

$z = \frac{x - \mu}{\sigma}$

i) Between 700 and 900 hours:

$x_1 = 700$, so $z_1 = \frac{700 - 800}{50} = \frac{-100}{50} = -2$

$x_2 = 900$, so $z_2 = \frac{900 - 800}{50} = \frac{100}{50} = 2$

We want to find $P(700 < X < 900) = P(-2 < Z < 2)$. From the standard normal table, the area between $Z=0$ and $Z=2$ is 0.4772. Due to symmetry, the area between $Z=0$ and $Z=-2$ is also 0.4772. Therefore:

$P(-2 < Z < 2) = 0.4772 + 0.4772 = 0.9544$

ii) Less than 750 hours:

$x = 750$, so $z = \frac{750 - 800}{50} = \frac{-50}{50} = -1$

We want to find $P(X < 750) = P(Z < -1)$. From the standard normal table, the area between $Z=0$ and $Z=-1$ is 0.3413. Since the total area to the left of $Z=0$ is 0.5, we have:

$P(Z < -1) = 0.5 - 0.3413 = 0.1587$

iii) More than 850 hours:

$x = 850$, so $z = \frac{850 - 800}{50} = \frac{50}{50} = 1$

We want to find $P(X > 850) = P(Z > 1)$. From the standard normal table, the area between $Z=0$ and $Z=1$ is 0.3413. Since the total area to the right of $Z=0$ is 0.5, we have:

$P(Z > 1) = 0.5 - 0.3413 = 0.1587$

Summary:

• $P(700 < X < 900) = 0.9544$
• $P(X < 750) = 0.1587$
• $P(X > 850) = 0.1587$

Mathematical Expectation (Mean):

The mathematical expectation, also known as the expected value or mean, of a discrete random variable X is a measure of the central tendency of the variable's distribution. It is a weighted average of the possible values of X, where the weights are the probabilities of each value.

If X is a discrete random variable with possible values $x_1, x_2, ..., x_n$ and corresponding probabilities $P(X = x_1), P(X = x_2), ..., P(X = x_n)$, then the expected value of X, denoted as E(X), is given by:

$E(X) = \sum_{i=1}^{n} x_i P(X = x_i)$

Variance:

The variance of a discrete random variable X is a measure of the spread or dispersion of the variable's values around its mean. It quantifies the average squared distance of the values from the mean.

The variance of X, denoted as Var(X), is given by:

$Var(X) = E[(X - E(X))^2] = \sum_{i=1}^{n} (x_i - E(X))^2 P(X = x_i)$

Proof that Var(X) = E($X^2$) - [E(X)]$^2$:

We start with the definition of variance:

$Var(X) = E[(X - E(X))^2]$

Expanding the square:

$Var(X) = E[X^2 - 2XE(X) + (E(X))^2]$

Using the linearity property of expectation, we can separate the terms:

$Var(X) = E(X^2) - 2E(XE(X)) + E[(E(X))^2]$

Since E(X) is a constant, we can take it out of the expectation:

$Var(X) = E(X^2) - 2E(X)E(X) + (E(X))^2$

$Var(X) = E(X^2) - 2[E(X)]^2 + [E(X)]^2$

$Var(X) = E(X^2) - [E(X)]^2$

Therefore, we have proven that:

$Var(X) = E(X^2) - [E(X)]^2$

Given a Poisson distribution with $\lambda = 2$, the probability mass function is:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$

$P(X = k) = \frac{e^{-2} (2)^k}{k!}$

i) There are exactly 3 accidents:

$P(X = 3) = \frac{e^{-2} (2)^3}{3!} = \frac{0.1353 \times 8}{6} = \frac{1.0824}{6} \approx 0.1804$

ii) There are more than 2 accidents:

$P(X > 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]$

$P(X = 0) = \frac{e^{-2} (2)^0}{0!} = e^{-2} = 0.1353$

$P(X = 1) = \frac{e^{-2} (2)^1}{1!} = 2e^{-2} = 2 \times 0.1353 = 0.2706$

$P(X = 2) = \frac{e^{-2} (2)^2}{2!} = \frac{4e^{-2}}{2} = 2e^{-2} = 2 \times 0.1353 = 0.2706$

$P(X > 2) = 1 - [0.1353 + 0.2706 + 0.2706] = 1 - 0.6765 = 0.3235$

iii) There is at least one accident:

$P(X \geq 1) = 1 - P(X = 0)$

$P(X = 0) = e^{-2} = 0.1353$

$P(X \geq 1) = 1 - 0.1353 = 0.8647$

Summary:

• $P(X = 3) \approx 0.1804$
• $P(X > 2) = 0.3235$
• $P(X \geq 1) = 0.8647$

We have a binomial distribution with $n = 15$ and $p = 0.10$. The probability mass function is given by:

$P(X = k) = {n \choose k} p^k (1-p)^{n-k}$

$P(X = k) = {15 \choose k} (0.1)^k (0.9)^{15-k}$

i) Exactly 2 are left-handed:

$P(X = 2) = {15 \choose 2} (0.1)^2 (0.9)^{15-2}$

${15 \choose 2} = \frac{15!}{2!13!} = \frac{15 \times 14}{2} = 105$

$P(X = 2) = 105 \times (0.01) \times (0.9)^{13} = 105 \times 0.01 \times 0.25418658258 = 0.2668959117$

$P(X=2) \approx 0.2669$

ii) At most 1 is left-handed:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = 0) = {15 \choose 0} (0.1)^0 (0.9)^{15-0} = 1 \times 1 \times (0.9)^{15} = 0.20589113209$

$P(X = 1) = {15 \choose 1} (0.1)^1 (0.9)^{15-1} = 15 \times 0.1 \times (0.9)^{14} = 1.5 \times 0.22876792455 = 0.34315188683$

$P(X \leq 1) = 0.20589113209 + 0.34315188683 = 0.54904301892$

$P(X \leq 1) \approx 0.5490$

Summary:

• $P(X = 2) \approx 0.2669$
• $P(X \leq 1) \approx 0.5490$

Given a Normal distribution with $\mu = 55$ and $\sigma = 5$, we first convert the weights to z-scores using the formula:

$z = \frac{x - \mu}{\sigma}$

i) Between 50 kg and 60 kg:

$x_1 = 50$, so $z_1 = \frac{50 - 55}{5} = \frac{-5}{5} = -1$

$x_2 = 60$, so $z_2 = \frac{60 - 55}{5} = \frac{5}{5} = 1$

We want to find $P(50 < X < 60) = P(-1 < Z < 1)$. From the standard normal table, the area between $Z=0$ and $Z=1$ is 0.3413. Due to symmetry, the area between $Z=0$ and $Z=-1$ is also 0.3413. Therefore:

$P(-1 < Z < 1) = 0.3413 + 0.3413 = 0.6826$

Since there are 1000 students, the estimated number of students weighing between 50 kg and 60 kg is:

$1000 \times 0.6826 = 682.6 \approx 683$ students.

ii) More than 65 kg:

$x = 65$, so $z = \frac{65 - 55}{5} = \frac{10}{5} = 2$

We want to find $P(X > 65) = P(Z > 2)$. From the standard normal table, the area between $Z=0$ and $Z=2$ is 0.4772. Since the total area to the right of $Z=0$ is 0.5, we have:

$P(Z > 2) = 0.5 - 0.4772 = 0.0228$

Since there are 1000 students, the estimated number of students weighing more than 65 kg is:

$1000 \times 0.0228 = 22.8 \approx 23$ students.

Summary:

• Number of students between 50 kg and 60 kg $\approx 683$
• Number of students more than 65 kg $\approx 23$

Standardizing a Normal Variable:

Standardizing a normal variable involves transforming a normal random variable $X$ with mean $\mu$ and standard deviation $\sigma$ into a standard normal random variable $Z$ with a mean of 0 and a standard deviation of 1. This transformation is done using the following formula:

$Z = \frac{X - \mu}{\sigma}$

Where:

• $X$ is the original normal random variable.
• $\mu$ is the mean of the original normal distribution.
• $\sigma$ is the standard deviation of the original normal distribution.
• $Z$ is the standardized normal random variable (also called the z-score).

Using the Standard Normal Table to Find Probabilities:

The standard normal table (also known as the Z-table) provides the cumulative probability (area under the standard normal curve) to the left of a given z-score. In other words, it gives the probability that a standard normal random variable Z is less than or equal to a specific value z, i.e., $P(Z \leq z)$.

Here's how the Standard Normal Table is used to find probabilities for any Normal distribution:

1. Standardize the variable: Convert the original normal random variable X to a standard normal random variable Z using the formula $Z = \frac{X - \mu}{\sigma}$.

2. Find the corresponding z-score(s): Identify the z-score(s) that correspond to the value(s) of interest in the original normal distribution.

3. Look up the probability in the Standard Normal Table: Use the Z-table to find the cumulative probability associated with the calculated z-score(s). The table typically provides the area under the standard normal curve to the left of the z-score.

4. Calculate the desired probability: Depending on the question, you may need to perform additional calculations using the table values. For example:

   • To find $P(X < x)$, look up the z-score corresponding to x in the Z-table. The table value directly gives you the probability.

   • To find $P(X > x)$, look up the z-score corresponding to x in the Z-table, and subtract the table value from 1 (since the total area under the curve is 1).

   • To find $P(a < X < b)$, look up the z-scores corresponding to a and b in the Z-table. Subtract the table value for a from the table value for b.

In summary, standardizing a normal variable allows us to use the Standard Normal Table to find probabilities for any normal distribution, regardless of its mean and standard deviation. This simplifies the calculation of probabilities and makes it easier to compare different normal distributions.

1. Find the value of c:

The sum of the probabilities must equal 1:

$c + 2c + 3c + 4c = 1$

$10c = 1$

$c = \frac{1}{10}$

2. Construct the cumulative distribution function (CDF) table:

The cumulative distribution function (CDF) is defined as $F(x) = P(X \leq x)$.

• $F(1) = P(X \leq 1) = P(X=1) = c = \frac{1}{10} = 0.1$
• $F(2) = P(X \leq 2) = P(X=1) + P(X=2) = c + 2c = 3c = \frac{3}{10} = 0.3$
• $F(3) = P(X \leq 3) = P(X=1) + P(X=2) + P(X=3) = c + 2c + 3c = 6c = \frac{6}{10} = 0.6$
• $F(4) = P(X \leq 4) = P(X=1) + P(X=2) + P(X=3) + P(X=4) = c + 2c + 3c + 4c = 10c = \frac{10}{10} = 1$

The CDF table is:

3. Find E(X):

$E(X) = \sum x P(X=x)$

$E(X) = (1 \times c) + (2 \times 2c) + (3 \times 3c) + (4 \times 4c)$

$E(X) = c + 4c + 9c + 16c = 30c$

$E(X) = 30 \times \frac{1}{10} = 3$

4. Find Var(X):

$Var(X) = E(X^2) - [E(X)]^2$

First, we need to calculate $E(X^2)$:

$E(X^2) = \sum x^2 P(X=x)$

$E(X^2) = (1^2 \times c) + (2^2 \times 2c) + (3^2 \times 3c) + (4^2 \times 4c)$

$E(X^2) = c + 8c + 27c + 64c = 100c$

$E(X^2) = 100 \times \frac{1}{10} = 10$

Now, we can calculate Var(X):

$Var(X) = E(X^2) - [E(X)]^2 = 10 - (3)^2 = 10 - 9 = 1$

Summary:

• $c = \frac{1}{10}$
• CDF Table (shown above)
• $E(X) = 3$
• $Var(X) = 1$

This is a binomial distribution problem. We are given:

• $n = 10$ (number of items in a box)
• $p = 0.05$ (probability that an item is defective)

The guarantee is fulfilled if there are not more than 1 defective item, which means 0 or 1 defective items. So we need to find $P(X \leq 1) = P(X=0) + P(X=1)$, where X is the number of defective items.

Using the binomial probability formula:

$P(X = k) = {n \choose k} p^k (1-p)^{n-k}$

1. Probability of 0 defective items:

$P(X = 0) = {10 \choose 0} (0.05)^0 (1-0.05)^{10-0}$

$P(X = 0) = {10 \choose 0} (0.05)^0 (0.95)^{10}$

$P(X = 0) = 1 \times 1 \times (0.95)^{10} = 0.59873693923$

2. Probability of 1 defective item:

$P(X = 1) = {10 \choose 1} (0.05)^1 (1-0.05)^{10-1}$

$P(X = 1) = {10 \choose 1} (0.05)^1 (0.95)^9$

$P(X = 1) = 10 \times 0.05 \times (0.95)^9 = 10 \times 0.05 \times 0.63024941 = 0.315124705$

Now, sum the probabilities:

$P(X \leq 1) = P(X=0) + P(X=1) = 0.59873693923 + 0.315124705 = 0.91386164423$

$P(X \leq 1) \approx 0.9139$

Therefore, the probability that the guarantee is fulfilled is approximately 0.9139.

Conditions for Normal Approximation of a Binomial Distribution:

A Binomial distribution can be approximated by a Normal distribution under the following conditions:

1. Large number of trials (n): The number of trials, $n$, should be sufficiently large. A common rule of thumb is that $n \geq 30$ is generally considered large enough.

2. Neither p nor (1-p) too close to 0: The probability of success, $p$, should not be too close to 0 or 1. The rule of thumb is that $np \geq 5$ and $n(1-p) \geq 5$.

When these conditions are met, the Binomial distribution can be approximated by a Normal distribution with:

• Mean: $\mu = np$
• Variance: $\sigma^2 = np(1-p)$
• Standard Deviation: $\sigma = \sqrt{np(1-p)}$

Given X ~ B(100, 0.5), use the Normal approximation to find P(X $\geq$ 60) with continuity correction:

Since $X \sim B(100, 0.5)$, we have $n = 100$ and $p = 0.5$. Both conditions are satisfied:

• $n = 100 \geq 30$
• $np = 100 \times 0.5 = 50 \geq 5$ and $n(1-p) = 100 \times 0.5 = 50 \geq 5$

So, we can approximate it with a Normal distribution with:

• Mean: $\mu = np = 100 \times 0.5 = 50$
• Variance: $\sigma^2 = np(1-p) = 100 \times 0.5 \times 0.5 = 25$
• Standard Deviation: $\sigma = \sqrt{25} = 5$

We want to find $P(X \geq 60)$. Using continuity correction, we adjust the value to $59.5$, since we want to include all values greater than or equal to 60. Therefore we actually look for $P(X > 59.5)$.

Now, we standardize the value:

$z = \frac{x - \mu}{\sigma} = \frac{59.5 - 50}{5} = \frac{9.5}{5} = 1.9$

We want to find $P(X > 59.5) = P(Z > 1.9)$. From the Standard Normal Table, the area between $Z=0$ and $Z=1.9$ is 0.4713. Since the total area to the right of $Z=0$ is 0.5, we have:

$P(Z > 1.9) = 0.5 - 0.4713 = 0.0287$

Therefore, $P(X \geq 60) \approx 0.0287$.

For a Poisson distribution, the probability mass function is given by:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the mean and $k$ is the number of events.

Given that $P(X=2) = P(X=3)$, we have:

$\frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-\lambda} \lambda^3}{3!}$

We can cancel out $e^{-\lambda}$ from both sides:

$\frac{\lambda^2}{2} = \frac{\lambda^3}{6}$

Multiplying both sides by 6, we get:

$3\lambda^2 = \lambda^3$

Rearranging the terms:

$\lambda^3 - 3\lambda^2 = 0$

$\lambda^2(\lambda - 3) = 0$

This gives us two possible values for $\lambda$: $\lambda = 0$ or $\lambda = 3$.

Since $\lambda$ represents the mean of the Poisson distribution, it must be a positive value. Therefore, $\lambda = 0$ is not a valid solution in this context.

Thus, the mean number of errors per page is $\lambda = 3$.

Now, we need to find the probability of finding at least one error on a page, which is $P(X \geq 1)$:

$P(X \geq 1) = 1 - P(X = 0)$

$P(X = 0) = \frac{e^{-3} (3)^0}{0!} = e^{-3}$

$P(X \geq 1) = 1 - e^{-3}$

Therefore, the mean number of errors per page is 3, and the probability of finding at least one error on a page is $1 - e^{-3} \approx 1 - 0.0498 = 0.9502$.

1. Verify if it is a valid probability distribution:

For a probability distribution to be valid, two conditions must be met:

1. $0 \leq P(X=x) \leq 1$ for all x.
2. $\sum P(X=x) = 1$

Let's check the given distribution:

• $P(X=1) = \frac{1}{10} = 0.1$
• $P(X=2) = \frac{2}{10} = 0.2$
• $P(X=3) = \frac{3}{10} = 0.3$
• $P(X=4) = \frac{4}{10} = 0.4$

All probabilities are between 0 and 1, so the first condition is met.

Now, let's check if the sum of the probabilities equals 1:

$P(X=1) + P(X=2) + P(X=3) + P(X=4) = \frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} = \frac{1+2+3+4}{10} = \frac{10}{10} = 1$

The sum of the probabilities is equal to 1, so the second condition is also met.

Therefore, the given distribution is a valid probability distribution.

2. Find E(X):

$E(X) = \sum x P(X=x)$

$E(X) = (1 \times \frac{1}{10}) + (2 \times \frac{2}{10}) + (3 \times \frac{3}{10}) + (4 \times \frac{4}{10})$

$E(X) = \frac{1}{10} + \frac{4}{10} + \frac{9}{10} + \frac{16}{10} = \frac{1+4+9+16}{10} = \frac{30}{10} = 3$

3. Find Var(X):

$Var(X) = E(X^2) - [E(X)]^2$

First, we need to calculate $E(X^2)$:

$E(X^2) = \sum x^2 P(X=x)$

$E(X^2) = (1^2 \times \frac{1}{10}) + (2^2 \times \frac{2}{10}) + (3^2 \times \frac{3}{10}) + (4^2 \times \frac{4}{10})$

$E(X^2) = \frac{1}{10} + \frac{8}{10} + \frac{27}{10} + \frac{64}{10} = \frac{1+8+27+64}{10} = \frac{100}{10} = 10$

Now, we can calculate Var(X):

$Var(X) = E(X^2) - [E(X)]^2 = 10 - (3)^2 = 10 - 9 = 1$

Summary:

• The distribution is a valid probability distribution.
• $E(X) = 3$
• $Var(X) = 1$

Given a Normal distribution with $\mu = 10$ and $\sigma = 0.1$, we first convert the diameters to z-scores using the formula:

$z = \frac{x - \mu}{\sigma}$

We want to find the probability that a shaft is usable, i.e., its diameter is between 9.8 cm and 10.2 cm. So we need to find $P(9.8 < X < 10.2)$.

$x_1 = 9.8$, so $z_1 = \frac{9.8 - 10}{0.1} = \frac{-0.2}{0.1} = -2$

$x_2 = 10.2$, so $z_2 = \frac{10.2 - 10}{0.1} = \frac{0.2}{0.1} = 2$

We want to find $P(9.8 < X < 10.2) = P(-2 < Z < 2)$. From the standard normal table, the area between $Z=0$ and $Z=2$ is 0.4772. Due to symmetry, the area between $Z=0$ and $Z=-2$ is also 0.4772. Therefore:

$P(-2 < Z < 2) = 0.4772 + 0.4772 = 0.9544$

This means that 95.44% of shafts are usable.

Therefore, the percentage of shafts that are usable is 95.44%.

For a Binomial distribution with parameters $n$ (number of trials) and $p$ (probability of success), the mean and variance are given by:

• Mean: $\mu = np$
• Variance: $\sigma^2 = np(1-p)$

Given that $X \sim B(10, p)$ and $E(X) = 4$, we have:

$np = 4$

$10p = 4$

$p = \frac{4}{10} = 0.4$

Now, we can find Var(X):

$Var(X) = np(1-p) = 10 \times 0.4 \times (1-0.4) = 10 \times 0.4 \times 0.6 = 2.4$

Next, we need to find $P(X=2)$ using the binomial probability formula:

$P(X = k) = {n \choose k} p^k (1-p)^{n-k}$

$P(X = 2) = {10 \choose 2} (0.4)^2 (0.6)^{10-2}$

$P(X = 2) = {10 \choose 2} (0.4)^2 (0.6)^8$

${10 \choose 2} = \frac{10!}{2!8!} = \frac{10 \times 9}{2} = 45$

$P(X = 2) = 45 \times (0.16) \times (0.01679616) = 45 \times 0.16 \times 0.01679616 = 0.120932352$

$P(X=2) \approx 0.1209$

Summary:

• $Var(X) = 2.4$
• $P(X=2) \approx 0.1209$

For a Poisson distribution, the probability mass function is given by:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the mean and $k$ is the number of events.

In this case, $\lambda = 5$ and we want to find $P(X < 3) = P(X=0) + P(X=1) + P(X=2)$.

$P(X = 0) = \frac{e^{-5} (5)^0}{0!} = e^{-5} \approx 0.0067$

$P(X = 1) = \frac{e^{-5} (5)^1}{1!} = 5e^{-5} \approx 5 \times 0.0067 = 0.0335$

$P(X = 2) = \frac{e^{-5} (5)^2}{2!} = \frac{25e^{-5}}{2} \approx \frac{25 \times 0.0067}{2} = \frac{0.1675}{2} = 0.08375$

$P(X < 3) = 0.0067 + 0.0335 + 0.08375 = 0.124 \approx 0.1240$

Therefore, the probability that on a randomly selected day, the number of complaints is less than 3 is approximately 0.1240.

Characteristics of a Normal Distribution:

1. Bell-Shaped: The normal distribution has a characteristic bell shape, which is symmetric and unimodal (has a single peak).

2. Symmetry: The normal distribution is perfectly symmetric around its mean ($\mu$). This means that the left and right halves of the curve are mirror images of each other.

3. Mean, Median, and Mode: In a normal distribution, the mean, median, and mode are all equal and located at the center of the distribution.

4. Asymptotic to the x-axis: The curve approaches the x-axis (horizontal axis) but never actually touches or crosses it. It extends infinitely in both directions.

5. Defined by $\mu$ and $\sigma$: The shape and position of the normal distribution are completely determined by its mean ($\mu$) and standard deviation ($\sigma$). The mean determines the center of the curve, and the standard deviation determines the spread or dispersion of the data.

6. Empirical Rule (68-95-99.7 Rule): Approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.

7. Continuous: The normal distribution is a continuous probability distribution, meaning that the random variable can take any value within a given range.

How the Area Under the Normal Curve is Related to Probability:

The area under the normal curve represents probability. Specifically:

• Total Area: The total area under the normal curve is equal to 1, representing the total probability of all possible outcomes.

• Area and Probability: The area under the normal curve between two points on the x-axis represents the probability that the random variable falls between those two points. For example, the area under the curve between $a$ and $b$ represents $P(a < X < b)$.

• Cumulative Probability: The area under the curve to the left of a point $x$ represents the cumulative probability $P(X \leq x)$. The area under the curve to the right of a point $x$ represents the probability $P(X > x)$.

Because the total area under the normal curve is 1, probabilities can be calculated by finding the appropriate area using a Standard Normal Table (Z-table) after standardizing the variable.

In summary, the area under the normal curve is directly related to probability. By finding the area under the curve within a certain range, we can determine the likelihood of the random variable falling within that range.

We are given the following information:

• $P(X=1) = a$
• $P(X=2) = b$
• $P(X=3) = c$
• $E(X) = 2$
• $Var(X) = 0.6$

We also know that the sum of the probabilities must equal 1:

The expected value E(X) is given by:

The variance Var(X) is given by:

$Var(X) = E(X^2) - [E(X)]^2$, so $E(X^2) = Var(X) + [E(X)]^2$

$E(X^2) = 0.6 + (2)^2 = 0.6 + 4 = 4.6$

Now we have a system of three linear equations with three variables:

Subtract equation (i) from equation (ii):

Subtract equation (i) from equation (iii):

Multiply equation (iv) by 3:

Subtract equation (vi) from equation (v):

$2c = 0.6$

$c = 0.3$

Substitute c = 0.3 into equation (iv):

$b + 2(0.3) = 1$

$b + 0.6 = 1$

$b = 0.4$

Substitute b = 0.4 and c = 0.3 into equation (i):

$a + 0.4 + 0.3 = 1$

$a + 0.7 = 1$

$a = 0.3$

Summary:

• $a = 0.3$
• $b = 0.4$
• $c = 0.3$

Given a Binomial distribution with parameters $n=2000$ and $p = \frac{1}{1000} = 0.001$, we can use the Poisson approximation because $n$ is large and $p$ is small.

The mean of the Poisson distribution is $\lambda = np = 2000 \times 0.001 = 2$

We want to find the probability of exactly 2 people having the disease, i.e., $P(X=2)$.

Using the Poisson probability mass function:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$

$P(X = 2) = \frac{e^{-2} (2)^2}{2!} = \frac{e^{-2} \cdot 4}{2} = 2e^{-2}$

Given that $e^{-2} \approx 0.1353$,

$P(X = 2) \approx 2 \times 0.1353 = 0.2706$

Therefore, the probability that exactly 2 people have the disease is approximately 0.2706.

Given a Normal distribution with $\mu = 40$ and $\sigma = 8$, we want to find the value of $X$ such that $P(X > x) = 0.10$. This means the area to the right of $x$ is 0.10.

Since the area to the right of the mean is 0.5, the area between the mean and $x$ is $0.5 - 0.10 = 0.40$.

We are given that the area between 0 and $Z = 1.28$ is approximately 0.40. Therefore, the z-score corresponding to $X$ is $z = 1.28$.

We can use the z-score formula to find $X$:

$z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 40}{8}$

$X - 40 = 1.28 \times 8 = 10.24$

$X = 40 + 10.24 = 50.24$

Therefore, the value of $X$ such that 10% of the area is to the right of $X$ is approximately 50.24.

Since the die is fair, the probability of rolling an even number (2, 4, or 6) on a single roll is $p = \frac{3}{6} = \frac{1}{2}$. The probability of not rolling an even number is $1-p = \frac{1}{2}$.

The number of times an even number appears in 5 rolls follows a Binomial distribution with $n=5$ and $p=\frac{1}{2}$. Therefore, $X \sim B(5, \frac{1}{2})$.

The probability distribution of X is given by:

$P(X = k) = {n \choose k} p^k (1-p)^{n-k} = {5 \choose k} (\frac{1}{2})^k (\frac{1}{2})^{5-k} = {5 \choose k} (\frac{1}{2})^5$ for $k = 0, 1, 2, 3, 4, 5$.

The probabilities are:

• $P(X=0) = {5 \choose 0} (\frac{1}{2})^5 = 1 \times \frac{1}{32} = \frac{1}{32}$
• $P(X=1) = {5 \choose 1} (\frac{1}{2})^5 = 5 \times \frac{1}{32} = \frac{5}{32}$
• $P(X=2) = {5 \choose 2} (\frac{1}{2})^5 = 10 \times \frac{1}{32} = \frac{10}{32}$
• $P(X=3) = {5 \choose 3} (\frac{1}{2})^5 = 10 \times \frac{1}{32} = \frac{10}{32}$
• $P(X=4) = {5 \choose 4} (\frac{1}{2})^5 = 5 \times \frac{1}{32} = \frac{5}{32}$
• $P(X=5) = {5 \choose 5} (\frac{1}{2})^5 = 1 \times \frac{1}{32} = \frac{1}{32}$

The probability distribution table is:

Now, let's calculate E(X) and Var(X):

For a Binomial distribution, $E(X) = np = 5 \times \frac{1}{2} = \frac{5}{2} = 2.5$

And $Var(X) = np(1-p) = 5 \times \frac{1}{2} \times \frac{1}{2} = \frac{5}{4} = 1.25$

Summary:

• Probability Distribution Table (shown above)
• $E(X) = 2.5$
• $Var(X) = 1.25$

Relationship Between Binomial, Poisson, and Normal Distributions:

These three distributions are related and can, under certain conditions, approximate each other. Here's a breakdown of their relationships and the conditions for approximation:

1. Binomial and Poisson Distributions:

• Relationship: The Poisson distribution can be used to approximate the Binomial distribution when the number of trials ($n$) is large, the probability of success ($p$) is small, and the mean ($np$) is moderate (typically less than 10).

• Conditions for Approximation:

  • $n$ is large (e.g., $n \geq 30$).
  • $p$ is small (e.g., $p \leq 0.1$).
  • $\lambda = np$ is moderate (e.g., $\lambda < 10$).

• Why it works: When $n$ is large and $p$ is small, the Binomial distribution becomes skewed, and the Poisson distribution provides a good approximation for modeling rare events in a large population.

2. Binomial and Normal Distributions:

• Relationship: The Normal distribution can be used to approximate the Binomial distribution when the number of trials ($n$) is large, and neither $p$ nor $(1-p)$ are too close to 0.

• Conditions for Approximation:

  • $n$ is large (e.g., $n \geq 30$).
  • $np \geq 5$ and $n(1-p) \geq 5$ (ensures that the distribution is not too skewed).

• Why it works: The Central Limit Theorem states that the sum (or average) of a large number of independent and identically distributed random variables tends to follow a normal distribution, regardless of the original distribution. As $n$ increases, the Binomial distribution becomes more and more bell-shaped, resembling a Normal distribution.

3. Poisson and Normal Distributions:

• Relationship: The Normal distribution can approximate the Poisson distribution when the mean ($\lambda$) is large (typically $\lambda \geq 10$).

• Conditions for Approximation:

  • $\lambda$ is large (e.g., $\lambda \geq 10$).

• Why it works: As the mean of the Poisson distribution increases, the distribution becomes more symmetric and bell-shaped, resembling a Normal distribution. This is also related to the Central Limit Theorem, as the Poisson distribution can be seen as the sum of a large number of independent Bernoulli trials with very small success probabilities.

Summary Table: