Chapter 15 Statistics (Additional Questions)

The range of a data set is the difference between the maximum and minimum values in the set.

Range = Maximum Value - Minimum Value

The given data set is: 10, 12, 15, 8, 20, 25.

Let's identify the minimum and maximum values:

The minimum value in the data set is 8.

The maximum value in the data set is 25.

Now, calculate the range:

Range = $25 - 8$

Range = $17$

The range of the given data set is 17.

Therefore, the correct option is (B) 17.

The range is defined as the difference between the highest and the lowest values in a data set.

Range = Highest Value - Lowest Value

Measures of central tendency (like mean, median, and mode) describe the center of a data set.

Measures of dispersion (like range, variance, and standard deviation) describe the spread or variability of a data set.

Frequency refers to how often a particular value appears in a data set.

Correlation measures the strength and direction of a linear relationship between two variables.

Since the range describes how spread out the data is, it is a measure of dispersion.

Therefore, the correct option is (B) Dispersion.

The given data set is: 4, 7, 8, 9, 10, 12, 13, 17.

The number of observations is $n = 8$.

First, we need to calculate the mean ($\bar{x}$) of the data.

The formula for the mean is:

$\bar{x} = \frac{\sum x_i}{n}$

Sum of observations ($\sum x_i$) = $4 + 7 + 8 + 9 + 10 + 12 + 13 + 17 = 80$

Mean ($\bar{x}$) = $\frac{80}{8} = 10$

The mean of the data is 10.

Next, we calculate the absolute deviation of each observation from the mean ($|x_i - \bar{x}|$).

For $x_i = 4$, $|4 - 10| = |-6| = 6$

For $x_i = 7$, $|7 - 10| = |-3| = 3$

For $x_i = 8$, $|8 - 10| = |-2| = 2$

For $x_i = 9$, $|9 - 10| = |-1| = 1$

For $x_i = 10$, $|10 - 10| = |0| = 0$

For $x_i = 12$, $|12 - 10| = |2| = 2$

For $x_i = 13$, $|13 - 10| = |3| = 3$

For $x_i = 17$, $|17 - 10| = |7| = 7$

Now, we calculate the sum of these absolute deviations.

$\sum |x_i - \bar{x}| = 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 = 24$

Finally, we calculate the mean deviation about the mean using the formula:

Mean Deviation = $\frac{\sum |x_i - \bar{x}|}{n}$

Mean Deviation = $\frac{24}{8} = 3$

The mean deviation about the mean for the given data is 3.

Therefore, the correct option is (A) 3.

The given data set is: 3, 4, 5, 6, 7, 8.

The data is already arranged in ascending order.

The number of observations is $n = 6$.

Since the number of observations ($n=6$) is even, the median ($M$) is the average of the $\left(\frac{n}{2}\right)$th and $\left(\frac{n}{2} + 1\right)$th observations.

$\frac{n}{2} = \frac{6}{2} = 3$. The 3rd observation is 5.

$\frac{n}{2} + 1 = 3 + 1 = 4$. The 4th observation is 6.

Median ($M$) = $\frac{5 + 6}{2} = \frac{11}{2} = 5.5$

The median of the data is 5.5.

Next, we calculate the absolute deviation of each observation from the median ($|x_i - M|$).

$|3 - 5.5| = |-2.5| = 2.5$

$|4 - 5.5| = |-1.5| = 1.5$

$|5 - 5.5| = |-0.5| = 0.5$

$|6 - 5.5| = |0.5| = 0.5$

$|7 - 5.5| = |1.5| = 1.5$

$|8 - 5.5| = |2.5| = 2.5$

Now, we calculate the sum of these absolute deviations ($\sum |x_i - M|$).

$\sum |x_i - M| = 2.5 + 1.5 + 0.5 + 0.5 + 1.5 + 2.5 = 9$

Finally, we calculate the mean deviation about the median using the formula:

Mean Deviation about Median = $\frac{\sum |x_i - M|}{n}$

Mean Deviation about Median = $\frac{9}{6}$

Mean Deviation about Median = $1.5$

The mean deviation about the median for the given data is 1.5.

Therefore, the correct option is (A) 1.5.

The question defines variance as the average of the squared differences from the mean.

Let's look at the symbols provided in the options:

$\sigma$ (sigma) typically represents the population standard deviation.

$\sigma^2$ (sigma squared) represents the population variance.

$\mu$ (mu) represents the population mean.

$MD$ often represents Mean Deviation.

The variance of a data set is conventionally denoted by $\sigma^2$ for a population variance or $s^2$ for a sample variance. Among the given options, $\sigma^2$ is the symbol representing variance.

Therefore, the correct option is (B) $\sigma^2$.

The question provides the relationship between standard deviation and variance: the standard deviation is the square root of the variance.

Given that the variance of the data set is 36.

We need to find the standard deviation.

Standard Deviation = $\sqrt{\text{Variance}}$

Substitute the given variance value into the formula:

Standard Deviation = $\sqrt{36}$

Calculate the square root of 36:

$\sqrt{36} = 6$

The standard deviation of the data set is 6.

Therefore, the correct option is (A) 6.

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

Assertion (A): Standard deviation is always non-negative.

The standard deviation is defined as the square root of the variance. Mathematically, $\sigma = \sqrt{\sigma^2}$ (for population standard deviation) or $s = \sqrt{s^2}$ (for sample standard deviation). The square root of any non-negative number is always taken as the principal (non-negative) root in statistics. Thus, standard deviation cannot be negative. So, Assertion (A) is true.

Reason (R): Standard deviation is the square root of variance, and variance is the average of squared deviations, which are always non-negative.

Let's break down the reason:

1. "Standard deviation is the square root of variance": This is true by definition.

2. "Variance is the average of squared deviations": The formula for variance involves summing the squared differences between each data point and the mean, and then dividing by the number of observations (or n-1 for sample variance). This is indeed the average of squared deviations.

3. "Squared deviations, which are always non-negative": The deviation of a data point $x_i$ from the mean $\bar{x}$ is $(x_i - \bar{x})$. Squaring this deviation gives $(x_i - \bar{x})^2$. The square of any real number (positive, negative, or zero) is always non-negative. So, $(x_i - \bar{x})^2 \geq 0$ for all $x_i$.

Since variance is the average of these non-negative squared deviations, the variance ($\sigma^2$ or $s^2$) must also be non-negative.

Because variance is non-negative, its square root (the standard deviation) must also be non-negative.

Thus, Reason (R) is also true, and it correctly explains why the standard deviation is always non-negative.

Both the Assertion (A) and the Reason (R) are true, and the Reason (R) provides a correct explanation for the Assertion (A).

Therefore, the correct option is (A) Both A and R are true and R is the correct explanation of A.

The Coefficient of Variation (CV) is a standardized measure of dispersion of a probability distribution or frequency distribution.

It is used to compare the degree of variation between data sets, even if their means are drastically different from each other.

The formula for the Coefficient of Variation is the ratio of the standard deviation to the mean, usually expressed as a percentage.

$\text{CV} = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100\%$

Let's examine the given options:

(A) $\frac{\text{Mean}}{\text{Standard Deviation}} \times 100$ - This is the reciprocal of the CV formula.

(B) $\frac{\text{Standard Deviation}}{\text{Mean}} \times 100$ - This matches the standard formula for CV.

(C) $\frac{\text{Variance}}{\text{Mean}} \times 100$ - This uses variance instead of standard deviation.

(D) $\frac{\text{Mean}}{\text{Variance}} \times 100$ - This uses variance and is the reciprocal ratio.

Based on the standard definition and formula, the Coefficient of Variation is given by the ratio of the standard deviation to the mean, multiplied by 100.

Therefore, the correct option is (B) $\frac{\text{Standard Deviation}}{\text{Mean}} \times 100$.

Let's match each measure of dispersion with its correct characteristic.

(i) Range: The range is the difference between the highest and lowest values. It is the simplest measure to calculate but is significantly influenced by extreme values (outliers).

This matches characteristic (b) Simplest measure, affected by extreme values.

(ii) Mean Deviation: The mean deviation is calculated by taking the average of the absolute differences between each data point and the mean or median. It uses $|x_i - \bar{x}|$ or $|x_i - M|$.

This matches characteristic (c) Uses absolute deviations from mean or median.

(iii) Variance: The variance is calculated as the average of the squared differences between each data point and the mean. It uses $(x_i - \bar{x})^2$.

This matches characteristic (a) Uses squared deviations from the mean.

(iv) Coefficient of Variation: The Coefficient of Variation (CV) is a relative measure of dispersion, calculated as $\frac{\text{Standard Deviation}}{\text{Mean}} \times 100$. It is a unitless measure (as the units cancel out) and is useful for comparing variability between data sets with different means or units.

This matches characteristic (d) Unitless measure, used for comparing variability.

So, the correct matching is:

(i) - (b)

(ii) - (c)

(iii) - (a)

(iv) - (d)

Let's check the options based on this matching:

(A) (i)-(a), (ii)-(c), (iii)-(b), (iv)-(d) - Incorrect

(B) (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d) - Correct

(C) (i)-(b), (ii)-(a), (iii)-(c), (iv)-(d) - Incorrect

(D) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d) - Incorrect

The correct option is (B) (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d).

The given information is:

• Mean Deviation about the Mean = 5
• Mean ($\bar{x}$) = 50

The coefficient of mean deviation is a relative measure of dispersion. It is calculated as the ratio of the mean deviation to the average (mean or median) from which the deviations are taken.

In this case, deviations are taken from the mean.

The formula for the coefficient of mean deviation about the mean is:

Coefficient of Mean Deviation = $\frac{\text{Mean Deviation about Mean}}{\text{Mean}}$

Substitute the given values into the formula:

Coefficient of Mean Deviation = $\frac{5}{50}$

Coefficient of Mean Deviation = $\frac{1}{10}$

Coefficient of Mean Deviation = $0.1$

The coefficient of mean deviation is often expressed as a percentage for easier comparison. To convert the ratio to a percentage, multiply by 100%:

Coefficient of Mean Deviation (%) = $\frac{5}{50} \times 100\%$

Coefficient of Mean Deviation (%) = $0.1 \times 100\%$

Coefficient of Mean Deviation (%) = $10\%$

We calculated the coefficient as $0.1$ (ratio) or $10\%$ (percentage).

Looking at the options:

• (A) 10%
• (B) 5%
• (C) 0.1
• (D) 0.05

Both 10% and 0.1 represent the same value. In the context of relative dispersion measures like the coefficient of variation, the percentage form is frequently used. Given the options, 10% is provided as option (A).

Therefore, the correct option is (A) 10%.

The first $n$ natural numbers are $1, 2, 3, \ldots, n$.

The mean of the first $n$ natural numbers is given by:

$\bar{x} = \frac{\sum_{i=1}^n i}{n} = \frac{1 + 2 + \ldots + n}{n} = \frac{n(n+1)/2}{n} = \frac{n+1}{2}$

The variance ($\sigma^2$) of the first $n$ natural numbers is a standard result given by the formula:

$\sigma^2 = \frac{n^2 - 1}{12}$

(This formula can be derived from the definition of variance $\sigma^2 = \frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n}$ or using $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$ with $\sum x_i^2 = \sum i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\bar{x} = \frac{n+1}{2}$)

The standard deviation ($\sigma$) is the square root of the variance.

$\sigma = \sqrt{\sigma^2}$

Substitute the formula for variance:

$\sigma = \sqrt{\frac{n^2 - 1}{12}}$

Comparing this result with the given options:

(A) $\sqrt{\frac{n^2 - 1}{12}}$

(B) $\frac{n^2 - 1}{12}$

(C) $\sqrt{\frac{n(n+1)}{2}}$

(D) $\frac{n(n+1)}{2}$

The calculated standard deviation matches option (A).

Therefore, the correct option is (A) $\sqrt{\frac{n^2 - 1}{12}}$.

We need to identify the measure of dispersion that is least affected by extreme values (outliers) in a data set.

Let's consider how each measure is calculated and its sensitivity to extreme values:

(A) Range: The range is the difference between the maximum and minimum values. If either the maximum or minimum value is an extreme outlier, the range will be significantly affected. It is the most affected by extreme values.

(B) Mean Deviation about Mean: This measure is based on the mean. The mean itself is sensitive to extreme values, as it is calculated by summing all data points. If an outlier is present, the mean shifts towards the outlier, which affects the deviations and thus the mean deviation. It is affected by outliers, but less so than the range.

(C) Standard Deviation: This measure is based on the mean and the squared deviations from the mean. Like the mean deviation about the mean, it is affected by the sensitivity of the mean to outliers. Furthermore, squaring the deviations gives more weight to larger deviations (those from extreme values), making the standard deviation even more sensitive to outliers than the mean deviation about the mean.

(D) Mean Deviation about Median: This measure is based on the median. The median is the middle value (or average of the two middle values) when the data is ordered. Unlike the mean, the median is resistant to extreme values because its value is not directly calculated from the magnitudes of all data points. Since the mean deviation about the median is calculated using deviations from the median, it is less affected by outliers compared to measures based on the mean (like mean deviation about mean or standard deviation).

Comparing the measures, the Range is most sensitive. Standard Deviation is generally more sensitive than Mean Deviation about Mean. Mean Deviation about Median is based on the median, which is resistant to outliers, making it the least affected measure among the given options.

Therefore, the measure of dispersion least affected by extreme values is the Mean Deviation about Median.

The correct option is (D) Mean Deviation about Median.

Let the data set consist of $n$ observations, where all observations are equal to a constant value, say $c$.

The data set is thus $c, c, c, \ldots, c$ ($n$ times).

First, we calculate the mean ($\bar{x}$) of the data set.

$\bar{x} = \frac{\sum x_i}{n} = \frac{c + c + \ldots + c \text{ ($n$ times)}}{n} = \frac{n \times c}{n} = c$

The mean is equal to the value of each observation.

Next, we calculate the deviations from the mean for each observation ($x_i - \bar{x}$).

For every observation $x_i = c$, the deviation is $x_i - \bar{x} = c - c = 0$.

Then, we calculate the squared deviations from the mean ($(x_i - \bar{x})^2$).

For every observation, the squared deviation is $(c - c)^2 = 0^2 = 0$.

The variance ($\sigma^2$ for population, $s^2$ for sample) is the average of the squared deviations.

Using the population variance formula:

$\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{0 + 0 + \ldots + 0 \text{ ($n$ times)}}{n} = \frac{0}{n} = 0$ (assuming $n>0$)

Using the sample variance formula (for $n > 1$):

$s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} = \frac{0}{n-1} = 0$

In either standard case (population or sample with $n>1$), the variance is 0.

Finally, the standard deviation ($\sigma$ or $s$) is the square root of the variance.

Standard Deviation = $\sqrt{\text{Variance}} = \sqrt{0} = 0$

If all observations in a data set are equal, there is no spread or variability among them, and the standard deviation is 0, indicating no dispersion.

Comparing this result with the given options:

• (A) 0
• (B) 1
• (C) 5
• (D) Undefined

The calculated standard deviation is 0.

Therefore, the correct option is (A) 0.

The time taken by Mr. Sharma on 5 different days is: 10, 12, 8, 15, 10 minutes.

The number of observations for Mr. Sharma is $n = 5$.

To find the mean time taken by Mr. Sharma, we sum the times and divide by the number of days.

Mean time ($\bar{x}_{\text{Sharma}}$) = $\frac{\text{Sum of times}}{\text{Number of days}}$

Sum of times = $10 + 12 + 8 + 15 + 10$

Sum of times = $55$ minutes

Mean time ($\bar{x}_{\text{Sharma}}$) = $\frac{55}{5}$

Mean time ($\bar{x}_{\text{Sharma}}$) = $11$ minutes

The mean time taken by Mr. Sharma is 11 minutes.

Option (A) shows the calculation $\frac{10+12+8+15+10}{5} = \frac{55}{5} = 11$ minutes, which matches our calculation and result.

Therefore, the correct option is (A) $\frac{10+12+8+15+10}{5} = \frac{55}{5} = 11$ minutes.

The time taken by Mr. Verma on 5 different days is: 11, 10, 12, 11, 11 minutes.

The number of observations for Mr. Verma is $n = 5$.

To find the mean time taken by Mr. Verma, we sum the times and divide by the number of days.

Mean time ($\bar{x}_{\text{Verma}}$) = $\frac{\text{Sum of times}}{\text{Number of days}}$

Sum of times = $11 + 10 + 12 + 11 + 11$

Sum of times = $55$ minutes

Mean time ($\bar{x}_{\text{Verma}}$) = $\frac{55}{5}$

Mean time ($\bar{x}_{\text{Verma}}$) = $11$ minutes

The mean time taken by Mr. Verma is 11 minutes.

Option (D) shows the calculation $\frac{11+10+12+11+11}{5} = \frac{55}{5} = 11$ minutes, which matches our calculation and result.

Therefore, the correct option is (D) $\frac{11+10+12+11+11}{5} = \frac{55}{5} = 11$ minutes.

The time taken by Mr. Sharma on 5 different days is: 10, 12, 8, 15, 10 minutes.

The number of observations for Mr. Sharma is $n = 5$.

First, we need the mean time taken by Mr. Sharma. From Question 14, we found the mean is 11 minutes.

Mean ($\bar{x}_{\text{Sharma}}$) = 11 minutes.

Next, we calculate the deviations of each observation from the mean ($x_i - \bar{x}$).

• $10 - 11 = -1$
• $12 - 11 = 1$
• $8 - 11 = -3$
• $15 - 11 = 4$
• $10 - 11 = -1$

Now, we calculate the squared deviations ($(x_i - \bar{x})^2$).

• $(-1)^2 = 1$
• $(1)^2 = 1$
• $(-3)^2 = 9$
• $(4)^2 = 16$
• $(-1)^2 = 1$

Calculate the sum of squared deviations ($\sum (x_i - \bar{x})^2$).

$\sum (x_i - \bar{x})^2 = 1 + 1 + 9 + 16 + 1 = 28$

We will calculate the sample standard deviation, which is common for small data sets. The formula for sample standard deviation ($s$) is:

$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$

Substitute the values:

$s = \sqrt{\frac{28}{5-1}}$

$s = \sqrt{\frac{28}{4}}$

$s = \sqrt{7}$

Now, we approximate the value of $\sqrt{7}$.

$\sqrt{7} \approx 2.646$

Comparing the approximate value 2.646 with the given options:

• (A) 1.2
• (B) 2.1
• (C) 2.5
• (D) 3.0

The value 2.646 is closest to 2.5.

The standard deviation of the time taken by Mr. Sharma is approximately 2.5 minutes.

Therefore, the correct option is (C) 2.5 minutes.

The time taken by Mr. Verma on 5 different days is: 11, 10, 12, 11, 11 minutes.

The number of observations for Mr. Verma is $n = 5$.

First, we need the mean time taken by Mr. Verma. From Question 15, we found the mean is 11 minutes.

Mean ($\bar{x}_{\text{Verma}}$) = 11 minutes.

Next, we calculate the deviations of each observation from the mean ($x_i - \bar{x}$).

• $11 - 11 = 0$
• $10 - 11 = -1$
• $12 - 11 = 1$
• $11 - 11 = 0$
• $11 - 11 = 0$

Now, we calculate the squared deviations ($(x_i - \bar{x})^2$).

• $(0)^2 = 0$
• $(-1)^2 = 1$
• $(1)^2 = 1$
• $(0)^2 = 0$
• $(0)^2 = 0$

Calculate the sum of squared deviations ($\sum (x_i - \bar{x})^2$).

$\sum (x_i - \bar{x})^2 = 0 + 1 + 1 + 0 + 0 = 2$

We will calculate the sample standard deviation ($s$) using the formula:

$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$

Substitute the values:

$s = \sqrt{\frac{2}{5-1}}$

$s = \sqrt{\frac{2}{4}}$

$s = \sqrt{0.5}$

Now, we approximate the value of $\sqrt{0.5}$.

$\sqrt{0.5} = \frac{1}{\sqrt{2}} \approx 0.707$

Comparing the approximate value 0.707 with the given options:

• (A) 0.4 minutes
• (B) 0.8 minutes
• (C) 1.0 minutes
• (D) 1.2 minutes

The value 0.707 is closest to 0.8.

The standard deviation of the time taken by Mr. Verma is approximately 0.8 minutes.

Therefore, the correct option is (B) 0.8 minutes.

Consistency in performance is typically measured by the inverse of variability or dispersion. A smaller standard deviation indicates less variability in the data, meaning the data points are clustered closer to the mean.

In the context of a worker completing a task, lower variability in time taken means the worker is more consistent in their performance.

From Question 16, the approximate standard deviation for Mr. Sharma's time is 2.5 minutes.

Standard Deviation for Mr. Sharma $\approx 2.5$ minutes.

From Question 17, the approximate standard deviation for Mr. Verma's time is 0.8 minutes.

Standard Deviation for Mr. Verma $\approx 0.8$ minutes.

Comparing the standard deviations:

$0.8 < 2.5$

The standard deviation of Mr. Verma's time (0.8 minutes) is less than the standard deviation of Mr. Sharma's time (2.5 minutes).

A smaller standard deviation implies less spread or variability in the time taken to complete the task. Therefore, Mr. Verma shows less variation in his time taken compared to Mr. Sharma.

This indicates that Mr. Verma is a more consistent worker than Mr. Sharma.

Therefore, the correct option is (B) Mr. Verma.

The given formula for the variance of ungrouped data is:

$\sigma^2 = \frac{1}{n} \sum\limits_{i=1}^n (x_i - \bar{x})^2$

where $\bar{x}$ is the mean of the data.

We can derive an alternative formula by expanding the term $(x_i - \bar{x})^2$:

$(x_i - \bar{x})^2 = x_i^2 - 2x_i\bar{x} + \bar{x}^2$

Substituting this into the variance formula:

$\sigma^2 = \frac{1}{n} \sum\limits_{i=1}^n (x_i^2 - 2x_i\bar{x} + \bar{x}^2)$

Using the properties of summation ($\sum (a_i + b_i) = \sum a_i + \sum b_i$ and $\sum c a_i = c \sum a_i$, $\sum c = nc$):

$\sigma^2 = \frac{1}{n} \left( \sum\limits_{i=1}^n x_i^2 - \sum\limits_{i=1}^n 2x_i\bar{x} + \sum\limits_{i=1}^n \bar{x}^2 \right)$

$\sigma^2 = \frac{1}{n} \left( \sum\limits_{i=1}^n x_i^2 - 2\bar{x} \sum\limits_{i=1}^n x_i + n\bar{x}^2 \right)$

Since $\bar{x} = \frac{1}{n} \sum\limits_{i=1}^n x_i$, we have $\sum\limits_{i=1}^n x_i = n\bar{x}$. Substitute this into the equation:

$\sigma^2 = \frac{1}{n} \left( \sum\limits_{i=1}^n x_i^2 - 2\bar{x}(n\bar{x}) + n\bar{x}^2 \right)$

$\sigma^2 = \frac{1}{n} \left( \sum\limits_{i=1}^n x_i^2 - 2n\bar{x}^2 + n\bar{x}^2 \right)$

$\sigma^2 = \frac{1}{n} \left( \sum\limits_{i=1}^n x_i^2 - n\bar{x}^2 \right)$

Distribute $\frac{1}{n}$:

$\sigma^2 = \frac{1}{n} \sum\limits_{i=1}^n x_i^2 - \frac{n\bar{x}^2}{n}$

So, the alternative formula is:

$\sigma^2 = \frac{1}{n} \sum\limits_{i=1}^n x_i^2 - \bar{x}^2$

This can also be written as:

$\sigma^2 = \frac{1}{n} (\sum\limits_{i=1}^n x_i^2) - (\bar{x})^2$

Comparing this result with the given options:

• (A) $\sigma^2 = \frac{1}{n} (\sum\limits_{i=1}^n x_i^2) - (\bar{x})^2$
• (B) $\sigma^2 = \frac{1}{n} \sum\limits_{i=1}^n x_i^2$
• (C) $\sigma^2 = (\sum\limits_{i=1}^n x_i^2) - (\bar{x})^2$
• (D) $\sigma^2 = \frac{1}{n} (\sum\limits_{i=1}^n x_i - \bar{x})^2$

Option (A) matches the derived alternative formula.

Therefore, the correct option is (A) $\sigma^2 = \frac{1}{n} (\sum\limits_{i=1}^n x_i^2) - (\bar{x})^2$.

The given data set is: 2, 4, 6, 8, 10.

The number of observations is $n = 5$.

First, we calculate the mean ($\bar{x}$) of the data.

$\bar{x} = \frac{\sum x_i}{n}$

Sum of observations ($\sum x_i$) = $2 + 4 + 6 + 8 + 10 = 30$

Mean ($\bar{x}$) = $\frac{30}{5} = 6$

The mean of the data is 6.

Next, we calculate the deviations of each observation from the mean ($x_i - \bar{x}$).

• $2 - 6 = -4$
• $4 - 6 = -2$
• $6 - 6 = 0$
• $8 - 6 = 2$
• $10 - 6 = 4$

Now, we calculate the squared deviations ($(x_i - \bar{x})^2$).

• $(-4)^2 = 16$
• $(-2)^2 = 4$
• $(0)^2 = 0$
• $(2)^2 = 4$
• $(4)^2 = 16$

Calculate the sum of squared deviations ($\sum (x_i - \bar{x})^2$).

$\sum (x_i - \bar{x})^2 = 16 + 4 + 0 + 4 + 16 = 40$

Finally, we calculate the variance ($\sigma^2$) using the formula for population variance (as it's a complete data set):

$\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$

Substitute the values:

$\sigma^2 = \frac{40}{5}$

$\sigma^2 = 8$

The variance of the data is 8.

Therefore, the correct option is (B) 8.

The question defines the mean deviation about the mean as the average of the absolute deviations from the mean.

Let the data set be $x_1, x_2, ..., x_n$. The mean is denoted by $\bar{x}$.

The deviation from the mean for each data point is $x_i - \bar{x}$.

The absolute deviation from the mean is $|x_i - \bar{x}|$.

The mean deviation about the mean is the average of these absolute deviations:

Mean Deviation $(\bar{x}) = \frac{\sum_{i=1}^{n} |x_i - \bar{x}|}{n}$

The symbol commonly used to represent the mean deviation about the mean is $MD(\bar{x})$ or sometimes $MD$.

Looking at the options:

(A) $\sigma$ represents the population standard deviation.

(B) $\bar{x}$ represents the sample mean.

(C) $MD(\bar{x})$ represents the mean deviation about the mean.

(D) $CV$ represents the coefficient of variation.

Therefore, the symbol that represents the mean deviation about the mean is $MD(\bar{x})$.

The correct option is (C).

Measures of dispersion are statistical values that describe the spread or variability of a data set.

Common measures of dispersion include Range, Variance, Standard Deviation, and Quartile Deviation.

Measures of central tendency are statistical values that describe the center or typical value of a data set.

Common measures of central tendency include Mean, Median, and Mode.

Let's examine the given options:

(A) Mean: This is the average value of the data set. It is a measure of central tendency.

(B) Range: This is the difference between the maximum and minimum values in the data set. It is a measure of dispersion.

(C) Standard Deviation: This measures the amount of variation or dispersion of a set of values. It is a measure of dispersion.

(D) Variance: This is the average of the squared differences from the Mean. It is a measure of dispersion.

The question asks which of the options is NOT a measure of dispersion.

Based on the definitions, the Mean is a measure of central tendency, not dispersion.

Therefore, the Mean is not a measure of dispersion.

The correct option is (A).

Let $\sigma$ denote the standard deviation of a set of data.

The question asks for the term used for the square of the standard deviation, which is $\sigma^2$.

In statistics, the square of the standard deviation is known as the variance.

Let's briefly define the other options for clarity:

• Mean Deviation: The average of the absolute deviations from the mean.
• Range: The difference between the maximum and minimum values in a data set.
• Coefficient of Variation: A relative measure of dispersion, calculated as the ratio of the standard deviation to the mean (usually expressed as a percentage).

Based on the definitions, the square of the standard deviation ($\sigma^2$) is called variance.

The correct answer is variance.

Thus, the correct option is (C) variance.

To find the coefficient of variation, we use the formula:

Coefficient of Variation (CV) $= \frac{\text{Standard Deviation}}{\text{Mean}} \times 100\% $

Given:

Mean $(\bar{x}) = 20$

Standard Deviation $(\sigma) = 4$

Substituting these values into the formula:

$CV = \frac{4}{20} \times 100\%$

Now, we simplify the expression:

$CV = \frac{\cancel{4}^1}{\cancel{20}_5} \times 100\%$

$CV = \frac{1}{5} \times 100\%$

$CV = 0.2 \times 100\%$

$CV = 20\%$

Thus, the coefficient of variation is $20\%$.

Comparing this result with the given options, we find that option (A) matches our result.

The coefficient of variation is $20\%$.

The correct option is (A) 20%.

The table presents the number of students organized according to specific age ranges or intervals, such as 10 - 12, 12 - 14, 14 - 16, and 16 - 18. For each interval, the corresponding frequency (number of students) is provided.

Let's consider the definitions of the given options:

(A) Ungrouped data: This refers to data that has not been classified or organized into groups or classes. It is the raw data set.

(B) Grouped data (with class intervals): This refers to data that has been organized into groups or classes, often defined by intervals, along with the frequency of occurrence within each group.

(C) Discrete data: This refers to data that can only take specific, distinct values and cannot take any value within a range. The number of students is a discrete variable (you can have 15 or 20 students, but not 15.5 students).

(D) Continuous data: This refers to data that can take any value within a given range. Age is generally considered a continuous variable.

In the given table, the age data is organized into classes defined by intervals (10-12, 12-14, etc.). The number of students is presented as the frequency for each of these intervals. This structure perfectly matches the definition of grouped data with class intervals.

Although age is a continuous variable and the number of students is discrete, the classification method shown in the table is what defines the data presentation type here. The data *itself* is presented in a grouped format using class intervals.

The data presented in the table is organized into groups defined by age intervals, with the count of students falling into each interval. This is characteristic of grouped data with class intervals.

The correct option is (B) Grouped data (with class intervals).

The mean deviation about the mean for grouped data is defined as the average of the absolute deviations of the class marks from the mean, weighted by their respective frequencies.

Let $x_i$ be the class marks (midpoints of the class intervals) and $f_i$ be the corresponding frequencies for $i = 1, 2, \dots, n$, where $n$ is the number of classes.

The mean of the grouped data, denoted by $\bar{x}$, is calculated as:

$\bar{x} = \frac{\sum\limits_{i=1}^n f_i x_i}{\sum\limits_{i=1}^n f_i}$

The absolute deviation of a class mark $x_i$ from the mean $\bar{x}$ is $|x_i - \bar{x}|$.

The weighted absolute deviation for the $i$-th class is $f_i |x_i - \bar{x}|$.

The sum of the weighted absolute deviations over all classes is $\sum\limits_{i=1}^n f_i |x_i - \bar{x}|$.

The total frequency is $N = \sum\limits_{i=1}^n f_i$.

The mean deviation about the mean, $MD(\bar{x})$, is the sum of the weighted absolute deviations divided by the total frequency:

$MD(\bar{x}) = \frac{\sum\limits_{i=1}^n f_i |x_i - \bar{x}|}{\sum\limits_{i=1}^n f_i}$

The formula provided in the question is exactly this standard formula used to calculate the mean deviation about the mean for grouped data, where $x_i$ represents the class marks and $f_i$ represents the frequencies.

Therefore, the statement is correct.

The formula provided is the standard formula for calculating the mean deviation about the mean for grouped data using class marks.

The correct option is (A) Yes.

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

Assertion (A): The range of a data set is heavily influenced by the presence of outliers.

The range is defined as the difference between the maximum and minimum values in a data set: $\text{Range} = \text{Maximum Value} - \text{Minimum Value}$.

Outliers are extreme values that lie far from the other data points. If a data set contains a very large outlier, this outlier will likely be the maximum value, significantly increasing the range. Similarly, if a data set contains a very small outlier, this outlier will likely be the minimum value, also significantly increasing the range.

For example, consider the data set $\{10, 12, 15, 18, 20\}$. The range is $20 - 10 = 10$.

Now, consider the data set $\{10, 12, 15, 18, 20, 100\}$. Here, $100$ is an outlier. The range is $100 - 10 = 90$. The presence of the outlier $100$ heavily influenced the range.

Thus, Assertion (A) is True.

Reason (R): The range only considers the two extreme values in the data.

As per the definition, the calculation of the range requires only the highest and lowest values in the data set. It does not use any of the intermediate values.

Thus, Reason (R) is True.

Now let's check if Reason (R) is the correct explanation for Assertion (A).

The fact that the range is calculated using only the maximum and minimum values means that these two extreme values determine the range. If these extreme values happen to be outliers, they will directly impact the range without being mitigated by other data points. This is precisely why the range is sensitive to outliers.

Therefore, Reason (R) correctly explains why Assertion (A) is true.

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

The standard deviation of a data set is the square root of its variance.

The relationship between standard deviation ($\sigma$) and variance ($\sigma^2$) is given by:

$\sigma = \sqrt{\sigma^2}$

Given that the variance of the data set is $100$.

Variance $(\sigma^2) = 100$

To find the standard deviation, we take the square root of the variance:

$\sigma = \sqrt{100}$

The square root of $100$ is $10$.

$\sigma = 10$

Note that standard deviation is a measure of dispersion and is always a non-negative value. Although the mathematical square root of $100$ is $\pm 10$, in the context of statistics, the standard deviation is taken as the positive square root.

Therefore, the standard deviation is $10$.

The standard deviation of the data set is $10$.

The correct option is (A) 10.

The coefficient of variation (CV) is a measure of relative variability. It is used to compare the degree of variation between different data sets, even if their means are different.

The formula for the coefficient of variation is:

$CV = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100\%$

Let $\bar{x}_1$ and $\sigma_1$ be the mean and standard deviation of the first data set, and $\bar{x}_2$ and $\sigma_2$ be the mean and standard deviation of the second data set.

According to the problem statement, the two data sets have the same mean. Let's denote this common mean by $\bar{x}$. So, $\bar{x}_1 = \bar{x}_2 = \bar{x}$.

The coefficient of variation for the first set is $20\%$.

$CV_1 = \frac{\sigma_1}{\bar{x}_1} \times 100\% = 20\%$

$\frac{\sigma_1}{\bar{x}} \times 100\% = 20\%$

This implies $\sigma_1 = \frac{20}{100} \times \bar{x} = 0.20 \times \bar{x}$.

The coefficient of variation for the second set is $30\%$.

$CV_2 = \frac{\sigma_2}{\bar{x}_2} \times 100\% = 30\%$

$\frac{\sigma_2}{\bar{x}} \times 100\% = 30\%$

This implies $\sigma_2 = \frac{30}{100} \times \bar{x} = 0.30 \times \bar{x}$.

Since $\bar{x}$ is the mean of a data set, it is typically a positive value (unless all data points are zero). Comparing the standard deviations, we have:

$\sigma_1 = 0.20 \times \bar{x}$

$\sigma_2 = 0.30 \times \bar{x}$

As $0.30 > 0.20$, and assuming $\bar{x} > 0$, we can conclude that $\sigma_2 > \sigma_1$.

Standard deviation is a measure of the absolute variability or dispersion of the data. A larger standard deviation indicates that the data points are, on average, farther from the mean, meaning the data is more spread out or more variable.

Since the second set has a larger standard deviation ($\sigma_2 > \sigma_1$) and the means are the same, the second set is more variable than the first set.

Alternatively, because the means are equal, we can directly compare the coefficients of variation. The coefficient of variation is the standard deviation relative to the mean. When the means are the same, a higher coefficient of variation directly implies a higher standard deviation.

Since $CV_2 (30\%) > CV_1 (20\%)$, the second set is more variable.

When comparing variability between data sets with the same mean, the data set with the higher coefficient of variation is considered more variable because it has a proportionally larger standard deviation.

Given that the second set has a higher coefficient of variation ($30\%$) compared to the first set ($20\%$), the second set is more variable.

The correct option is (B) The second set.

For grouped data, the range is defined as the difference between the upper limit of the highest class and the lower limit of the lowest class.

From the given table, the classes are 10 - 20, 20 - 30, and 30 - 40.

The lowest class is 10 - 20.

The lower limit of the lowest class is $10$.

The highest class is 30 - 40.

The upper limit of the highest class is $40$.

The formula for the range of grouped data is:

Range = (Upper limit of the highest class) - (Lower limit of the lowest class)

Substituting the values, we get:

Range = $40 - 10$

Range = $30$

Let's look at the given options:

(A) $30$: This is the calculated range.

(B) $40$: This is the upper limit of the highest class.

(C) $40 - 10 = 30$: This shows the calculation and the result.

(D) $20$: This is the class width of the individual classes or the difference between the upper limit of the lowest class and the lower limit of the highest class (40-20=20). It is not the range of the entire distribution.

Option (C) provides the calculation method derived from the definition of range for grouped data and the resulting value. This option demonstrates a clear understanding of how the range is obtained from the given grouped data.

The range of the distribution is calculated as the difference between the upper limit of the highest class and the lower limit of the lowest class, which is $40 - 10 = 30$.

The correct option is (C) $40 - 10 = 30$.

Let's examine each statement regarding measures of dispersion:

(A) A larger standard deviation indicates greater variability.

The standard deviation ($\sigma$) measures the average distance of each data point from the mean. A higher standard deviation means the data points are more spread out, indicating greater variability or dispersion in the data. This statement is correct.

(B) The range is easy to calculate but sensitive to outliers.

The range is the difference between the highest and lowest values in a data set. It is simple to compute. However, since it only considers the two extreme values, a single outlier (an extremely high or low value) can drastically change the range, making it sensitive to outliers. This statement is correct.

(C) Mean deviation is based on absolute values of deviations.

The mean deviation is calculated as the average of the absolute differences between each data point and the mean (or median). Using absolute values ($|x_i - \bar{x}|$) ensures that deviations above and below the mean do not cancel each other out, which would result in a mean deviation of zero for any data set if non-absolute differences were used. This statement is correct.

(D) Coefficient of variation allows comparison of variability between data sets with different units or means.

The coefficient of variation (CV) is a relative measure of dispersion, calculated as $\frac{\text{Standard Deviation}}{\text{Mean}} \times 100\%$. It is a unitless percentage. Because it is a relative measure (standard deviation relative to the mean) and unitless, it is particularly useful for comparing the variability of data sets that have different units of measurement or significantly different means, where comparing standard deviations directly would be misleading. This statement is correct.

All the given statements accurately describe properties of the respective measures of dispersion.

The correct options are (A), (B), (C), and (D).

Let a data set consist of $n$ observations: $x_1, x_2, \dots, x_n$.

The mean of the data set, denoted by $\bar{x}$, is calculated as:

$\bar{x} = \frac{\sum\limits_{i=1}^n x_i}{n}$

The deviation of an individual observation $x_i$ from the mean $\bar{x}$ is given by $(x_i - \bar{x})$.

We want to find the sum of these deviations, which is $\sum\limits_{i=1}^n (x_i - \bar{x})$.

Using the properties of summation, we can expand the sum:

$\sum\limits_{i=1}^n (x_i - \bar{x}) = \sum\limits_{i=1}^n x_i - \sum\limits_{i=1}^n \bar{x}$

The term $\sum\limits_{i=1}^n \bar{x}$ represents adding the constant value $\bar{x}$, $n$ times. So, $\sum\limits_{i=1}^n \bar{x} = n \times \bar{x}$.

Therefore, the sum of deviations becomes:

$\sum\limits_{i=1}^n (x_i - \bar{x}) = \sum\limits_{i=1}^n x_i - n\bar{x}$

From the definition of the mean, we know that $n\bar{x} = \sum\limits_{i=1}^n x_i$.

Substituting this back into the equation for the sum of deviations:

$\sum\limits_{i=1}^n (x_i - \bar{x}) = \sum\limits_{i=1}^n x_i - \sum\limits_{i=1}^n x_i$

$\sum\limits_{i=1}^n (x_i - \bar{x}) = 0$

Thus, the sum of the deviations of individual observations from their mean is always zero, regardless of the data set.

The sum of deviations from the mean is always zero.

The correct option is (C) Zero.

Let the original data set be $x_1, x_2, \dots, x_n$.

The mean of the original data set is $\bar{x} = \frac{1}{n} \sum\limits_{i=1}^n x_i$.

The variance of the original data set (for a sample) is $\sigma^2 = \frac{1}{n-1} \sum\limits_{i=1}^n (x_i - \bar{x})^2$.

The standard deviation of the original data set is $\sigma = \sqrt{\frac{1}{n-1} \sum\limits_{i=1}^n (x_i - \bar{x})^2}$.

Now, each observation is increased by 5. Let the new data set be $y_1, y_2, \dots, y_n$, where $y_i = x_i + 5$ for each $i$.

The mean of the new data set, $\bar{y}$, is:

$\bar{y} = \frac{1}{n} \sum\limits_{i=1}^n y_i = \frac{1}{n} \sum\limits_{i=1}^n (x_i + 5)$

Using the property of summation, $\sum (a_i + b) = \sum a_i + \sum b$:

$\bar{y} = \frac{1}{n} \left( \sum\limits_{i=1}^n x_i + \sum\limits_{i=1}^n 5 \right)$

$\bar{y} = \frac{1}{n} \left( \sum\limits_{i=1}^n x_i + 5n \right)$

$\bar{y} = \frac{\sum\limits_{i=1}^n x_i}{n} + \frac{5n}{n}$

$\bar{y} = \bar{x} + 5$

So, the new mean is the original mean plus 5.

Now, let's find the deviation of a new observation $y_i$ from the new mean $\bar{y}$:

$y_i - \bar{y} = (x_i + 5) - (\bar{x} + 5)$

$y_i - \bar{y} = x_i + 5 - \bar{x} - 5$

$y_i - \bar{y} = x_i - \bar{x}$

The deviation of each new observation from the new mean is exactly the same as the deviation of the corresponding original observation from the original mean.

Now let's calculate the variance of the new data set, $\sigma_{new}^2$:

$\sigma_{new}^2 = \frac{1}{n-1} \sum\limits_{i=1}^n (y_i - \bar{y})^2$

Substituting $(y_i - \bar{y}) = (x_i - \bar{x})$:

$\sigma_{new}^2 = \frac{1}{n-1} \sum\limits_{i=1}^n (x_i - \bar{x})^2$

This is the formula for the original variance, $\sigma^2$.

$\sigma_{new}^2 = \sigma^2$

Finally, the new standard deviation, $\sigma_{new}$, is the square root of the new variance:

$\sigma_{new} = \sqrt{\sigma_{new}^2} = \sqrt{\sigma^2} = \sigma$

Adding a constant value to each observation shifts the entire distribution but does not change its spread or variability. Therefore, the standard deviation remains unchanged.

The new standard deviation will be the same as the original standard deviation.

The correct option is (B) $\sigma$.

Let the original data set be $x_1, x_2, \dots, x_n$.

The mean of the original data set is $\bar{x} = \frac{1}{n} \sum\limits_{i=1}^n x_i$.

The standard deviation of the original data set (using the sample standard deviation formula) is $\sigma = \sqrt{\frac{1}{n-1} \sum\limits_{i=1}^n (x_i - \bar{x})^2}$.

Now, each observation is multiplied by 5. Let the new data set be $y_1, y_2, \dots, y_n$, where $y_i = 5 x_i$ for each $i$.

The mean of the new data set, $\bar{y}$, is:

$\bar{y} = \frac{1}{n} \sum\limits_{i=1}^n y_i = \frac{1}{n} \sum\limits_{i=1}^n (5 x_i)$

Using the property of summation $\sum c a_i = c \sum a_i$:

$\bar{y} = \frac{5}{n} \sum\limits_{i=1}^n x_i = 5 \left(\frac{1}{n} \sum\limits_{i=1}^n x_i\right) = 5 \bar{x}$

So, the new mean is 5 times the original mean.

Now, let's find the deviation of a new observation $y_i$ from the new mean $\bar{y}$:

$y_i - \bar{y} = (5 x_i) - (5 \bar{x})$

$y_i - \bar{y} = 5(x_i - \bar{x})$

Now, let's calculate the variance of the new data set, $\sigma_{new}^2$:

$\sigma_{new}^2 = \frac{1}{n-1} \sum\limits_{i=1}^n (y_i - \bar{y})^2$

Substitute $(y_i - \bar{y}) = 5(x_i - \bar{x})$:

$\sigma_{new}^2 = \frac{1}{n-1} \sum\limits_{i=1}^n (5(x_i - \bar{x}))^2$

$\sigma_{new}^2 = \frac{1}{n-1} \sum\limits_{i=1}^n 25 (x_i - \bar{x})^2$

$\sigma_{new}^2 = 25 \left(\frac{1}{n-1} \sum\limits_{i=1}^n (x_i - \bar{x})^2\right)$

The expression in the parenthesis is the original variance, $\sigma^2$.

$\sigma_{new}^2 = 25 \sigma^2$

Finally, the new standard deviation, $\sigma_{new}$, is the square root of the new variance:

$\sigma_{new} = \sqrt{25 \sigma^2}$

Using the property $\sqrt{ab} = \sqrt{a}\sqrt{b}$ and $\sqrt{c^2} = |c|$:

$\sigma_{new} = \sqrt{25} \sqrt{\sigma^2} = 5 \sigma$ (since standard deviation is always non-negative, $\sqrt{\sigma^2} = \sigma$).

In general, if each observation in a data set is multiplied by a constant $k$, the new standard deviation is $|k|$ times the original standard deviation. In this case, $k=5$, so the new standard deviation is $|5|\sigma = 5\sigma$.

The new standard deviation is $5\sigma$.

Looking at the options, both (A) and (D) state that the result is $5\sigma$. Option (D) explicitly shows the step involving the absolute value, which represents the general rule for scaling by a constant. Therefore, (D) is a more complete representation of the result derived from the property of standard deviation when data is scaled.

The correct option is (D) $|5|\sigma = 5\sigma$.

The mean deviation about a central value 'a' for a data set $x_1, x_2, \dots, x_n$ is defined as:

$MD(a) = \frac{1}{n} \sum\limits_{i=1}^n |x_i - a|$

Here, 'a' can be the mean, median, or mode of the data set.

A fundamental property in statistics states that the sum of the absolute deviations of a set of numbers from a point 'a', i.e., $\sum\limits_{i=1}^n |x_i - a|$, is minimized when 'a' is equal to the median of the data set.

Since the mean deviation is simply the sum of absolute deviations divided by the number of observations ($n$), and $n$ is a positive constant, the mean deviation $MD(a)$ will also be minimum when the sum $\sum\limits_{i=1}^n |x_i - a|$ is minimum.

This minimum sum occurs when 'a' is the median.

Therefore, the mean deviation is minimum when calculated about the median.

The mean deviation is minimum when calculated about the Median.

The correct option is (B) Median.

The statement consists of two parts:

Part 1: "For symmetrical distributions, the mean, median, and mode are equal."

For perfectly symmetrical, unimodal distributions (like the normal distribution), the mean, median, and mode all coincide. This is a well-established property of symmetrical distributions. This part of the statement is generally true.

Part 2: "In such cases, the mean deviation about the mean is equal to the mean deviation about the median."

The mean deviation about a value 'a' is calculated using the formula $MD(a) = \frac{1}{n} \sum\limits_{i=1}^n |x_i - a|$ for ungrouped data or $MD(a) = \frac{\sum\limits_{i=1}^n f_i |x_i - a|}{\sum\limits_{i=1}^n f_i}$ for grouped data.

A key property of mean deviation is that it is minimized when calculated about the median. This means that $\sum |x_i - a|$ (or $\sum f_i |x_i - a|$) is smallest when $a = \text{Median}$.

For symmetrical distributions, we know that Mean = Median.

Let $\bar{x}$ be the mean and $M$ be the median. For symmetrical distributions, $\bar{x} = M$.

The mean deviation about the mean is $MD(\bar{x}) = \frac{1}{n} \sum\limits_{i=1}^n |x_i - \bar{x}|$.

The mean deviation about the median is $MD(M) = \frac{1}{n} \sum\limits_{i=1}^n |x_i - M|$.

Since $\bar{x} = M$ in a symmetrical distribution, the formulas become identical:

$MD(\bar{x}) = \frac{1}{n} \sum\limits_{i=1}^n |x_i - \bar{x}|$

$MD(M) = \frac{1}{n} \sum\limits_{i=1}^n |x_i - \bar{x}|$ (because $M = \bar{x}$)

Therefore, $MD(\bar{x}) = MD(M)$ for a symmetrical distribution.

This also aligns with the property that MD is minimum at the median; since the mean coincides with the median, the MD about the mean is equal to the minimum possible MD, which is the MD about the median.

Both parts of the statement are true, and the conclusion logically follows from the first part and the definition/property of mean deviation.

The statement is true for symmetrical distributions where Mean = Median = Mode.

The correct option is (A) Yes.

To find the mean of a set of ungrouped data, we use the formula:

Mean $(\bar{x}) = \frac{\text{Sum of all observations}}{\text{Number of observations}}$

The given data set is: 10, 20, 30, 40, 50.

First, let's find the sum of all observations:

Sum $= 10 + 20 + 30 + 40 + 50$

Sum $= 150$

Next, count the number of observations. There are 5 observations in the data set.

Number of observations $(n) = 5$

Now, apply the formula for the mean:

$\bar{x} = \frac{150}{5}$

Performing the division:

$\bar{x} = 30$

The mean of the given data set is 30.

Comparing this result with the given options, we see that option (B) is 30.

The mean of the data is 30.

The correct option is (B) 30.

The range of a data set is defined as the difference between the maximum value and the minimum value in the data set.

Range = Maximum Value - Minimum Value

The given data set is: 5, 5, 5, 5, 5.

The maximum value in this data set is 5.

The minimum value in this data set is 5.

Now, calculate the range:

Range = $5 - 5$

Range = $0$

The range of the data set is 0.

The range of the data set is 0.

The correct option is (A) 0.

Measures of dispersion quantify the spread or variability of a data set. When a data set contains extreme values, also known as outliers, some measures of dispersion can be significantly affected, while others are more resistant.

Let's analyze how each of the given measures is affected by outliers:

(A) Range: The range is the difference between the maximum and minimum values. Since outliers are extreme values, they will directly become the maximum or minimum value, thus heavily influencing and often inflating the range. Therefore, the range is highly sensitive to outliers and is not suitable when outliers are present.

(B) Standard Deviation and (D) Variance: Variance ($\sigma^2$) and Standard Deviation ($\sigma$) are calculated based on the deviations from the mean. The mean itself is influenced by outliers. Furthermore, the calculation involves squaring the deviations, which gives more weight to larger deviations (those from outliers). This makes variance and standard deviation sensitive to outliers.

The formulas typically involve terms like $(x_i - \bar{x})^2$. An outlier $x_i$ far from the mean $\bar{x}$ will result in a large $|x_i - \bar{x}|$, and squaring this difference leads to a very large contribution to the sum of squared deviations, significantly affecting the variance and standard deviation.

(C) Mean Deviation about Median: The mean deviation about a central value is calculated as the average of the absolute deviations from that value, i.e., $\frac{1}{n}\sum |x_i - a|$. As discussed in Question 35, the mean deviation is minimized when calculated about the median. The median is a measure of central tendency that is less affected by extreme values compared to the mean. When calculating the mean deviation about the median, we use absolute differences $|x_i - \text{Median}|$ instead of squared differences. While outliers still contribute to the sum of absolute deviations, their impact is less magnified compared to squaring the deviations in variance/standard deviation calculations.

Since the median is a robust measure of central tendency (less sensitive to outliers), and the use of absolute deviations (instead of squared deviations) makes the measure less sensitive to large differences, the mean deviation about the median is more resistant to the influence of outliers compared to the range, variance, and standard deviation.

Considering the sensitivity to extreme values, the measure that is most suitable (i.e., least affected) among the given options is the Mean Deviation about the Median.

The correct option is (C) Mean Deviation about Median.

The coefficient of variation (CV) is a measure of relative variability. It is defined as the ratio of the standard deviation ($\sigma$) to the mean ($\bar{x}$), usually expressed as a percentage:

$CV = \frac{\sigma}{\bar{x}} \times 100\%$

The CV allows us to compare the variability of different data sets, even if they have different units or significantly different means. It expresses the standard deviation relative to the size of the mean.

Let's analyze what a high or low CV signifies:

• A high value of the CV indicates that the standard deviation is large relative to the mean. This means the data points are widely spread out or dispersed around the mean compared to the magnitude of the mean itself. Therefore, a high CV indicates **high variability relative to the mean**.
• A low value of the CV indicates that the standard deviation is small relative to the mean. This means the data points are clustered closely around the mean. Therefore, a low CV indicates **low variability relative to the mean** and suggests that the data is more consistent or less dispersed.

Based on this interpretation, if the coefficient of variation is high, it means there is high variability in the data relative to its mean.

Let's evaluate the options:

(A) High variability relative to the mean: This aligns with the definition and interpretation of a high CV.

(B) Low variability relative to the mean: This is incorrect; low variability relative to the mean is indicated by a low CV.

(C) The data is consistent: Consistency implies low variability, which is indicated by a low CV, not a high one.

(D) The mean is very large: The size of the mean alone does not determine if the CV is high. A large mean combined with a proportionally even larger standard deviation would result in a high CV. A large mean combined with a relatively small standard deviation results in a low CV.

A high coefficient of variation indicates high variability relative to the mean.

The correct option is (A) High variability relative to the mean.

Let's consider the definitions of the measures of dispersion listed in the options:

(A) Standard Deviation: The standard deviation ($\sigma$) is calculated using the square root of the variance. The variance is the average of the squared deviations from the mean, i.e., $\frac{1}{n-1}\sum (x_i - \bar{x})^2$ (for a sample). The calculation explicitly involves the deviations $(x_i - \bar{x})$ and their magnitudes (by squaring). A larger deviation $(x_i - \bar{x})$ results in a larger contribution to the sum of squares, reflecting its magnitude.

(B) Mean Deviation: The mean deviation about a central value (mean or median) is calculated as the average of the absolute deviations from that value, i.e., $\frac{1}{n}\sum |x_i - a|$. The calculation explicitly involves the absolute deviations $|x_i - a|$, which are the magnitudes of the deviations of each data point from the central value. A larger absolute deviation $|x_i - a|$ contributes more to the sum, reflecting its magnitude.

(D) Range: The range is calculated as the difference between the maximum and minimum values in the data set. While it indicates the overall spread, it does not take into account the deviations of all individual data points from a central value or from each other. It only uses the two extreme values.

Both Standard Deviation and Mean Deviation are calculated by considering the deviations of each data point from a central value. Standard Deviation uses the square of the deviations, and Mean Deviation uses the absolute value of the deviations. In both cases, the magnitude of these deviations directly influences the calculated value of the measure.

The question asks for the measure of dispersion that takes into account the magnitude of deviations. Both "Standard deviation" and "Mean deviation" fit this description.

Therefore, the most appropriate option is the one that includes both.

Both Standard Deviation and Mean Deviation are calculated using deviations from a central value, and their values depend on the magnitudes of these deviations.

The correct option is (C) Both Standard and Mean.

Let's examine each option to identify the formula for standard deviation.

(A) $\sqrt{\frac{1}{n} \sum\limits_{i=1}^n (x_i - \bar{x})^2}$

This formula calculates the square root of the average of the squared deviations of each observation ($x_i$) from the mean ($\bar{x}$). This is the standard formula for the population standard deviation, where $n$ is the size of the population. It is a measure of the typical distance of data points from the mean.

(B) $\frac{1}{n} \sum\limits_{i=1}^n |x_i - \bar{x}|$

This formula calculates the average of the absolute deviations of each observation ($x_i$) from the mean ($\bar{x}$). This is the formula for the mean deviation about the mean.

(C) $(\frac{1}{n} \sum\limits_{i=1}^n (x_i - \bar{x})^2)^2$

The expression inside the parenthesis, $\frac{1}{n} \sum\limits_{i=1}^n (x_i - \bar{x})^2$, is the formula for the population variance ($\sigma^2$). Squaring the variance gives $(\sigma^2)^2 = \sigma^4$, which is not standard deviation.

(D) $\text{Highest value} - \text{Lowest value}$

This formula calculates the difference between the maximum and minimum values in a data set. This is the formula for the range.

Comparing the options with the standard statistical formulas, option (A) correctly represents the formula for standard deviation (specifically, the population standard deviation).

The correct option is (A) $\sqrt{\frac{1}{n} \sum\limits_{i=1}^n (x_i - \bar{x})^2}$.

The standard deviation of a data set is the square root of its variance.

The relationship between standard deviation ($\sigma$) and variance ($\sigma^2$) is given by:

$\sigma = \sqrt{\text{Variance}}$

Given that the variance of the data set is $25$.

Variance = $25$

To find the standard deviation, we take the square root of the variance:

$\sigma = \sqrt{25}$

The square root operation $\sqrt{25}$ yields a principal (non-negative) root. In statistics, the standard deviation is always a non-negative value representing the spread. Therefore, the standard deviation is the positive square root of the variance.

$\sigma = 5$

So, the standard deviation is $5$.

Let's examine the given options:

(A) $5$: This is the calculated value of the standard deviation.

(B) $\sqrt{25}$: This is the expression representing the calculation of the standard deviation from the variance.

(C) Both (A) and (B): This option suggests that both the value $5$ and the expression $\sqrt{25}$ are correct representations of the standard deviation in this context. Since $\sqrt{25} = 5$, both are indeed correct ways to state the standard deviation.

(D) $\pm 5$: This is incorrect because standard deviation must be non-negative.

Given that option (C) includes both (A) and (B), and both $5$ and $\sqrt{25}$ are valid representations of the standard deviation (with $5$ being the simplified value), option (C) is the most comprehensive correct answer.

The standard deviation is $5$, which is also $\sqrt{25}$.

The correct option is (C) Both (A) and (B).

Let the data set be $x_1, x_2, \dots, x_n$. We are considering the sum of absolute deviations from a value 'a', given by $\sum\limits_{i=1}^n |x_i - a|$.

There is a mathematical property that states the sum of absolute deviations, $\sum\limits_{i=1}^n |x_i - a|$, is minimized when 'a' is the median of the data set.

The statement in the question claims that this sum is minimum when calculated about the mean ($\bar{x}$). This is incorrect.

It is important to distinguish this from another related property: the sum of squared deviations, $\sum\limits_{i=1}^n (x_i - a)^2$, is minimized when 'a' is the mean ($\bar{x}$) of the data set.

The question specifically asks about the sum of absolute deviations.

Based on the property that the sum of absolute deviations is minimized about the median, the statement "The sum of absolute deviations from the mean is minimum compared to the sum of absolute deviations from any other value" is false.

Let's look at the options:

(A) Yes, the sum of absolute deviations is minimum about the mean. - This is false.

(B) No, the sum of absolute deviations is minimum about the median. - This correctly states that the sum of absolute deviations is minimized about the median, thereby indicating that the original statement (about the mean) is false.

(C) Yes, the sum of squared deviations is minimum about the mean. - This statement is true, but it answers a different question (about squared deviations, not absolute deviations).

(D) No, the sum of squared deviations is minimum about the median. - This statement is false; the sum of squared deviations is minimized about the mean.

The sum of absolute deviations is minimum when calculated about the median, not the mean.

The correct option is (B) No, the sum of absolute deviations is minimum about the median.

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

Assertion (A): Coefficient of variation is a relative measure of dispersion.

Measures of dispersion can be absolute (like range, standard deviation, variance) or relative (like coefficient of variation). Absolute measures are expressed in the same units as the data, while relative measures are dimensionless or expressed as a percentage. Relative measures are useful for comparing variability between data sets that have different means or are measured in different units. The coefficient of variation (CV) serves this purpose by providing a measure of variability relative to the mean.

Thus, Assertion (A) is True.

Reason (R): It expresses the standard deviation as a percentage of the mean.

The formula for the coefficient of variation is given by:

$CV = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100\%$

This formula shows that the coefficient of variation is indeed calculated by taking the ratio of the standard deviation to the mean and multiplying by 100% to express it as a percentage. This is the standard definition of the coefficient of variation.

Thus, Reason (R) is True.

Now, let's determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that CV is a relative measure. Reason (R) states how CV is calculated: standard deviation as a percentage of the mean. The calculation described in Reason (R) inherently makes the CV a relative measure because it normalizes the standard deviation (an absolute measure of spread) by dividing it by the mean (a measure of the location or magnitude of the data). This process removes the units and expresses the spread relative to the average value of the data, which is the definition of a relative measure of dispersion.

Therefore, Reason (R) correctly explains why Coefficient of Variation is a relative measure of dispersion.

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

To find the coefficient of variation, we use the formula:

Coefficient of Variation (CV) $= \frac{\text{Standard Deviation}}{\text{Mean}} \times 100\% $

Given:

Mean $(\bar{x}) = 10$

Standard Deviation $(\sigma) = 2.5$

Substitute these values into the formula:

$CV = \frac{2.5}{10} \times 100\%$

Now, perform the calculation:

$CV = 0.25 \times 100\%$

$CV = 25\%$

Thus, the coefficient of variation is $25\%$.

Compare this result with the given options.

The coefficient of variation is $25\%$.

The correct option is (B) 25%.

To find the mean deviation about the mean for ungrouped data, we follow these steps:

Step 1: Calculate the mean ($\bar{x}$) of the data.

The data set is: $x_1 = 10, x_2 = 20, x_3 = 30, x_4 = 40, x_5 = 50$.

The number of observations is $n = 5$.

Mean ($\bar{x}$) $= \frac{\sum\limits_{i=1}^n x_i}{n} = \frac{10 + 20 + 30 + 40 + 50}{5}$

$\bar{x} = \frac{150}{5}$

$\bar{x} = 30$

The mean of the data is $30$.

Step 2: Calculate the absolute deviation of each observation from the mean, $|x_i - \bar{x}|$.

$|x_1 - \bar{x}| = |10 - 30| = |-20| = 20$

$|x_2 - \bar{x}| = |20 - 30| = |-10| = 10$

$|x_3 - \bar{x}| = |30 - 30| = |0| = 0$

$|x_4 - \bar{x}| = |40 - 30| = |10| = 10$

$|x_5 - \bar{x}| = |50 - 30| = |20| = 20$

Step 3: Calculate the sum of the absolute deviations, $\sum\limits_{i=1}^n |x_i - \bar{x}|$.

$\sum\limits_{i=1}^n |x_i - \bar{x}| = 20 + 10 + 0 + 10 + 20$

$\sum\limits_{i=1}^n |x_i - \bar{x}| = 60$

Step 4: Calculate the mean deviation about the mean ($MD(\bar{x})$).

The formula for mean deviation about the mean for ungrouped data is:

$MD(\bar{x}) = \frac{\sum\limits_{i=1}^n |x_i - \bar{x}|}{n}$

Substitute the sum of absolute deviations and the number of observations:

$MD(\bar{x}) = \frac{60}{5}$

$MD(\bar{x}) = 12$

The mean deviation about the mean is $12$.

The mean deviation about the mean for the given data is $12$.

Comparing this result with the given options, we see that option (B) is 12.

The correct option is (B) 12.

Let the data set be $x_1, x_2, \dots, x_n$. The mean of the data is $\bar{x}$.

The deviation of an observation $x_i$ from the mean is $(x_i - \bar{x})$.

The square of the deviation is $(x_i - \bar{x})^2$.

The sum of the squares of the deviations from the mean is $\sum\limits_{i=1}^n (x_i - \bar{x})^2$.

A fundamental property in statistics states that the sum of the squares of the deviations of a set of numbers from a value 'a', given by $\sum\limits_{i=1}^n (x_i - a)^2$, is minimized when 'a' is equal to the mean ($\bar{x}$) of the data set.

This means that for any value $a \neq \bar{x}$, the sum of squared deviations from 'a' is greater than the sum of squared deviations from the mean:

The equality holds only when $a = \bar{x}$.

Therefore, the sum of the squares of the deviations from the mean is the smallest possible value for the sum of squared deviations from any point. In other words, it is the minimum.

Let's consider the options:

(A) 0: The sum of squared deviations is 0 only if all observations are equal to the mean (i.e., all observations are the same), which is not generally true for any data set.

(B) Minimum: As explained above, the sum of squared deviations is minimized when calculated about the mean.

(C) Maximum: This is incorrect; the sum of squared deviations has a minimum value, not a maximum, at the mean.

(D) Equal to the variance: The population variance is $\sigma^2 = \frac{1}{n} \sum\limits_{i=1}^n (x_i - \bar{x})^2$. Thus, the sum of squared deviations is $n \times \sigma^2$. The sample variance is $s^2 = \frac{1}{n-1} \sum\limits_{i=1}^n (x_i - \bar{x})^2$. Thus, the sum of squared deviations is $(n-1) \times s^2$. The sum of squared deviations is not equal to the variance itself unless $n=1$ (for population variance) or $n-1=1$ (for sample variance), which are trivial or special cases.

The sum of the squares of the deviations from the mean is always the minimum possible sum of squared deviations from any value.

The correct option is (B) Minimum.

The coefficient of variation (CV) is a measure of relative variability. Its formula relates an absolute measure of dispersion to the mean.

The standard formula for the coefficient of variation is:

$CV = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100\%$

From this formula, we can see that the measure of dispersion used in the numerator is the Standard Deviation.

Let's briefly look at the other options:

• Range: The range is the difference between the maximum and minimum values. It is not used in the calculation of the standard CV.
• Mean Deviation: Mean deviation is the average of absolute deviations from a central value (usually the mean or median). It is not used in the standard CV formula.
• Variance: Variance is the square of the standard deviation ($\sigma^2$). While standard deviation is derived from variance ($\sigma = \sqrt{\sigma^2}$), the CV formula explicitly uses standard deviation ($\sigma$), not variance ($\sigma^2$), in the numerator.

The coefficient of variation is calculated using the standard deviation divided by the mean.

The correct option is (C) Standard Deviation.

To find the median of ungrouped data, we first need to arrange the data in ascending or descending order.

The given data set is: 1, 5, 8, 3, 10, 6, 2.

Arranging the data in ascending order:

1, 2, 3, 5, 6, 8, 10

Next, we need to determine the number of observations in the data set.

Number of observations ($n$) = 7.

Since the number of observations ($n=7$) is an odd number, the median is the value of the $\left(\frac{n+1}{2}\right)$-th observation in the ordered data.

Position of the median = $\left(\frac{7+1}{2}\right)$-th observation = $\left(\frac{8}{2}\right)$-th observation = 4th observation.

Now, we locate the 4th observation in the ordered data set (1, 2, 3, 5, 6, 8, 10).

The 1st observation is 1.

The 2nd observation is 2.

The 3rd observation is 3.

The 4th observation is 5.

So, the median of the data set is 5.

The median of the data set is 5.

Comparing this result with the given options, we see that option (B) is 5.

The correct option is (B) 5.

To calculate the mean runs scored by Batsman A, we need to find the sum of the runs scored in all matches and divide by the number of matches.

The runs scored by Batsman A in 5 matches are: 25, 50, 45, 60, 30.

Step 1: Sum the runs scored by Batsman A.

Sum of runs for Batsman A = $25 + 50 + 45 + 60 + 30$

Sum of runs = $15 + 50 + 45 + 60$ (adding 25 and 30 first)

Sum of runs = $75 + 45 + 60$

Sum of runs = $120 + 60$

Sum of runs = $180$

Step 2: Determine the number of matches.

Number of matches played by Batsman A = $5$.

Step 3: Calculate the mean runs using the formula:

Mean $(\bar{x}_A) = \frac{\text{Sum of runs}}{\text{Number of matches}}$

$\bar{x}_A = \frac{180}{5}$

Performing the division:

$\bar{x}_A = 36$

Wait, let me recheck the sum.

Sum $= 25 + 50 + 45 + 60 + 30$

Sum $= 75 + 45 + 60 + 30$ (Adding 25 and 50)

Sum $= 120 + 60 + 30$ (Adding 75 and 45)

Sum $= 180 + 30$ (Adding 120 and 60)

Sum $= 210$

Okay, the sum is 210.

Step 3 (Corrected): Calculate the mean runs using the formula:

Mean $(\bar{x}_A) = \frac{\text{Sum of runs}}{\text{Number of matches}}$

$\bar{x}_A = \frac{210}{5}$

Performing the division:

$\bar{x}_A = 42$

The mean runs scored by Batsman A is 42.

The mean runs scored by Batsman A is 42.

Comparing this result with the given options, we see that option (B) is 42.

The correct option is (B) 42.

To calculate the mean runs scored by Batsman B, we need to find the sum of the runs scored in all matches and divide by the number of matches.

The runs scored by Batsman B in 5 matches are: 40, 45, 55, 50, 60.

Step 1: Sum the runs scored by Batsman B.

Sum of runs for Batsman B = $40 + 45 + 55 + 50 + 60$

Sum of runs = $85 + 55 + 50 + 60$ (Adding 40 and 45)

Sum of runs = $140 + 50 + 60$ (Adding 85 and 55)

Sum of runs = $190 + 60$ (Adding 140 and 50)

Sum of runs = $250$

Step 2: Determine the number of matches.

Number of matches played by Batsman B = $5$.

Step 3: Calculate the mean runs using the formula:

Mean $(\bar{x}_B) = \frac{\text{Sum of runs}}{\text{Number of matches}}$

$\bar{x}_B = \frac{250}{5}$

Performing the division:

$\bar{x}_B = 50$

The mean runs scored by Batsman B is 50.

The mean runs scored by Batsman B is 50.

Comparing this result with the given options, we see that option (B) is 50.

The correct option is (B) 50.

To calculate the standard deviation, we first need the mean of the data, which we calculated in Question 51.

The runs scored by Batsman A are: 25, 50, 45, 60, 30.

Number of matches ($n$) = 5.

Mean runs for Batsman A ($\bar{x}_A$) = 42.

We will calculate the sample standard deviation, which is typically used for data sets representing a sample from a larger population (in this case, these 5 matches are a sample of the batsman's performance).

The formula for the sample standard deviation ($s$) is:

$s = \sqrt{\frac{\sum\limits_{i=1}^n (x_i - \bar{x})^2}{n-1}}$

Step 1: Calculate the deviation of each score from the mean ($x_i - \bar{x}_A$).

$25 - 42 = -17$

$50 - 42 = 8$

$45 - 42 = 3$

$60 - 42 = 18$

$30 - 42 = -12$

Step 2: Square each deviation $(x_i - \bar{x}_A)^2$.

$(-17)^2 = 289$

$(8)^2 = 64$

$(3)^2 = 9$

$(18)^2 = 324$

$(-12)^2 = 144$

Step 3: Sum the squared deviations $\sum\limits_{i=1}^n (x_i - \bar{x}_A)^2$.

Sum of squared deviations = $289 + 64 + 9 + 324 + 144 = 830$

Step 4: Divide the sum of squared deviations by $(n-1)$. This gives the variance ($s^2$).

Variance ($s^2$) $= \frac{830}{5 - 1} = \frac{830}{4} = 207.5$

Step 5: Take the square root of the variance to find the standard deviation ($s$).

$s_A = \sqrt{207.5}$

Now, we need to approximate the value of $\sqrt{207.5}$.

$14^2 = 196$

$15^2 = 225$

So, $\sqrt{207.5}$ is between 14 and 15.

Let's check the options:

(A) $10.1^2 \approx 102.01$ (Too low)

(B) $12.2^2 \approx 148.84$ (Too low)

(C) $13.4^2 \approx 179.56$ (Too low)

(D) $14.5^2 \approx 210.25$ (Closest to 207.5)

Calculating precisely:

$\sqrt{207.5} \approx 14.4048...$

Rounding to one decimal place, we get approximately $14.4$. The closest option is $14.5$.

The standard deviation of runs scored by Batsman A is approximately $14.4$. The closest option is $14.5$.

The correct option is (D) 14.5.

To calculate the standard deviation, we first need the mean of the data, which we calculated in Question 52.

The runs scored by Batsman B are: 40, 45, 55, 50, 60.

Number of matches ($n$) = 5.

Mean runs for Batsman B ($\bar{x}_B$) = 50.

We will calculate the standard deviation. For a small dataset like this, the sample standard deviation formula ($s$) is often used, which divides by $n-1$. However, sometimes the population standard deviation formula ($\sigma$), which divides by $n$, is used, especially in textbook examples or if the dataset is considered the entire population of interest. Given the options provided, it appears the population standard deviation formula was likely used.

The formula for the population standard deviation ($\sigma$) is:

$\sigma = \sqrt{\frac{\sum\limits_{i=1}^n (x_i - \bar{x})^2}{n}}$

Step 1: Calculate the deviation of each score from the mean ($x_i - \bar{x}_B$).

$40 - 50 = -10$

$45 - 50 = -5$

$55 - 50 = 5$

$50 - 50 = 0$

$60 - 50 = 10$

Step 2: Square each deviation $(x_i - \bar{x}_B)^2$.

$(-10)^2 = 100$

$(-5)^2 = 25$

$(5)^2 = 25$

$(0)^2 = 0$

$(10)^2 = 100$

Step 3: Sum the squared deviations $\sum\limits_{i=1}^n (x_i - \bar{x}_B)^2$.

Sum of squared deviations = $100 + 25 + 25 + 0 + 100 = 250$

Step 4: Calculate the population variance ($\sigma_B^2$).

Population Variance ($\sigma_B^2$) $= \frac{\sum (x_i - \bar{x}_B)^2}{n} = \frac{250}{5} = 50$

Step 5: Take the square root of the population variance to find the population standard deviation ($\sigma_B$).

$\sigma_B = \sqrt{50}$

Now, we approximate the value of $\sqrt{50}$.

$7^2 = 49$ and $8^2 = 64$. So, $\sqrt{50}$ is between 7 and 8, and closer to 7.

Let's check the options by squaring them:

(A) $6.3^2 = 39.69$

(B) $7.1^2 = 50.41$

(C) $8.9^2 = 79.21$

(D) $9.5^2 = 90.25$

The value closest to 50 is $50.41$, which is $7.1^2$.

Calculating $\sqrt{50}$ precisely gives approximately 7.071. The closest option to 7.071 is 7.1.

(Note: If we were to calculate the sample standard deviation $s_B = \sqrt{\frac{250}{5-1}} = \sqrt{\frac{250}{4}} = \sqrt{62.5} \approx 7.906$. Comparing 7.906 to the options, 7.1 (difference 0.806) and 8.9 (difference 0.994) are the closest, but 7.1 is slightly closer. However, 7.1 is significantly closer to the population standard deviation $\sqrt{50} \approx 7.07$. This confirms the options are likely based on the population standard deviation.)

The standard deviation of runs scored by Batsman B, calculated using the population formula, is $\sqrt{50} \approx 7.07$. The closest option is 7.1.

The correct option is (B) 7.1.

In statistics, consistency of a dataset (like batting scores) is measured by its variability. A batsman is considered more consistent if their scores show less variation from their average score.

The standard deviation ($\sigma$) is a common measure of the spread or dispersion of a set of values. A lower standard deviation generally indicates less variability.

However, comparing the consistency of two datasets using only their standard deviations can be misleading if their means (average scores) are significantly different.

For example, a standard deviation of 10 on an average score of 50 represents a larger relative variation than a standard deviation of 10 on an average score of 100.

The Coefficient of Variation (CV) is a standardized measure of relative variability. It is calculated as the ratio of the standard deviation to the mean:

$CV = \frac{\sigma}{\mu} \times 100\%$

where $\sigma$ is the standard deviation and $\mu$ is the mean.

The Coefficient of Variation allows for a better comparison of the degree of variation between datasets, even if their means are vastly different. A lower Coefficient of Variation indicates less relative variability and thus greater consistency.

The question asks who is a more consistent batsman and option (D) specifically points out that consistency cannot be determined from standard deviation alone and the Coefficient of Variation is needed. This suggests that the means of the scores for Batsman A and Batsman B are likely different, making the Coefficient of Variation the appropriate measure for comparing consistency.

Without the actual data (mean and standard deviation for both batsmen) from the preceding Data Interpretation section, or at least the calculated Coefficient of Variation for each, we cannot definitively determine which batsman is more consistent by merely comparing standard deviations (if they were provided). The comparison requires calculating and comparing their Coefficient of Variations.

Based on the structure of the options, especially option (D), the problem intends to highlight the importance of the Coefficient of Variation for comparing consistency when the means might differ. Since the data is missing, the determination is not possible with just standard deviation.

Therefore, consistency cannot be determined from standard deviation alone; the Coefficient of Variation is needed.

The correct option is (D) Cannot be determined from standard deviation alone; need Coefficient of Variation.

Measures of dispersion quantify the spread or variability of a dataset. They can be classified into two types:

1. Absolute Measures of Dispersion: These measures are expressed in the same units as the original data. They describe the actual amount of variation. Examples include Range, Quartile Deviation, Mean Deviation, Variance, and Standard Deviation.

2. Relative Measures of Dispersion: These measures are unitless and are used for comparing the variability of two or more datasets, especially when they have different units or significantly different means. They are usually expressed as ratios or percentages. Examples include Coefficient of Range, Coefficient of Quartile Deviation, Coefficient of Mean Deviation, and Coefficient of Variation.

Let's examine the given options:

(A) Coefficient of Variation: This is defined as the ratio of the standard deviation to the mean, usually expressed as a percentage. It is a relative measure of dispersion.

(B) Z-score: A Z-score (or standard score) measures how many standard deviations a data point is from the mean. It is a standardized score and is unitless, but it is not typically classified as an overall measure of dispersion for the entire dataset itself; rather, it describes the position of a single value relative to the mean in terms of standard deviation units.

(C) Standard Deviation: This is the square root of the variance and measures the average distance of data points from the mean. It is expressed in the same units as the original data. Therefore, it is an absolute measure of dispersion.

(D) All of the above: This is incorrect as Coefficient of Variation and Z-score are not absolute measures of dispersion.

Based on the definitions, Standard Deviation is an absolute measure of dispersion.

The correct option is (C) Standard Deviation.

The given formula is for the variance ($\sigma^2$) of grouped data:

$\sigma^2 = \frac{\sum\limits_{i=1}^n f_i (x_i - \bar{x})^2}{\sum\limits_{i=1}^n f_i}$

In this formula, the symbols represent the following:

$\sigma^2$ represents the variance of the grouped data.

$\sum\limits_{i=1}^n$ indicates the sum over all classes, from the first class ($i=1$) to the last class ($i=n$).

$x_i$ represents the class mark (or midpoint) of the $i^{th}$ class.

$\bar{x}$ represents the mean of the grouped data.

$(x_i - \bar{x})$ represents the deviation of the class mark of the $i^{th}$ class from the mean.

$(x_i - \bar{x})^2$ represents the squared deviation of the class mark of the $i^{th}$ class from the mean.

$f_i$ represents the frequency of the $i^{th}$ class. This is the number of observations that fall within the range of the $i^{th}$ class.

$f_i (x_i - \bar{x})^2$ represents the squared deviation of the $i^{th}$ class mark multiplied by its frequency.

$\sum\limits_{i=1}^n f_i (x_i - \bar{x})^2$ represents the sum of the squared deviations of all class marks from the mean, weighted by their respective frequencies.

$\sum\limits_{i=1}^n f_i$ represents the sum of all frequencies, which is the total number of observations in the dataset.

The question asks what $f_i$ represents. Based on the standard notation for grouped data formulas in statistics, $f_i$ stands for the frequency of the $i^{th}$ class.

Let's check the given options:

(A) Class mark of the $i^{th}$ class is represented by $x_i$.

(B) Frequency of the $i^{th}$ class is represented by $f_i$. This matches our identification.

(C) Number of observations is the total frequency, $\sum f_i$.

(D) Mean of the data is represented by $\bar{x}$.

Therefore, $f_i$ represents the frequency of the $i^{th}$ class.

The correct option is (B) Frequency of the $i^{th}$ class.

The variance ($\sigma^2$ for population, $s^2$ for sample) is a measure of the spread or dispersion of a set of data points around their mean. It is calculated as the average of the squared differences from the mean.

For a population, the variance is given by the formula:

$\sigma^2 = \frac{\sum (x_i - \mu)^2}{N}$

where $x_i$ are the individual data points, $\mu$ is the population mean, and $N$ is the number of observations.

We are given that the variance of the data set is 0.

So, $\sigma^2 = 0$.

This means $\frac{\sum (x_i - \mu)^2}{N} = 0$.

For this fraction to be zero, the numerator must be zero, assuming $N > 0$.

$\sum (x_i - \mu)^2 = 0$

The term $(x_i - \mu)^2$ is the squared difference between each observation $x_i$ and the mean $\mu$. Since it is a squared term, $(x_i - \mu)^2 \geq 0$ for all $i$.

The sum of non-negative terms can only be zero if and only if each individual term is zero.

Therefore, $(x_i - \mu)^2 = 0$ for all $i = 1, 2, ..., N$.

Taking the square root of both sides, we get $x_i - \mu = 0$ for all $i$.

This implies $x_i = \mu$ for all $i = 1, 2, ..., N$.

So, if the variance of a data set is 0, it means that every observation in the data set is equal to the mean. In other words, all data points are identical.

Let's evaluate the options based on this finding:

(A) All observations are equal to the mean. This directly follows from our derivation.

(B) The range is very small. If all observations are equal ($x_i = \mu$), then the minimum value is $\mu$ and the maximum value is $\mu$. The range is Max - Min = $\mu - \mu = 0$. A range of 0 is the smallest possible range, not just "very small". Option (A) provides the fundamental reason for the range being 0.

(C) The mean deviation is very small. The mean deviation is $\frac{\sum |x_i - \mu|}{N}$. If $x_i = \mu$ for all $i$, then $|x_i - \mu| = |\mu - \mu| = 0$ for all $i$. So, the mean deviation is $\frac{\sum 0}{N} = 0$. A mean deviation of 0 is the smallest possible, not just "very small". Again, option (A) is the fundamental reason.

(D) There is high variability. A variance of 0 indicates no variability, as all data points are the same. This option is incorrect.

Option (A) is the most precise and correct statement describing the situation when the variance of a data set is 0.

The correct option is (A) All observations are equal to the mean.

The Coefficient of Variation (CV) is a measure of relative variability. It is defined as the ratio of the standard deviation ($\sigma$ or $s$) to the mean ($\mu$ or $\bar{x}$), usually expressed as a percentage:

$CV = \frac{\sigma}{\mu} \times 100\%$ (for population data)

or

$CV = \frac{s}{\bar{x}} \times 100\%$ (for sample data)

Since the mean and standard deviation are in the same units, the units cancel out, making the Coefficient of Variation a unitless measure. This is its primary advantage for comparing variability across different datasets.

Let's consider the utility of CV in the scenarios given in the options:

(A) Data sets with the same mean: If two datasets have the same mean ($\mu_1 = \mu_2$), then $CV_1 = \frac{\sigma_1}{\mu_1} \times 100\%$ and $CV_2 = \frac{\sigma_2}{\mu_2} \times 100\%$. Since $\mu_1 = \mu_2$, comparing $CV_1$ and $CV_2$ is equivalent to comparing $\sigma_1$ and $\sigma_2$. In this specific case, comparing standard deviations is sufficient to compare relative variability. However, CV still provides a relative measure and is therefore useful.

(B) Data sets with the same standard deviation: If two datasets have the same standard deviation ($\sigma_1 = \sigma_2$) but different means ($\mu_1 \neq \mu_2$), then $CV_1 = \frac{\sigma_1}{\mu_1} \times 100\%$ and $CV_2 = \frac{\sigma_2}{\mu_2} \times 100\%$. Since $\sigma_1 = \sigma_2$, the CV is inversely proportional to the mean. The dataset with the larger mean will have a smaller CV, indicating less relative variability despite having the same absolute standard deviation. Comparing standard deviations alone would be misleading regarding relative consistency. CV is very useful here.

(C) Data sets measured in different units: If two datasets are measured in entirely different units (e.g., height in cm and weight in kg), their standard deviations cannot be directly compared because they have different units. Since the Coefficient of Variation is unitless, it allows for a direct comparison of the relative variability between such datasets. This is a key scenario where CV is indispensable.

Based on the analysis, the Coefficient of Variation is particularly useful when comparing variability of datasets with different means or datasets measured in different units. It provides a standardized, relative measure of dispersion that is valid for comparison regardless of the scale or units of the original data. Therefore, it is useful in all the scenarios listed.

The correct option is (D) All of the above.

Measures of dispersion quantify the spread or variability of a dataset. Some measures of dispersion are based on the deviations of individual data points from a central value. The central value used can be the mean, median, or mode.

Let's consider the given options:

(A) Range: The Range is the difference between the maximum and minimum values in a dataset ($Range = Maximum - Minimum$). This measure uses the two extreme values as points of reference, not the mean.

(B) Mean Deviation about Mean: The Mean Deviation about Mean is calculated as the average of the absolute deviations of each observation from the mean ($\bar{x}$). The formula is:

$\text{MD}_{\bar{x}} = \frac{\sum |x_i - \bar{x}|}{N}$ (for ungrouped data)

or

$\text{MD}_{\bar{x}} = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$ (for grouped data)

This measure explicitly uses the mean ($\bar{x}$) as the reference point from which deviations are calculated.

(C) Standard Deviation: The Standard Deviation ($\sigma$ or $s$) is the square root of the variance. Variance is calculated as the average of the squared deviations of each observation from the mean ($\bar{x}$). The formula for sample standard deviation is:

$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$ (for ungrouped data)

or

$\sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{N}}$ (for population data)

This measure explicitly uses the mean ($\bar{x}$ or $\mu$) as the reference point from which deviations are calculated and squared.

Both the Mean Deviation about Mean and the Standard Deviation are calculated based on deviations from the mean. Therefore, they both use the mean as their point of reference.

Option (D) states "Both (B) and (C)". Since both Mean Deviation about Mean and Standard Deviation use the mean as the point of reference, this option is correct.

The correct option is (D) Both (B) and (C).

The Range is the simplest measure of dispersion. It is defined as the difference between the highest (maximum) and lowest (minimum) values in a data set.

The given data set is: 5, 12, 18, 6, 25, 3, 15, 20.

To find the range, we first need to identify the maximum and minimum values in the data set.

Arranging the data in ascending order (optional but helpful): 3, 5, 6, 12, 15, 18, 20, 25.

The maximum value in the data set is 25.

The minimum value in the data set is 3.

The formula for the Range is:

Range = Maximum Value - Minimum Value

Calculating the range for the given data set:

Range = 25 - 3

Range = 22

Comparing the calculated range with the given options:

(A) 22

(B) 20

(C) 18

(D) 25

The calculated range is 22, which matches option (A).

The correct option is (A) 22.

The Standard Deviation is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean (and to each other), while a high standard deviation indicates that the values are spread out over a wider range.

The given data set is: 7, 7, 7, 7.

In this data set, all the observations are identical.

When all values in a data set are the same, there is absolutely no variation or dispersion among the values.

Let's calculate the mean ($\bar{x}$) of the data set:

$\bar{x} = \frac{7 + 7 + 7 + 7}{4} = \frac{28}{4} = 7$

The standard deviation is calculated based on the deviations of each data point from the mean.

The deviations from the mean for each data point are:

$7 - 7 = 0$

$7 - 7 = 0$

$7 - 7 = 0$

$7 - 7 = 0$

The squared deviations are $(0)^2 = 0$ for all data points.

The sum of squared deviations is $\sum (x_i - \bar{x})^2 = 0 + 0 + 0 + 0 = 0$.

The variance ($\sigma^2$ or $s^2$) is the average of the squared deviations. Whether we divide by $N$ or $n-1$, if the sum of squared deviations is 0, the variance is 0.

Variance $= \frac{\sum (x_i - \bar{x})^2}{\text{denominator}} = \frac{0}{\text{denominator}} = 0$.

The standard deviation ($\sigma$ or $s$) is the square root of the variance.

Standard Deviation $= \sqrt{\text{Variance}} = \sqrt{0} = 0$.

When all observations in a data set are identical, the standard deviation is always 0, indicating no dispersion.

Let's check the given options:

(A) 0

(B) 1

(C) 7

(D) Undefined

Our calculated standard deviation is 0, which matches option (A). The standard deviation is not undefined unless the denominator is zero, which is not the case here as we have more than one data point.

The correct option is (A) 0.

Given:

Coefficient of Variation (CV) = 10%

Standard Deviation ($\sigma$ or $s$) = 5

To Find:

The Mean ($\mu$ or $\bar{x}$)

Solution:

The formula for the Coefficient of Variation (CV) is given by:

$CV = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100\%$

Using symbols, this can be written as:

$CV = \frac{\sigma}{\mu} \times 100\%$ (for population data)

or

$CV = \frac{s}{\bar{x}} \times 100\%$ (for sample data)

We are given CV = 10% and the standard deviation is 5. Let the mean be $\bar{x}$.

Substituting the given values into the formula:

$10\% = \frac{5}{\bar{x}} \times 100\%$

To solve for $\bar{x}$, we can remove the percentage signs from both sides, or convert 10% to a decimal (0.10). Let's remove the percentage signs for simplicity in this setup:

$10 = \frac{5}{\bar{x}} \times 100$

Now, we rearrange the equation to isolate $\bar{x}$:

Multiply both sides by $\bar{x}$:

$10 \times \bar{x} = 5 \times 100$

$10 \bar{x} = 500$

Divide both sides by 10:

$\bar{x} = \frac{500}{10}$

$\bar{x} = 50$

So, the mean of the data set is 50.

Let's check the given options:

(A) 0.5

(B) 5

(C) 50

(D) 100

Our calculated mean is 50, which matches option (C).

The correct option is (C) 50.

Measures of dispersion are used to quantify the spread or variability of a dataset. They can be classified into two main types:

1. Absolute Measures of Dispersion: These measures are expressed in the same units as the original data. They provide the actual amount of variation. Examples include Range, Quartile Deviation, Mean Deviation, Variance, and Standard Deviation.

2. Relative Measures of Dispersion: These measures are dimensionless (unitless) and are typically expressed as a ratio or percentage. They are used to compare the variability of two or more datasets, especially when their means or units are different. Examples include Coefficient of Range, Coefficient of Quartile Deviation, Coefficient of Mean Deviation, and Coefficient of Variation.

Let's examine the given options:

(A) Range: Calculated as Maximum value - Minimum value. It is an absolute measure, expressed in the units of the data.

(B) Mean Deviation: Calculated as the average of the absolute deviations from a central value (mean, median, or mode). It is an absolute measure, expressed in the units of the data.

(C) Variance: Calculated as the average of the squared deviations from the mean. It is an absolute measure, expressed in the square of the units of the data ($\text{units}^2$).

(D) Coefficient of Variation: Calculated as $\frac{\text{Standard Deviation}}{\text{Mean}} \times 100\%$. Since the standard deviation and the mean are in the same units, the ratio is unitless. It is a relative measure of dispersion.

Among the given options, the Coefficient of Variation is the only relative measure of dispersion.

The correct option is (D) Coefficient of Variation.

The Variance ($\sigma^2$) is a measure of the average of the squared differences from the mean.

The Standard Deviation ($\sigma$) is a measure of the amount of variation or dispersion of a set of values. It is defined as the positive square root of the variance.

The relationship between standard deviation and variance is:

Standard Deviation = $\sqrt{\text{Variance}}$

Using symbols:

$\sigma = \sqrt{\sigma^2}$

Given:

Variance ($\sigma^2$) = 81

To Find:

Standard Deviation ($\sigma$)

Solution:

Using the formula, we take the square root of the variance to find the standard deviation:

$\sigma = \sqrt{81}$

Calculating the square root:

$\sigma = 9$

The standard deviation is always a non-negative value, as it measures the magnitude of dispersion. Therefore, we take the positive square root.

Comparing our result with the given options:

(A) 9

(B) $\sqrt{81}$

(C) Both (A) and (B)

(D) $\pm 9$

Our calculated value is 9, which is option (A). Also, $\sqrt{81}$ is the expression that results in 9, which is option (B). Since 9 is equal to $\sqrt{81}$, both (A) and (B) correctly represent the standard deviation.

Option (D) $\pm 9$ is incorrect because standard deviation is defined as the positive square root, hence it cannot be -9.

Therefore, both options (A) and (B) are correct.

The correct option is (C) Both (A) and (B).

Given:

Mean Deviation about Median = 6

Median = 30

To Find:

The Coefficient of Mean Deviation about Median.

Solution:

The Coefficient of Mean Deviation is a relative measure of dispersion. It is calculated by dividing the Mean Deviation by the average used for its calculation (mean or median) and is often expressed as a percentage.

The formula for the Coefficient of Mean Deviation about the Median is:

$ \text{Coefficient of MD}_{\text{Median}} = \frac{\text{Mean Deviation about Median}}{\text{Median}} $

Substitute the given values into the formula:

$ \text{Coefficient of MD}_{\text{Median}} = \frac{6}{30} $

Calculate the value:

$ \frac{6}{30} = \frac{1}{5} = 0.2 $

The coefficient can also be expressed as a percentage by multiplying by 100%:

$ 0.2 \times 100\% = 20\% $

So, the Coefficient of Mean Deviation about the Median is 0.2 or 20%.

Let's check the given options:

(A) 0.2

(B) 0.1

(C) 20%

(D) Both (A) and (C)

Our calculated value matches both option (A) (as a ratio) and option (C) (as a percentage).

Therefore, both (A) and (C) are correct representations of the coefficient of mean deviation.

The correct option is (D) Both (A) and (C).

Inferential Statistics involves using data from a sample to make inferences, estimates, or predictions about a larger population. Measures of dispersion play a crucial role in inferential statistics as they help in understanding the variability within the sample and the uncertainty in estimating population parameters.

Let's consider the suitability of each measure in the context of inferential statistics:

(A) Range: The range is the difference between the maximum and minimum values. It is simple to calculate but only uses two values from the dataset and is highly sensitive to outliers. It does not provide information about the dispersion of values between the extremes. Due to these limitations, it is rarely used in formal inferential procedures.

(B) Mean Deviation: The mean deviation (about mean or median) considers the absolute deviations from a central value. While it uses all observations, the use of absolute values makes it mathematically difficult to handle in theoretical derivations and complex statistical models required for inference.

(C) Standard Deviation: The standard deviation is the square root of the variance, which is the average of the squared deviations from the mean. The standard deviation (and variance) uses all observations and has desirable mathematical properties. Squaring the deviations makes the measure mathematically tractable, allowing for the use of calculus and integration which are fundamental to statistical theory and inferential techniques (like deriving probability distributions, standard errors, and test statistics). The standard deviation is directly used in many key inferential statistics formulas, such as those for calculating standard errors, constructing confidence intervals, and performing hypothesis tests (e.g., t-tests, ANOVA, Z-tests).

(D) Coefficient of Variation: The coefficient of variation is a relative measure ($\frac{\text{Standard Deviation}}{\text{Mean}} \times 100\%$). While useful for comparing variability between datasets with different means or units, it is not the primary measure of dispersion used *within* the core formulas of most inferential statistical methods. The standard deviation (or variance) is the absolute measure of variability typically used in these formulas.

Based on their mathematical properties and their fundamental role in the formulas used for sampling distributions, standard errors, hypothesis testing, and confidence intervals, the Standard Deviation (and Variance) is the most commonly used measure of dispersion in inferential statistics.

The correct option is (C) Standard Deviation.

Given:

Mean of the data set = $\bar{x}$

Standard deviation of the data set = $\sigma$

To Find:

The Variance of the data set.

Solution:

The Standard Deviation ($\sigma$) is a measure of the dispersion of a data set and is defined as the positive square root of the Variance ($\sigma^2$).

The relationship between standard deviation and variance is:

Standard Deviation = $\sqrt{\text{Variance}}$

To find the variance from the standard deviation, we need to square the standard deviation.

Variance = $(\text{Standard Deviation})^2$

Using the given symbols:

Variance = $(\sigma)^2 = \sigma^2$

So, if the standard deviation is $\sigma$, the variance is $\sigma^2$. The mean ($\bar{x}$) is not directly involved in this definition relating standard deviation and variance.

Let's examine the given options:

(A) $\bar{x}^2$: This is the square of the mean, not the variance.

(B) $\sigma^2$: This is the square of the standard deviation, which is the definition of variance.

(C) $\bar{x} \sigma$: This is the product of the mean and standard deviation, not the variance.

(D) $\sqrt{\sigma}$: This is the square root of the standard deviation, not the variance.

The correct representation of the variance, given the standard deviation $\sigma$, is $\sigma^2$.

The correct option is (B) $\sigma^2$.

Let the data set be $x_1, x_2, ..., x_n$. Let the mean of the data set be $\bar{x}$ (for a sample) or $\mu$ (for a population).

The deviation of an observation $x_i$ from the mean is $(x_i - \bar{x})$.

The square of the deviation for an observation $x_i$ is $(x_i - \bar{x})^2$.

The sum of the squares of deviations from the mean is $\sum_{i=1}^n (x_i - \bar{x})^2$.

The question asks what is obtained when this sum is divided by the number of observations (let's denote the number of observations by $n$). The calculation is:

$\frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n}$

This formula represents the average of the squared deviations from the mean. This measure is known as the Variance.

Specifically, for a sample, the sample variance ($s^2$) is often calculated using $n-1$ in the denominator ($s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}$), but the definition of population variance ($\sigma^2 = \frac{\sum (x_i - \mu)^2}{N}$) uses the total number of observations $N$. The phrasing in the question ("divided by the number of observations") directly corresponds to the conceptual average of squared deviations or the population variance calculation.

Let's consider the other options:

(A) Mean Deviation: The Mean Deviation is the average of the absolute deviations from the mean (or median), $\frac{\sum |x_i - \bar{x}|}{n}$. It uses absolute values, not squared values.

(B) Standard Deviation: The Standard Deviation ($\sigma$ or $s$) is the square root of the variance. It is $\sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$ or $\sqrt{\frac{\sum (x_i - \mu)^2}{N}}$.

(D) Range: The Range is the difference between the maximum and minimum values, and does not involve summing or squaring deviations from the mean.

Therefore, dividing the sum of the squares of deviations from the mean by the number of observations gives the Variance.

The correct option is (C) Variance.

A dimensionless measure of dispersion is one that does not have any units associated with it. These measures are typically ratios of two quantities that have the same units, causing the units to cancel out. This makes them useful for comparing the variability of datasets that are measured in different units or have significantly different scales.

Let's examine the dimensions (units) of each given measure of dispersion, assuming the original data is measured in some unit (e.g., cm, kg, $\textsf{₹}$).

(A) Range: Calculated as the difference between the maximum and minimum values. If the data is in units of 'U', the range is in units of 'U'. It is not dimensionless.

(B) Standard Deviation: Calculated as the square root of the variance. The variance is the average of squared deviations, so its units are 'U$^2$'. The standard deviation, being the square root of variance, has units of $\sqrt{\text{U}^2} =$ 'U'. It is not dimensionless.

(C) Variance: Calculated as the average of the squared deviations from the mean. If the data is in units of 'U', the deviations $(x_i - \bar{x})$ are in units of 'U', and the squared deviations $(x_i - \bar{x})^2$ are in units of 'U$^2$'. The variance is the average of these squared deviations, so its units are 'U$^2$'. It is not dimensionless.

(D) Coefficient of Variation: Calculated as the ratio of the Standard Deviation to the Mean, often multiplied by 100%. The formula is $CV = \frac{\sigma}{\mu} \times 100\%$. Both the standard deviation ($\sigma$) and the mean ($\mu$) have the same units as the original data ('U').

$ CV = \frac{\text{Units of U}}{\text{Units of U}} \times 100\% $

The units in the numerator and denominator cancel out, leaving the coefficient of variation as a ratio or percentage, which is unitless or dimensionless.

Therefore, the Coefficient of Variation is a dimensionless measure of dispersion.

The correct option is (D) Coefficient of Variation.

Given:

The given data set is: 15, 20, 30, 25, 10.

To Find:

1. Range

2. Coefficient of Range

Solution:

First, let's arrange the data in ascending order: 10, 15, 20, 25, 30.

The largest value (L) in the data set is 30.

The smallest value (S) in the data set is 10.

Range:

The range is the difference between the largest and smallest values in the data set.

Range = L - S

Range = $30 - 10$

Range = $20$

Coefficient of Range:

The coefficient of range is given by the formula:

Coefficient of Range = $\frac{L - S}{L + S}$

Coefficient of Range = $\frac{30 - 10}{30 + 10}$

Coefficient of Range = $\frac{20}{40}$

Coefficient of Range = $\frac{\cancel{20}^{1}}{\cancel{40}_{2}}$

Coefficient of Range = $\frac{1}{2}$

Coefficient of Range = $0.5$

Therefore, the range is $20$ and the coefficient of range is $0.5$.

Given:

The given data set is: 6, 7, 10, 12, 13, 4, 8, 12.

Number of observations, $n = 8$.

To Find:

The mean deviation about the mean.

Solution:

First, we need to calculate the mean ($\bar{x}$) of the given data.

Mean ($\bar{x}$) = $\frac{\text{Sum of observations}}{\text{Number of observations}}$

$\bar{x} = \frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8}$

$\bar{x} = \frac{72}{8}$

$\bar{x} = 9$

Now, we calculate the absolute deviation of each observation from the mean, i.e., $|x_i - \bar{x}|$.

Next, we find the sum of the absolute deviations, $\sum |x_i - \bar{x}|$.

$\sum |x_i - \bar{x}| = 3 + 2 + 1 + 3 + 4 + 5 + 1 + 3 = 22$

Finally, we calculate the mean deviation about the mean using the formula:

Mean Deviation about Mean = $\frac{\sum |x_i - \bar{x}|}{n}$

Mean Deviation about Mean = $\frac{22}{8}$

Mean Deviation about Mean = $\frac{\cancel{22}^{11}}{\cancel{8}_{4}}$

Mean Deviation about Mean = $\frac{11}{4}$

Mean Deviation about Mean = $2.75$

The mean deviation about the mean for the given data is $2.75$.

Given:

The given data set is: 13, 17, 16, 14, 11, 13, 10, 16, 11, 10, 16.

Number of observations, $n = 11$.

To Find:

The mean deviation about the median.

Solution:

First, we need to arrange the data in ascending order to find the median.

Arranged data: 10, 10, 11, 11, 13, 13, 14, 16, 16, 16, 17.

Since the number of observations ($n = 11$) is odd, the median (M) is the $\left(\frac{n+1}{2}\right)^{\text{th}}$ observation.

Median = $\left(\frac{11+1}{2}\right)^{\text{th}}$ observation

Median = $\left(\frac{12}{2}\right)^{\text{th}}$ observation

Median = $6^{\text{th}}$ observation

The $6^{\text{th}}$ observation in the arranged data is 13.

So, the median (M) is $13$.

Now, we calculate the absolute deviation of each observation from the median, i.e., $|x_i - M|$.

Next, we find the sum of the absolute deviations, $\sum |x_i - M|$.

$\sum |x_i - M| = 3 + 3 + 2 + 2 + 0 + 0 + 1 + 3 + 3 + 3 + 4 = 24$

Finally, we calculate the mean deviation about the median using the formula:

Mean Deviation about Median = $\frac{\sum |x_i - M|}{n}$

Mean Deviation about Median = $\frac{24}{11}$

Mean Deviation about Median $\approx 2.18$

The mean deviation about the median for the given data is $\frac{24}{11}$ or approximately $2.18$.

Given:

The given discrete frequency distribution:

To Find:

The mean deviation about the mean.

Solution:

First, we need to calculate the mean ($\bar{x}$) of the data. We add columns for $f_i x_i$ and $|x_i - \bar{x}|$ and $f_i |x_i - \bar{x}|$.

Calculate the mean ($\bar{x}$):

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{225}{20}$

$\bar{x} = \frac{225}{20} = \frac{\cancel{225}^{45}}{\cancel{20}_{4}}$

$\bar{x} = \frac{45}{4} = 11.25$

Now, we calculate $|x_i - \bar{x}|$ and $f_i |x_i - \bar{x}|$ for each observation.

Finally, calculate the mean deviation about the mean:

Mean Deviation about Mean = $\frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$

Mean Deviation about Mean = $\frac{97.50}{20}$

Mean Deviation about Mean = $\frac{97.5}{20} = \frac{975}{200}$

Mean Deviation about Mean = $\frac{\cancel{975}^{195}}{\cancel{200}_{40}} = \frac{\cancel{195}^{39}}{\cancel{40}_{8}}$

Mean Deviation about Mean = $\frac{39}{8} = 4.875$

The mean deviation about the mean is $4.875$.

Given:

The given data set is: 5, 10, 15, 20, 25.

Number of observations, $n = 5$.

To Find:

1. Variance ($\sigma^2$)

2. Standard Deviation ($\sigma$)

Solution:

First, we need to calculate the mean ($\bar{x}$) of the given data.

Mean ($\bar{x}$) = $\frac{\text{Sum of observations}}{\text{Number of observations}}$

$\bar{x} = \frac{5 + 10 + 15 + 20 + 25}{5}$

$\bar{x} = \frac{75}{5}$

$\bar{x} = 15$

Now, we calculate the deviation of each observation from the mean ($x_i - \bar{x}$) and the square of these deviations ($(x_i - \bar{x})^2$).

Next, we find the sum of the squared deviations, $\sum (x_i - \bar{x})^2$.

$\sum (x_i - \bar{x})^2 = 100 + 25 + 0 + 25 + 100 = 250$

Variance ($\sigma^2$):

The variance is the average of the squared deviations from the mean.

Variance ($\sigma^2$) = $\frac{\sum (x_i - \bar{x})^2}{n}$

$\sigma^2 = \frac{250}{5}$

$\sigma^2 = 50$

Standard Deviation ($\sigma$):

The standard deviation is the square root of the variance.

Standard Deviation ($\sigma$) = $\sqrt{\text{Variance}}$

$\sigma = \sqrt{50}$

$\sigma = \sqrt{25 \times 2}$

$\sigma = 5\sqrt{2}$

Using the approximate value $\sqrt{2} \approx 1.414$:

$\sigma \approx 5 \times 1.414 = 7.07$

The variance is $50$ and the standard deviation is $5\sqrt{2}$ (approximately $7.07$).

Given:

The given discrete frequency distribution:

To Find:

The standard deviation ($\sigma$).

Solution:

First, we need to calculate the mean ($\bar{x}$) of the data. We find $\sum f_i$ and $\sum f_i x_i$.

Calculate the mean ($\bar{x}$):

$\bar{x} = \frac{\sum\limits f_i x_i}{\sum\limits f_i} = \frac{228}{20}$

$\bar{x} = \frac{\cancel{228}^{114}}{\cancel{20}_{10}} = \frac{114}{10} = 11.4$

Next, we calculate the deviations from the mean $(x_i - \bar{x})$, their squares $(x_i - \bar{x})^2$, and the product of the squared deviations with the frequencies $f_i (x_i - \bar{x})^2$.

Calculate the variance ($\sigma^2$):

Variance ($\sigma^2$) = $\frac{\sum\limits f_i (x_i - \bar{x})^2}{\sum\limits f_i}$

$\sigma^2 = \frac{328.8}{20}$

$\sigma^2 = \frac{3288}{200} = \frac{1644}{100} = 16.44$

Calculate the standard deviation ($\sigma$):

Standard Deviation ($\sigma$) = $\sqrt{\text{Variance}}$

$\sigma = \sqrt{16.44}$

$\sigma \approx 4.05$ (approximately)

The standard deviation for the given data is $\sqrt{16.44}$ or approximately $4.05$.

Given:

Number of observations, $n = 100$.

Incorrect Mean, $\bar{x}_{incorrect} = 40$.

Incorrect Standard Deviation, $\sigma_{incorrect} = 10$.

Incorrect value taken = 50.

Correct value = 40.

To Find:

1. Corrected Mean

2. Corrected Standard Deviation

Solution:

The formula for mean is $\bar{x} = \frac{\sum x_i}{n}$.

From the incorrect mean, we can find the incorrect sum of observations:

Substituting the given values into equation (i):

$\sum x_{incorrect} = 100 \times 40$

$\sum x_{incorrect} = 4000$

Corrected Mean:

To find the corrected sum of observations, we subtract the incorrect value and add the correct value to the incorrect sum.

Substituting the values into equation (ii):

$\sum x_{correct} = 4000 - 50 + 40$

$\sum x_{correct} = 4000 - 10$

$\sum x_{correct} = 3990$

Now, we can calculate the corrected mean:

Substituting the values into equation (iii):

$\bar{x}_{correct} = \frac{3990}{100}$

$\bar{x}_{correct} = 39.9$

Corrected Standard Deviation:

The formula for standard deviation is $\sigma = \sqrt{\frac{\sum x_i^2}{n} - \bar{x}^2}$.

Squaring both sides, we get $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$.

Rearranging the formula to find the sum of squares:

Using the incorrect values in equation (iv), we find the incorrect sum of squares:

Substituting the incorrect values into equation (v):

$\sum x^2_{incorrect} = 100(10^2 + 40^2)$

$\sum x^2_{incorrect} = 100(100 + 1600)$

$\sum x^2_{incorrect} = 100(1700)$

$\sum x^2_{incorrect} = 170000$

To find the corrected sum of squares, we subtract the square of the incorrect value and add the square of the correct value to the incorrect sum of squares.

Substituting the values into equation (vi):

$\sum x^2_{correct} = 170000 - (50)^2 + (40)^2$

$\sum x^2_{correct} = 170000 - 2500 + 1600$

$\sum x^2_{correct} = 170000 - 900$

$\sum x^2_{correct} = 169100$

Now, we calculate the corrected variance using the corrected sum of squares and the corrected mean.

Substituting the values into equation (vii):

$\sigma^2_{correct} = \frac{169100}{100} - (39.9)^2$

$\sigma^2_{correct} = 1691 - (39.9)^2$

Calculate $(39.9)^2$:

$39.9 \times 39.9 = 1592.01$

$\sigma^2_{correct} = 1691 - 1592.01$

$\sigma^2_{correct} = 98.99$

Finally, we calculate the corrected standard deviation by taking the square root of the corrected variance.

Substituting the value into equation (viii):

$\sigma_{correct} = \sqrt{98.99}$

$\sigma_{correct} \approx 9.95$ (approximately)

The corrected mean is $39.9$ and the corrected standard deviation is $\sqrt{98.99}$ (approximately $9.95$).

Given:

The data set of wickets taken by a bowler in 10 matches is: 2, 6, 4, 5, 0, 2, 1, 3, 2, 3.

Number of observations, $n = 10$.

To Find:

1. Mean ($\bar{x}$)

2. Standard Deviation ($\sigma$)

Solution:

First, we need to calculate the mean ($\bar{x}$) of the given data.

Mean ($\bar{x}$) = $\frac{\text{Sum of observations}}{\text{Number of observations}}$

$\bar{x} = \frac{2 + 6 + 4 + 5 + 0 + 2 + 1 + 3 + 2 + 3}{10}$

$\bar{x} = \frac{28}{10}$

$\bar{x} = 2.8$

Now, we calculate the deviation of each observation from the mean ($x_i - \bar{x}$) and the square of these deviations ($(x_i - \bar{x})^2$).

Next, we find the sum of the squared deviations, $\sum (x_i - \bar{x})^2$.

$\sum (x_i - \bar{x})^2 = 0.64 + 10.24 + 1.44 + 4.84 + 7.84 + 0.64 + 3.24 + 0.04 + 0.64 + 0.04 = 29.60$

Variance ($\sigma^2$):

The variance is the average of the squared deviations from the mean.

Variance ($\sigma^2$) = $\frac{\sum (x_i - \bar{x})^2}{n}$

$\sigma^2 = \frac{29.60}{10}$

$\sigma^2 = 2.96$

Standard Deviation ($\sigma$):

The standard deviation is the square root of the variance.

Standard Deviation ($\sigma$) = $\sqrt{\text{Variance}}$

$\sigma = \sqrt{2.96}$

Using a calculator, $\sqrt{2.96} \approx 1.72$ (approximately).

The mean is $2.8$ and the standard deviation is $\sqrt{2.96}$ (approximately $1.72$).

Given:

The given grouped frequency distribution:

To Find:

The mean deviation about the mean.

Solution:

First, we need to find the midpoint ($x_i$) of each class interval. The midpoint is calculated as $\frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

Then, we calculate $f_i x_i$ for each class and their sum ($\sum f_i x_i$). We also find the sum of frequencies ($\sum f_i$).

Calculate the mean ($\bar{x}$):

$\bar{x} = \frac{\sum\limits f_i x_i}{\sum\limits f_i} = \frac{1350}{50}$

$\bar{x} = \frac{135}{5}$

$\bar{x} = 27$

Now, we calculate the absolute deviation $|x_i - \bar{x}|$ for each midpoint and the product of the frequencies with these absolute deviations $f_i |x_i - \bar{x}|$.

Finally, calculate the mean deviation about the mean:

Mean Deviation about Mean = $\frac{\sum\limits f_i |x_i - \bar{x}|}{\sum\limits f_i}$

Mean Deviation about Mean = $\frac{472}{50}$

Mean Deviation about Mean = $\frac{\cancel{472}^{236}}{\cancel{50}_{25}}$

Mean Deviation about Mean = $\frac{236}{25} = 9.44$

The mean deviation about the mean for the given data is $9.44$.

Given:

The given data set is: 6, 8, 10, 12, 14.

Number of observations, $n = 5$.

Use $\sqrt{2} = 1.414$.

To Find:

The Coefficient of Variation (CV).

Solution:

First, we need to calculate the mean ($\bar{x}$) of the given data.

Mean ($\bar{x}$) = $\frac{\text{Sum of observations}}{\text{Number of observations}}$

Sum of observations = $6 + 8 + 10 + 12 + 14 = 50$

$\bar{x} = \frac{50}{5}$

$\bar{x} = 10$

Next, we need to calculate the standard deviation ($\sigma$). We calculate the deviations from the mean ($x_i - \bar{x}$) and their squares ($(x_i - \bar{x})^2$).

Next, we find the sum of the squared deviations, $\sum (x_i - \bar{x})^2$.

$\sum (x_i - \bar{x})^2 = 16 + 4 + 0 + 4 + 16 = 40$

Now, we calculate the variance ($\sigma^2$).

Variance ($\sigma^2$) = $\frac{\sum (x_i - \bar{x})^2}{n}$

$\sigma^2 = \frac{40}{5}$

$\sigma^2 = 8$

The standard deviation ($\sigma$) is the square root of the variance.

Standard Deviation ($\sigma$) = $\sqrt{\text{Variance}}$

$\sigma = \sqrt{8}$

$\sigma = \sqrt{4 \times 2}$

$\sigma = 2\sqrt{2}$

Using the given value $\sqrt{2} = 1.414$:

$\sigma = 2 \times 1.414$

$\sigma = 2.828$

Finally, we calculate the coefficient of variation (CV) using the formula:

CV = $\frac{\sigma}{\bar{x}} \times 100$

CV = $\frac{2.828}{10} \times 100$

CV = $0.2828 \times 100$

CV = $28.28\%$

The coefficient of variation for the given data is $28.28\%$.

Given:

Number of observations, $n = 20$.

Original Variance, $\sigma^2_{old} = 5$.

Each observation is multiplied by a constant, $k = 2$.

To Find:

1. New Variance ($\sigma^2_{new}$)

2. New Standard Deviation ($\sigma_{new}$)

Solution:

Let the original observations be $x_1, x_2, \dots, x_{20}$.

The original variance is given by:

$\sigma^2_{old} = \frac{1}{n} \sum\limits_{i=1}^{n} (x_i - \bar{x}_{old})^2 = 5$

where $\bar{x}_{old}$ is the mean of the original observations.

When each observation is multiplied by 2, the new observations are $y_i = 2x_i$.

The new mean will be $\bar{y}_{new} = 2\bar{x}_{old}$.

The new variance ($\sigma^2_{new}$) is given by:

$\sigma^2_{new} = \frac{1}{n} \sum\limits_{i=1}^{n} (y_i - \bar{y}_{new})^2$

Substitute $y_i = 2x_i$ and $\bar{y}_{new} = 2\bar{x}_{old}$:

$\sigma^2_{new} = \frac{1}{n} \sum\limits_{i=1}^{n} (2x_i - 2\bar{x}_{old})^2$

$\sigma^2_{new} = \frac{1}{n} \sum\limits_{i=1}^{n} (2(x_i - \bar{x}_{old}))^2$

$\sigma^2_{new} = \frac{1}{n} \sum\limits_{i=1}^{n} 4(x_i - \bar{x}_{old})^2$

$\sigma^2_{new} = 4 \times \frac{1}{n} \sum\limits_{i=1}^{n} (x_i - \bar{x}_{old})^2$

Recognize that $\frac{1}{n} \sum\limits_{i=1}^{n} (x_i - \bar{x}_{old})^2$ is the original variance, $\sigma^2_{old}$.

$\sigma^2_{new} = 4 \times \sigma^2_{old}$

Substitute the given original variance:

$\sigma^2_{new} = 4 \times 5$

$\sigma^2_{new} = 20$

The new standard deviation ($\sigma_{new}$) is the square root of the new variance.

$\sigma_{new} = \sqrt{\sigma^2_{new}}$

$\sigma_{new} = \sqrt{20}$

$\sigma_{new} = \sqrt{4 \times 5}$

$\sigma_{new} = 2\sqrt{5}$

Alternatively, if each observation is multiplied by a constant $k$, the new standard deviation is $|k|$ times the original standard deviation.

Original standard deviation, $\sigma_{old} = \sqrt{5}$.

New standard deviation, $\sigma_{new} = |2| \times \sigma_{old} = 2 \times \sqrt{5} = 2\sqrt{5}$.

The new variance is $20$ and the new standard deviation is $2\sqrt{5}$.

Given:

The first $n$ natural numbers: $1, 2, 3, \dots, n$.

To Find:

1. Mean

2. Variance

Solution:

Let the set of the first $n$ natural numbers be $X = \{1, 2, 3, \dots, n\}$.

The number of observations is $n$.

Mean ($\bar{x}$):

The sum of the first $n$ natural numbers is given by the formula:

$\sum\limits_{i=1}^{n} x_i = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$

The mean is calculated as:

$\bar{x} = \frac{\sum\limits_{i=1}^{n} x_i}{n}$

Substitute the sum of observations:

$\bar{x} = \frac{\frac{n(n+1)}{2}}{n}$

$\bar{x} = \frac{n(n+1)}{2n}$

$\bar{x} = \frac{n+1}{2}$

Variance ($\sigma^2$):

The sum of the squares of the first $n$ natural numbers is given by the formula:

$\sum\limits_{i=1}^{n} x_i^2 = 1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$

The variance can be calculated using the formula:

$\sigma^2 = \frac{\sum\limits_{i=1}^{n} x_i^2}{n} - (\bar{x})^2$

Substitute the values of $\sum\limits_{i=1}^{n} x_i^2$ and $\bar{x}$:

$\sigma^2 = \frac{\frac{n(n+1)(2n+1)}{6}}{n} - \left(\frac{n+1}{2}\right)^2$

$\sigma^2 = \frac{n(n+1)(2n+1)}{6n} - \frac{(n+1)^2}{4}$

$\sigma^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$

Find a common denominator (12):

$\sigma^2 = \frac{2(n+1)(2n+1)}{12} - \frac{3(n+1)^2}{12}$

$\sigma^2 = \frac{(n+1)[2(2n+1) - 3(n+1)]}{12}$

$\sigma^2 = \frac{(n+1)[4n+2 - 3n-3]}{12}$

$\sigma^2 = \frac{(n+1)(n-1)}{12}$

$\sigma^2 = \frac{n^2 - 1}{12}$

The mean of the first $n$ natural numbers is $\frac{n+1}{2}$.

The variance of the first $n$ natural numbers is $\frac{n^2 - 1}{12}$.

Given:

The given grouped frequency distribution:

To Find:

The mean deviation about the median.

Solution:

First, we find the cumulative frequency ($cf$) and total frequency ($\sum f_i$).

To find the median, we calculate $\frac{N}{2}$.

$\frac{N}{2} = \frac{44}{2} = 22$

The cumulative frequency just greater than 22 is 30, which corresponds to the class interval 12-18. This is the median class.

Now, we calculate the median (M) using the formula for grouped data:

$M = L + \frac{\frac{N}{2} - cf}{f} \times h$

Where:

$L$ = Lower limit of the median class = 12

$N$ = Total frequency = 44

$cf$ = Cumulative frequency of the class preceding the median class = 18

$f$ = Frequency of the median class = 12

$h$ = Class size = $18 - 12 = 6$

Substitute the values into the formula:

$M = 12 + \frac{22 - 18}{12} \times 6$

$M = 12 + \frac{4}{12} \times 6$

$M = 12 + \frac{\cancel{4}^{1}}{\cancel{12}_{3}} \times 6$

$M = 12 + \frac{1}{3} \times 6$

$M = 12 + \frac{\cancel{6}^{2}}{\cancel{3}_{1}}$

$M = 12 + 2$

$M = 14$

The median (M) is $14$.

Next, we find the midpoint ($x_i$) of each class interval and calculate the absolute deviation $|x_i - M|$ and the product $f_i |x_i - M|$.

Finally, calculate the mean deviation about the median:

Mean Deviation about Median = $\frac{\sum\limits f_i |x_i - M|}{\sum\limits f_i}$

Mean Deviation about Median = $\frac{278}{44}$

Mean Deviation about Median = $\frac{\cancel{278}^{139}}{\cancel{44}_{22}}$

Mean Deviation about Median = $\frac{139}{22}$

Mean Deviation about Median $\approx 6.318$

The mean deviation about the median for the given data is $\frac{139}{22}$ or approximately $6.318$.

Given:

The numbers are 2, 3, x, 11.

The standard deviation ($\sigma$) is 3.5.

To Find:

The value of x.

Solution:

The formula for standard deviation is:

Where $x_i$ are the individual numbers, $\bar{x}$ is the mean of the numbers, and $n$ is the number of observations.

First, calculate the mean ($\bar{x}$):

$\bar{x} = \frac{2 + 3 + x + 11}{4}$

$\bar{x} = \frac{16 + x}{4}$

Now, calculate the squared difference of each number from the mean:

$(2 - \bar{x})^2 = (2 - \frac{16+x}{4})^2 = (\frac{8 - (16+x)}{4})^2 = (\frac{8 - 16 - x}{4})^2 = (\frac{-8 - x}{4})^2 = \frac{(x+8)^2}{16}$

$(3 - \bar{x})^2 = (3 - \frac{16+x}{4})^2 = (\frac{12 - (16+x)}{4})^2 = (\frac{12 - 16 - x}{4})^2 = (\frac{-4 - x}{4})^2 = \frac{(x+4)^2}{16}$

$(x - \bar{x})^2 = (x - \frac{16+x}{4})^2 = (\frac{4x - (16+x)}{4})^2 = (\frac{4x - 16 - x}{4})^2 = (\frac{3x - 16}{4})^2 = \frac{(3x-16)^2}{16}$

$(11 - \bar{x})^2 = (11 - \frac{16+x}{4})^2 = (\frac{44 - (16+x)}{4})^2 = (\frac{44 - 16 - x}{4})^2 = (\frac{28 - x}{4})^2 = \frac{(28-x)^2}{16}$

Now, sum these squared differences:

$\sum(x_i - \bar{x})^2 = \frac{(x+8)^2}{16} + \frac{(x+4)^2}{16} + \frac{(3x-16)^2}{16} + \frac{(28-x)^2}{16}$

$\sum(x_i - \bar{x})^2 = \frac{1}{16} [ (x^2 + 16x + 64) + (x^2 + 8x + 16) + (9x^2 - 96x + 256) + (784 - 56x + x^2) ]$

$\sum(x_i - \bar{x})^2 = \frac{1}{16} [ x^2 + 16x + 64 + x^2 + 8x + 16 + 9x^2 - 96x + 256 + 784 - 56x + x^2 ]$

Combine like terms:

$\sum(x_i - \bar{x})^2 = \frac{1}{16} [ (1+1+9+1)x^2 + (16+8-96-56)x + (64+16+256+784) ]$

$\sum(x_i - \bar{x})^2 = \frac{1}{16} [ 12x^2 - 128x + 1120 ]$

Now, substitute this into the standard deviation formula (i):

$3.5 = \sqrt{\frac{\frac{1}{16} [ 12x^2 - 128x + 1120 ]}{4}}$

$3.5 = \sqrt{\frac{12x^2 - 128x + 1120}{64}}$

Square both sides:

$3.5^2 = \frac{12x^2 - 128x + 1120}{64}$

$12.25 = \frac{12x^2 - 128x + 1120}{64}$

$12.25 \times 64 = 12x^2 - 128x + 1120$

$784 = 12x^2 - 128x + 1120$

Rearrange the equation to form a quadratic equation:

$12x^2 - 128x + 1120 - 784 = 0$

$12x^2 - 128x + 336 = 0$

Divide the entire equation by 4:

$3x^2 - 32x + 84 = 0$

Now, solve this quadratic equation for x. We can use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$a=3$, $b=-32$, $c=84$

$x = \frac{-(-32) \pm \sqrt{(-32)^2 - 4(3)(84)}}{2(3)}$

$x = \frac{32 \pm \sqrt{1024 - 1008}}{6}$

$x = \frac{32 \pm \sqrt{16}}{6}$

$x = \frac{32 \pm 4}{6}$

We have two possible values for x:

Case 1: $x = \frac{32 + 4}{6} = \frac{36}{6} = 6$

Case 2: $x = \frac{32 - 4}{6} = \frac{28}{6} = \frac{14}{3}$

Let's check both values:

If x = 6:

Numbers are 2, 3, 6, 11.

Mean $\bar{x} = \frac{2+3+6+11}{4} = \frac{22}{4} = 5.5$

Squared differences from the mean:

$(2-5.5)^2 = (-3.5)^2 = 12.25$

$(3-5.5)^2 = (-2.5)^2 = 6.25$

$(6-5.5)^2 = (0.5)^2 = 0.25$

$(11-5.5)^2 = (5.5)^2 = 30.25$

Sum of squared differences = $12.25 + 6.25 + 0.25 + 30.25 = 49$

Variance = $\frac{49}{4} = 12.25$

Standard Deviation = $\sqrt{12.25} = 3.5$. This matches the given value.

If x = 14/3:

Numbers are 2, 3, 14/3, 11.

Mean $\bar{x} = \frac{2+3+\frac{14}{3}+11}{4} = \frac{16 + \frac{14}{3}}{4} = \frac{\frac{48+14}{3}}{4} = \frac{62/3}{4} = \frac{62}{12} = \frac{31}{6}$

Squared differences from the mean:

$(2 - \frac{31}{6})^2 = (\frac{12-31}{6})^2 = (\frac{-19}{6})^2 = \frac{361}{36}$

$(3 - \frac{31}{6})^2 = (\frac{18-31}{6})^2 = (\frac{-13}{6})^2 = \frac{169}{36}$

$(\frac{14}{3} - \frac{31}{6})^2 = (\frac{28-31}{6})^2 = (\frac{-3}{6})^2 = (\frac{-1}{2})^2 = \frac{1}{4} = \frac{9}{36}$

$(11 - \frac{31}{6})^2 = (\frac{66-31}{6})^2 = (\frac{35}{6})^2 = \frac{1225}{36}$

Sum of squared differences = $\frac{361+169+9+1225}{36} = \frac{1764}{36} = 49$

Variance = $\frac{49}{4} = 12.25$

Standard Deviation = $\sqrt{12.25} = 3.5$. This also matches the given value.

Therefore, the possible values of x are 6 and 14/3.

Given:

Plant A:

Number of workers ($n_A$) = 5000

Average daily wage ($\bar{x}_A$) = ₹250

Variance of wages ($\sigma_A^2$) = 100

Plant B:

Number of workers ($n_B$) = 6000

Average daily wage ($\bar{x}_B$) = ₹200

Variance of wages ($\sigma_B^2$) = 144

To Find:

Which plant pays more variable wages.

Solution:

Variability of wages can be determined by comparing the standard deviations of the wages for each plant. A higher standard deviation indicates greater variability.

The standard deviation is the square root of the variance.

For Plant A:

Standard Deviation ($\sigma_A$) = $\sqrt{\text{Variance}_A}$

The standard deviation of wages for Plant A is ₹10.

For Plant B:

Standard Deviation ($\sigma_B$) = $\sqrt{\text{Variance}_B}$

The standard deviation of wages for Plant B is ₹12.

Comparing the standard deviations:

$\sigma_B = 12$ and $\sigma_A = 10$

Since $\sigma_B > \sigma_A$ (12 > 10), Plant B pays more variable wages than Plant A.

Given:

Frequency distribution:

Class Intervals: 10-20, 20-30, 30-40, 40-50

Frequencies: 5, 12, 15, 8

To Find:

Range and Coefficient of Range.

Solution:

The range of a frequency distribution is the difference between the upper boundary of the highest class and the lower boundary of the lowest class.

From the given data:

The lowest class is 10-20, so the lower boundary of the lowest class is 10.

The highest class is 40-50, so the upper boundary of the highest class is 50.

Range (R):

Range = Upper boundary of the highest class - Lower boundary of the lowest class

The range is 40.

Coefficient of Range (CR):

The formula for the coefficient of range is:

CR = $\frac{R}{L + H}$ where H is the upper boundary of highest class and L is the lower boundary of lowest class.

CR = $\frac{50 - 10}{50 + 10}$

CR = $\frac{40}{60}$

CR $\approx$ 0.667

The coefficient of range is $\frac{2}{3}$ or approximately 0.667.

Given:

The data is: 3, 6, 9, 12, 15.

To Find:

Mean Deviation about the Mean (MD).

Solution:

Step 1: Calculate the Mean ($\bar{x}$).

The mean is the sum of all observations divided by the number of observations.

$\bar{x} = \frac{3 + 6 + 9 + 12 + 15}{5}$

$\bar{x} = \frac{45}{5}$

Step 2: Calculate the absolute deviations of each observation from the mean ($|x_i - \bar{x}|$).

For the observation 3: $|3 - 9| = |-6| = 6$

For the observation 6: $|6 - 9| = |-3| = 3$

For the observation 9: $|9 - 9| = |0| = 0$

For the observation 12: $|12 - 9| = |3| = 3$

For the observation 15: $|15 - 9| = |6| = 6$

Step 3: Calculate the Mean Deviation (MD).

The Mean Deviation is the average of these absolute deviations.

MD = $\frac{\sum |x_i - \bar{x}|}{n}$

MD = $\frac{6 + 3 + 0 + 3 + 6}{5}$

MD = $\frac{18}{5}$

Therefore, the mean deviation about the mean for the given data is 3.6.

Given:

The data is: 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.

To Find:

Variance ($\sigma^2$).

Solution:

Step 1: Calculate the Mean ($\bar{x}$).

The mean is the sum of all observations divided by the number of observations.

The given data is an arithmetic progression with first term $a = 6$, common difference $d = 2$, and number of terms $n = 10$.

Sum of an AP = $\frac{n}{2} [2a + (n-1)d]$

Sum = $\frac{10}{2} [2(6) + (10-1)2]$

Sum = $5 [12 + 9 \times 2]$

Sum = $5 [12 + 18]$

Sum = $5 [30]$

Sum = 150

Mean $\bar{x} = \frac{\text{Sum}}{\text{Number of terms}} = \frac{150}{10}$

Step 2: Calculate the squared deviations of each observation from the mean ($(x_i - \bar{x})^2$).

For 6: $(6 - 15)^2 = (-9)^2 = 81$

For 8: $(8 - 15)^2 = (-7)^2 = 49$

For 10: $(10 - 15)^2 = (-5)^2 = 25$

For 12: $(12 - 15)^2 = (-3)^2 = 9$

For 14: $(14 - 15)^2 = (-1)^2 = 1$

For 16: $(16 - 15)^2 = (1)^2 = 1$

For 18: $(18 - 15)^2 = (3)^2 = 9$

For 20: $(20 - 15)^2 = (5)^2 = 25$

For 22: $(22 - 15)^2 = (7)^2 = 49$

For 24: $(24 - 15)^2 = (9)^2 = 81$

Step 3: Calculate the sum of squared deviations ($\sum (x_i - \bar{x})^2$).

$\sum (x_i - \bar{x})^2 = 81 + 49 + 25 + 9 + 1 + 1 + 9 + 25 + 49 + 81$

$\sum (x_i - \bar{x})^2 = 2 \times (81 + 49 + 25 + 9 + 1)$

$\sum (x_i - \bar{x})^2 = 2 \times (165)$

Step 4: Calculate the Variance ($\sigma^2$).

The variance is the average of the squared deviations.

$\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$

$\sigma^2 = \frac{330}{10}$

Therefore, the variance of the given data is 33.

Alternative method using formula for variance of AP:

For an arithmetic progression $a, a+d, a+2d, \dots, a+(n-1)d$, the variance is given by:

Here, $a=6$, $d=2$, $n=10$.

$\sigma^2 = \frac{2^2 (10^2 - 1)}{12}$

$\sigma^2 = \frac{4 (100 - 1)}{12}$

$\sigma^2 = \frac{4 (99)}{12}$

$\sigma^2 = \frac{396}{12}$

This matches the result obtained by the direct method.

Given:

Number of observations ($n$) = 5

Mean ($\bar{x}$) = 4.4

Variance ($\sigma^2$) = 8.24

Three of the observations are 1, 2, and 6.

To Find:

The other two observations.

Solution:

Let the other two observations be $x$ and $y$.

The five observations are 1, 2, 6, $x$, and $y$.

Step 1: Use the mean to form an equation.

The formula for the mean is $\bar{x} = \frac{\sum x_i}{n}$.

$4.4 = \frac{1 + 2 + 6 + x + y}{5}$

$4.4 \times 5 = 9 + x + y$

$22 = 9 + x + y$

$x + y = 22 - 9$

Step 2: Use the variance to form another equation.

The formula for variance is $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$.

First, calculate the squared deviations for the known observations from the mean (4.4):

For 1: $(1 - 4.4)^2 = (-3.4)^2 = 11.56$

For 2: $(2 - 4.4)^2 = (-2.4)^2 = 5.76$

For 6: $(6 - 4.4)^2 = (1.6)^2 = 2.56$

The squared deviations for the unknown observations are $(x - 4.4)^2$ and $(y - 4.4)^2$.

Now, substitute these into the variance formula:

$8.24 = \frac{11.56 + 5.76 + 2.56 + (x - 4.4)^2 + (y - 4.4)^2}{5}$

$8.24 \times 5 = 11.56 + 5.76 + 2.56 + (x - 4.4)^2 + (y - 4.4)^2$

$41.2 = 19.88 + (x - 4.4)^2 + (y - 4.4)^2$

$(x - 4.4)^2 + (y - 4.4)^2 = 41.2 - 19.88$

Step 3: Solve the system of equations (i) and (ii).

From equation (i), $y = 13 - x$. Substitute this into equation (ii):

$(x - 4.4)^2 + ((13 - x) - 4.4)^2 = 21.32$

$(x - 4.4)^2 + (8.6 - x)^2 = 21.32$

Expand the terms:

$(x^2 - 8.8x + 19.36) + (73.96 - 17.2x + x^2) = 21.32$

Combine like terms:

$2x^2 - 26x + 93.32 = 21.32$

$2x^2 - 26x + 93.32 - 21.32 = 0$

$2x^2 - 26x + 72 = 0$

Divide the equation by 2:

$x^2 - 13x + 36 = 0$

Factor the quadratic equation:

We need two numbers that multiply to 36 and add up to -13. These numbers are -4 and -9.

$(x - 4)(x - 9) = 0$

This gives two possible values for $x$: $x = 4$ or $x = 9$.

Case 1: If $x = 4$

Substitute $x=4$ into equation (i): $4 + y = 13 \implies y = 9$.

The other two observations are 4 and 9.

Case 2: If $x = 9$

Substitute $x=9$ into equation (i): $9 + y = 13 \implies y = 4$.

The other two observations are 9 and 4.

In both cases, the other two observations are 4 and 9.

Verification:

Observations: 1, 2, 6, 4, 9.

Mean = $\frac{1+2+6+4+9}{5} = \frac{22}{5} = 4.4$ (Correct)

Deviations from the mean: (1-4.4), (2-4.4), (6-4.4), (4-4.4), (9-4.4)

Deviations: -3.4, -2.4, 1.6, -0.4, 4.6

Squared Deviations: 11.56, 5.76, 2.56, 0.16, 21.16

Sum of Squared Deviations = $11.56 + 5.76 + 2.56 + 0.16 + 21.16 = 41.2$

Variance = $\frac{41.2}{5} = 8.24$ (Correct)

Therefore, the other two observations are 4 and 9.

To Find:

Mean and Standard Deviation of the first $n$ even natural numbers.

Solution:

The first $n$ even natural numbers are 2, 4, 6, 8, ..., 2n.

This is an arithmetic progression (AP) with:

First term ($a$) = 2

Common difference ($d$) = 2

Number of terms ($N$) = $n$

Step 1: Find the Mean.

The sum of an AP is given by $S_N = \frac{N}{2} [2a + (N-1)d]$.

Sum of the first $n$ even natural numbers = $\frac{n}{2} [2(2) + (n-1)2]$

Sum = $\frac{n}{2} [4 + 2n - 2]$

Sum = $\frac{n}{2} [2n + 2]$

Sum = $\frac{n}{2} \times 2(n+1)$

The Mean ($\bar{x}$) is the sum divided by the number of terms ($n$):

$\bar{x} = \frac{n(n+1)}{n}$

The mean of the first $n$ even natural numbers is $n+1$.

Step 2: Find the Standard Deviation.

The variance ($\sigma^2$) of an AP is given by the formula $\sigma^2 = \frac{d^2 (N^2 - 1)}{12}$.

In this case, $d = 2$ and $N = n$.

$\sigma^2 = \frac{2^2 (n^2 - 1)}{12}$

$\sigma^2 = \frac{4 (n^2 - 1)}{12}$

The standard deviation ($\sigma$) is the square root of the variance:

$\sigma = \sqrt{\frac{n^2 - 1}{3}}$

Therefore:

Mean = $n+1$

Standard Deviation = $\sqrt{\frac{n^2 - 1}{3}}$

Given:

Data 1: 6, 7, 8, 9, 10. Its standard deviation is $\sigma$.

Data 2: 12, 14, 16, 18, 20.

To Find:

The standard deviation of the second dataset.

Solution:

Part 1: Calculate the standard deviation ($\sigma$) of the first dataset (6, 7, 8, 9, 10).

The data set 6, 7, 8, 9, 10 is an arithmetic progression with $a=6$, $d=1$, and $n=5$.

First, find the mean ($\bar{x}_1$):

$\bar{x}_1 = \frac{6+7+8+9+10}{5} = \frac{40}{5} = 8$.

Next, calculate the squared deviations from the mean:

$(6-8)^2 = (-2)^2 = 4$

$(7-8)^2 = (-1)^2 = 1$

$(8-8)^2 = (0)^2 = 0$

$(9-8)^2 = (1)^2 = 1$

$(10-8)^2 = (2)^2 = 4$

Sum of squared deviations ($\sum (x_i - \bar{x}_1)^2$) = $4 + 1 + 0 + 1 + 4 = 10$.

Variance ($\sigma^2$) = $\frac{\sum (x_i - \bar{x}_1)^2}{n} = \frac{10}{5} = 2$.

Standard Deviation ($\sigma$) = $\sqrt{2}$.

Part 2: Analyze the relationship between the two datasets.

Let the first dataset be $X = \{x_1, x_2, x_3, x_4, x_5\} = \{6, 7, 8, 9, 10\}$.

Let the second dataset be $Y = \{y_1, y_2, y_3, y_4, y_5\} = \{12, 14, 16, 18, 20\}$.

Observe the relationship between corresponding elements:

$y_1 = 12 = 2 \times 6 = 2x_1$

$y_2 = 14 = 2 \times 7 = 2x_2$

$y_3 = 16 = 2 \times 8 = 2x_3$

$y_4 = 18 = 2 \times 9 = 2x_4$

$y_5 = 20 = 2 \times 10 = 2x_5$

It can be seen that each element in the second dataset is obtained by multiplying the corresponding element in the first dataset by 2. So, $y_i = 2x_i$.

Property of Standard Deviation:

If a dataset $X = \{x_1, x_2, \dots, x_n\}$ has a standard deviation $\sigma_X$, then a new dataset $Y = \{ax_1, ax_2, \dots, ax_n\}$, where $a$ is a constant, will have a standard deviation $\sigma_Y = |a|\sigma_X$.

In this case, $a = 2$.

Therefore, the standard deviation of the second dataset ($\sigma_Y$) will be $2 \times \sigma_X$.

Since the standard deviation of the first dataset is $\sigma$, the standard deviation of the second dataset is $2\sigma$.

We already found $\sigma = \sqrt{2}$.

So, the standard deviation of the second dataset is $2\sqrt{2}$.

Alternative Calculation for the second dataset:

Dataset 2: 12, 14, 16, 18, 20.

Mean ($\bar{x}_2$) = $\frac{12+14+16+18+20}{5} = \frac{80}{5} = 16$.

Squared deviations from the mean:

$(12-16)^2 = (-4)^2 = 16$

$(14-16)^2 = (-2)^2 = 4$

$(16-16)^2 = (0)^2 = 0$

$(18-16)^2 = (2)^2 = 4$

$(20-16)^2 = (4)^2 = 16$

Sum of squared deviations ($\sum (x_i - \bar{x}_2)^2$) = $16 + 4 + 0 + 4 + 16 = 40$.

Variance ($\sigma_2^2$) = $\frac{\sum (x_i - \bar{x}_2)^2}{n} = \frac{40}{5} = 8$.

Standard Deviation ($\sigma_2$) = $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$.

This confirms that the standard deviation of the second dataset is $2\sigma$.

The standard deviation of the data 12, 14, 16, 18, 20 is $2\sigma$.

Given:

The data is: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.

To Find:

Mean Deviation about the Mean (MD).

Solution:

Step 1: Calculate the Mean ($\bar{x}$).

First, let's arrange the data in ascending order to make calculations easier:

38, 40, 42, 44, 46, 48, 54, 55, 63, 70.

Number of observations ($n$) = 10.

Sum of observations = $38 + 40 + 42 + 44 + 46 + 48 + 54 + 55 + 63 + 70 = 490$.

Mean ($\bar{x}$) = $\frac{\text{Sum of observations}}{\text{Number of observations}}$

$\bar{x} = \frac{490}{10}$

Step 2: Calculate the absolute deviations of each observation from the mean ($|x_i - \bar{x}|$).

|38 - 49| = |-11| = 11

|40 - 49| = |-9| = 9

|42 - 49| = |-7| = 7

|44 - 49| = |-5| = 5

|46 - 49| = |-3| = 3

|48 - 49| = |-1| = 1

|54 - 49| = |5| = 5

|55 - 49| = |6| = 6

|63 - 49| = |14| = 14

|70 - 49| = |21| = 21

Step 3: Calculate the sum of the absolute deviations ($\sum |x_i - \bar{x}|$).

Sum of absolute deviations = $11 + 9 + 7 + 5 + 3 + 1 + 5 + 6 + 14 + 21 = 82$.

Step 4: Calculate the Mean Deviation (MD).

MD = $\frac{\sum |x_i - \bar{x}|}{n}$

MD = $\frac{82}{10}$

Therefore, the mean deviation about the mean for the given data is 8.2.

Given:

Frequency distribution:

To Find:

Mean Deviation about the Mean (MD).

Solution:

To find the mean deviation from the mean for a frequency distribution, we follow these steps:

1. Find the class marks ($x_i$) for each class interval.

2. Calculate the mean ($\bar{x}$) of the distribution using the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

3. Calculate the absolute deviation of each class mark from the mean, $|x_i - \bar{x}|$.

4. Calculate the product of each absolute deviation and its corresponding frequency, $f_i |x_i - \bar{x}|$.

5. Calculate the Mean Deviation using the formula MD = $\frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$.

Let's create a table to organize the calculations:

Calculate the Mean ($\bar{x}$):

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1860}{50}$

Now, we complete the table with the absolute deviations and their products with frequencies:

Calculate the Mean Deviation (MD):

MD = $\frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$

MD = $\frac{558.8}{50}$

Therefore, the mean deviation about the mean for the given frequency distribution is 11.176.

Given:

Frequency distribution:

To Find:

Variance ($\sigma^2$) and Standard Deviation ($\sigma$).

Solution:

To calculate the variance and standard deviation for a frequency distribution, we use the following steps:

1. Find the class mark ($x_i$) for each class interval.

2. Calculate the mean ($\bar{x}$) of the distribution using $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

3. Calculate the squared deviation of each class mark from the mean, $(x_i - \bar{x})^2$.

4. Calculate the product of each squared deviation and its corresponding frequency, $f_i (x_i - \bar{x})^2$.

5. Calculate the Variance using the formula $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$.

6. Calculate the Standard Deviation, which is the square root of the variance ($\sigma = \sqrt{\sigma^2}$).

Let's create a table to organize the calculations:

Calculate the Mean ($\bar{x}$):

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1320}{40}$

Now, we calculate $(x_i - \bar{x})^2$ and $f_i (x_i - \bar{x})^2$:

Calculate the Variance ($\sigma^2$):

$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} = \frac{6640}{40}$

Calculate the Standard Deviation ($\sigma$):

$\sigma = \sqrt{\sigma^2} = \sqrt{166}$

Therefore:

Variance ($\sigma^2$) = 166

Standard Deviation ($\sigma$) $\approx$ 12.884

Given:

Frequency distribution:

To Find:

Mean Deviation about the Median (MD).

Solution:

To find the mean deviation about the median, we need to follow these steps:

1. Find the class marks ($x_i$) for each class interval.

2. Calculate the cumulative frequencies (c.f.).

3. Determine the median class by finding the class interval corresponding to $\frac{N}{2}$, where $N$ is the total frequency.

4. Calculate the median using the formula: Median = $L + \frac{\frac{N}{2} - C}{f} \times h$, where:

• $L$ is the lower limit of the median class.
• $N$ is the total frequency.
• $C$ is the cumulative frequency of the class preceding the median class.
• $f$ is the frequency of the median class.
• $h$ is the class width.

5. Calculate the absolute deviation of each class mark from the median, $|x_i - \text{Median}|$.

6. Calculate the product of each absolute deviation and its corresponding frequency, $f_i |x_i - \text{Median}|$.

7. Calculate the Mean Deviation about the Median using the formula: MD = $\frac{\sum f_i |x_i - \text{Median}|}{\sum f_i}$.

Let's construct a table for these calculations:

Calculate the Median:

The total frequency $N = 50$.

The position of the median is $\frac{N}{2} = \frac{50}{2} = 25$.

We look for the class interval where the cumulative frequency is just greater than or equal to 25. This is the 30-40 class, as its cumulative frequency is 40.

Median Class = 30-40.

Using the median formula:

Median = $L + \frac{\frac{N}{2} - C}{f} \times h$

Here, $L = 30$ (lower limit of the median class)

$N/2 = 25$

$C = 25$ (cumulative frequency of the class preceding the median class, i.e., 20-30 class)

$f = 15$ (frequency of the median class, i.e., 30-40 class)

$h = 10$ (class width)

Median = $30 + \frac{25 - 25}{15} \times 10$

Median = $30 + \frac{0}{15} \times 10$

Median = $30 + 0 = 30$.

Now, we calculate the absolute deviations from the median ($|x_i - \text{Median}| = |x_i - 30|$) and their products with frequencies ($f_i |x_i - 30|$):

Calculate the Mean Deviation about the Median (MD):

MD = $\frac{\sum f_i |x_i - \text{Median}|}{\sum f_i}$

MD = $\frac{570}{50}$

Therefore, the mean deviation about the median for the given frequency distribution is 11.4.

Given:

Number of observations ($n$) = 7

Mean ($\bar{x}$) = 8

Variance ($\sigma^2$) = 16

Five of the observations are 2, 4, 10, 12, 14.

To Find:

The remaining two observations.

Solution:

Let the remaining two observations be $x$ and $y$.

The seven observations are 2, 4, 10, 12, 14, $x$, and $y$.

Step 1: Use the mean to form an equation.

The formula for the mean is $\bar{x} = \frac{\sum x_i}{n}$.

$8 = \frac{2 + 4 + 10 + 12 + 14 + x + y}{7}$

$8 \times 7 = 42 + x + y$

$56 = 42 + x + y$

$x + y = 56 - 42$

Step 2: Use the variance to form another equation.

The formula for variance is $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$.

First, calculate the squared deviations for the known observations from the mean (8):

For 2: $(2 - 8)^2 = (-6)^2 = 36$

For 4: $(4 - 8)^2 = (-4)^2 = 16$

For 10: $(10 - 8)^2 = (2)^2 = 4$

For 12: $(12 - 8)^2 = (4)^2 = 16$

For 14: $(14 - 8)^2 = (6)^2 = 36$

The squared deviations for the unknown observations are $(x - 8)^2$ and $(y - 8)^2$.

Now, substitute these into the variance formula:

$16 = \frac{36 + 16 + 4 + 16 + 36 + (x - 8)^2 + (y - 8)^2}{7}$

$16 \times 7 = 108 + (x - 8)^2 + (y - 8)^2$

$112 = 108 + (x - 8)^2 + (y - 8)^2$

$(x - 8)^2 + (y - 8)^2 = 112 - 108$

Step 3: Solve the system of equations (i) and (ii).

From equation (i), $y = 14 - x$. Substitute this into equation (ii):

$(x - 8)^2 + ((14 - x) - 8)^2 = 4$

$(x - 8)^2 + (6 - x)^2 = 4$

Expand the terms:

$(x^2 - 16x + 64) + (36 - 12x + x^2) = 4$

Combine like terms:

$2x^2 - 28x + 100 = 4$

$2x^2 - 28x + 100 - 4 = 0$

$2x^2 - 28x + 96 = 0$

Divide the equation by 2:

$x^2 - 14x + 48 = 0$

Factor the quadratic equation:

We need two numbers that multiply to 48 and add up to -14. These numbers are -6 and -8.

$(x - 6)(x - 8) = 0$

This gives two possible values for $x$: $x = 6$ or $x = 8$.

Case 1: If $x = 6$

Substitute $x=6$ into equation (i): $6 + y = 14 \implies y = 8$.

The remaining two observations are 6 and 8.

Case 2: If $x = 8$

Substitute $x=8$ into equation (i): $8 + y = 14 \implies y = 6$.

The remaining two observations are 8 and 6.

In both cases, the remaining two observations are 6 and 8.

Verification:

Observations: 2, 4, 10, 12, 14, 6, 8.

Mean = $\frac{2+4+10+12+14+6+8}{7} = \frac{56}{7} = 8$ (Correct)

Deviations from the mean: (2-8), (4-8), (10-8), (12-8), (14-8), (6-8), (8-8)

Deviations: -6, -4, 2, 4, 6, -2, 0

Squared Deviations: 36, 16, 4, 16, 36, 4, 0

Sum of Squared Deviations = $36 + 16 + 4 + 16 + 36 + 4 + 0 = 112$

Variance = $\frac{112}{7} = 16$ (Correct)

Therefore, the remaining two observations are 6 and 8.

Given:

Group A:

Mean ($\bar{x}_A$) = 525

Standard Deviation ($\sigma_A$) = 100

Group B:

Mean ($\bar{x}_B$) = 480

Standard Deviation ($\sigma_B$) = 90

To Determine:

Which group is more variable.

Solution:

To compare the variability of two groups with different means, we use the coefficient of variation (CV).

The formula for the coefficient of variation is:

A higher coefficient of variation indicates greater relative variability.

Calculate the Coefficient of Variation for Group A:

CV$_A$ = $\frac{\sigma_A}{\bar{x}_A} \times 100\%$

CV$_A$ = $\frac{100}{525} \times 100\%$

CV$_A$ = $\frac{10000}{525} \%$

Calculate the Coefficient of Variation for Group B:

CV$_B$ = $\frac{\sigma_B}{\bar{x}_B} \times 100\%$

CV$_B$ = $\frac{90}{480} \times 100\%$

CV$_B$ = $\frac{9000}{480} \%$

CV$_B$ = $\frac{900}{48} \%$

CV$_B$ = $\frac{150}{8} \%$

CV$_B$ = $\frac{75}{4} \%$

Comparison:

CV$_A \approx 19.05\%$

CV$_B = 18.75\%$

Since CV$_A >$ CV$_B$ (19.05% > 18.75%), Group A is more variable than Group B in terms of relative dispersion.

Therefore, Group A is more variable.

Given:

Frequency distribution:

To Find:

Mean ($\bar{x}$), Variance ($\sigma^2$), and Standard Deviation ($\sigma$).

Solution:

To calculate the mean, variance, and standard deviation for a frequency distribution, we follow these steps:

1. Find the class mark ($x_i$) for each class interval.

2. Calculate the mean ($\bar{x}$) using the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

3. Calculate the squared deviation of each class mark from the mean, $(x_i - \bar{x})^2$.

4. Calculate the product of each squared deviation and its corresponding frequency, $f_i (x_i - \bar{x})^2$.

5. Calculate the Variance using the formula $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$.

6. Calculate the Standard Deviation, which is the square root of the variance ($\sigma = \sqrt{\sigma^2}$).

Let's create a table for these calculations:

Calculate the Mean ($\bar{x}$):

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3080}{50}$

Now, we calculate $(x_i - \bar{x})^2$ and $f_i (x_i - \bar{x})^2$:

Calculate the Variance ($\sigma^2$):

$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} = \frac{8922}{50}$

Calculate the Standard Deviation ($\sigma$):

$\sigma = \sqrt{\sigma^2} = \sqrt{178.44}$

Therefore:

Mean ($\bar{x}$) = 61.6

Variance ($\sigma^2$) = 178.44

Standard Deviation ($\sigma$) $\approx$ 13.358

Given:

Frequency distribution of diameters of circles:

Note: The class intervals are inclusive. We need to convert them to exclusive class intervals for calculation.

To Find:

Mean diameter ($\bar{x}$) and Standard Deviation ($\sigma$).

Solution:

Step 1: Adjust Class Intervals to Exclusive Form.

The given class intervals are inclusive (e.g., 33-36 includes both 33 and 36). To convert them to exclusive class intervals, we adjust the boundaries by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class.

Step 2: Calculate Class Marks ($x_i$).

The class mark is the midpoint of each class interval: $x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$.

Step 3: Calculate Mean ($\bar{x}$).

The formula for the mean is $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

Step 4: Calculate Variance ($\sigma^2$) and Standard Deviation ($\sigma$).

The formula for variance is $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$.

Let's create a table for these calculations:

Calculate the Mean Diameter ($\bar{x}$):

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3821}{90}$

Now, we calculate $(x_i - \bar{x})^2$ and $f_i (x_i - \bar{x})^2$. We will use the mean $\bar{x} \approx 42.456$ for these calculations.

Calculate the Variance ($\sigma^2$):

$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} = \frac{2255.69}{90}$

Calculate the Standard Deviation ($\sigma$):

$\sigma = \sqrt{\sigma^2} = \sqrt{25.063}$

Therefore:

Mean diameter ($\bar{x}$) $\approx$ 42.456 mm

Standard Deviation ($\sigma$) $\approx$ 5.006 mm

Given:

Firm A:

Number of workers ($n_A$) = 586

Average monthly wage ($\bar{x}_A$) = ₹5253

Variance ($\sigma_A^2$) = 100

Firm B:

Number of workers ($n_B$) = 648

Average monthly wage ($\bar{x}_B$) = ₹5253

Variance ($\sigma_B^2$) = 121

To Find:

(i) Which firm pays out a larger amount as monthly wages?

(ii) Which firm shows greater variability in individual wages?

Solution:

Part (i): Which firm pays out larger amount as monthly wages?

To find the total amount paid as monthly wages, we multiply the average monthly wage by the number of workers.

Total monthly wages for Firm A = Number of workers in A $\times$ Average monthly wage in A

Total wages (A) = $586 \times 5253$

Total wages (A) = $3077718$

Total monthly wages for Firm B = Number of workers in B $\times$ Average monthly wage in B

Total wages (B) = $648 \times 5253$

Total wages (B) = $3403844$

Comparing the total monthly wages:

Total wages (B) > Total wages (A) ($3403844 > 3077718$)

Therefore, Firm B pays out a larger amount as monthly wages.

Part (ii): Which firm shows greater variability in individual wages?

Variability in individual wages can be compared using the standard deviation. A higher standard deviation indicates greater variability.

Standard Deviation for Firm A ($\sigma_A$) = $\sqrt{\text{Variance}_A} = \sqrt{100} = 10$.

Standard Deviation for Firm B ($\sigma_B$) = $\sqrt{\text{Variance}_B} = \sqrt{121} = 11$.

Comparing the standard deviations:

$\sigma_B = 11$ and $\sigma_A = 10$.

Since $\sigma_B > \sigma_A$ (11 > 10), Firm B shows greater variability in individual wages.

To be more precise when means are equal, we can also use the coefficient of variation. However, in this case, since the means are equal, the standard deviation directly indicates the variability.

Therefore, Firm B shows greater variability in individual wages.

Given:

Frequency distribution of daily wages:

Total number of workers ($N$) = 50.

To Find:

Mean Deviation about the Mean (MD).

Solution:

To calculate the mean deviation about the mean for a frequency distribution, we follow these steps:

1. Find the class marks ($x_i$) for each class interval.

2. Calculate the mean ($\bar{x}$) of the distribution using the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

3. Calculate the absolute deviation of each class mark from the mean, $|x_i - \bar{x}|$.

4. Calculate the product of each absolute deviation and its corresponding frequency, $f_i |x_i - \bar{x}|$.

5. Calculate the Mean Deviation using the formula MD = $\frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$.

Let's construct a table for these calculations:

Calculate the Mean ($\bar{x}$):

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{7420}{50}$

Now, we calculate the absolute deviations from the mean ($|x_i - \bar{x}|$) and their products with frequencies ($f_i |x_i - \bar{x}|$):

Calculate the Mean Deviation about the Mean (MD):

MD = $\frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$

MD = $\frac{1283.2}{50}$

Therefore, the mean deviation about the mean for the given data is 25.664.

Given:

Number of observations ($n$) = 100

Incorrect Mean ($\bar{x}_{inc}$) = 20

Incorrect Standard Deviation ($\sigma_{inc}$) = 3

Wrongly recorded observations: 21, 21, 18

Correct observations: 18, 8, 12

To Find:

Corrected Mean ($\bar{x}_{cor}$) and Corrected Standard Deviation ($\sigma_{cor}$).

Solution:

Part 1: Calculate the Corrected Mean.

The formula for the mean is $\bar{x} = \frac{\sum x_i}{n}$.

From the incorrect mean, we can find the sum of the incorrect observations:

$\sum x_{inc} = \bar{x}_{inc} \times n = 20 \times 100 = 2000$.

Now, we need to adjust this sum by removing the wrongly recorded observations and adding the correct ones.

Sum of wrongly recorded observations = $21 + 21 + 18 = 60$.

Sum of correct observations = $18 + 8 + 12 = 38$.

The corrected sum of observations ($\sum x_{cor}$) is:

$\sum x_{cor} = \sum x_{inc} - (\text{Sum of wrong observations}) + (\text{Sum of correct observations})$

$\sum x_{cor} = 2000 - 60 + 38$

$\sum x_{cor} = 1940 + 38$

The corrected mean ($\bar{x}_{cor}$) is:

$\bar{x}_{cor} = \frac{\sum x_{cor}}{n} = \frac{1978}{100}$

Part 2: Calculate the Corrected Standard Deviation.

The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.

From the incorrect variance, we can find the sum of the squares of the incorrect observations:

$\sigma_{inc}^2 = 3^2 = 9$.

$9 = \frac{\sum x_{inc}^2}{100} - (20)^2$

$9 = \frac{\sum x_{inc}^2}{100} - 400$

$9 + 400 = \frac{\sum x_{inc}^2}{100}$

$409 = \frac{\sum x_{inc}^2}{100}$

$\sum x_{inc}^2 = 409 \times 100 = 40900$.

Now, we need to adjust the sum of squares by removing the squares of the wrongly recorded observations and adding the squares of the correct ones.

Square of wrongly recorded observations: $21^2 = 441$, $21^2 = 441$, $18^2 = 324$.

Sum of squares of wrong observations = $441 + 441 + 324 = 1206$.

Square of correct observations: $18^2 = 324$, $8^2 = 64$, $12^2 = 144$.

Sum of squares of correct observations = $324 + 64 + 144 = 532$.

The corrected sum of squares of observations ($\sum x_{cor}^2$) is:

$\sum x_{cor}^2 = \sum x_{inc}^2 - (\text{Sum of squares of wrong observations}) + (\text{Sum of squares of correct observations})$

$\sum x_{cor}^2 = 40900 - 1206 + 532$

$\sum x_{cor}^2 = 39694 + 532$

Now we can calculate the corrected variance ($\sigma_{cor}^2$):

$\sigma_{cor}^2 = \frac{\sum x_{cor}^2}{n} - (\bar{x}_{cor})^2$

$\sigma_{cor}^2 = \frac{40226}{100} - (19.78)^2$

$\sigma_{cor}^2 = 402.26 - 391.2484$

The corrected standard deviation ($\sigma_{cor}$) is:

$\sigma_{cor} = \sqrt{11.0116}$

Therefore:

Corrected Mean = 19.78

Corrected Standard Deviation $\approx$ 3.318

Given:

Frequency distribution:

To Find:

Coefficient of Variation (CV).

Solution:

To calculate the coefficient of variation, we need to find the mean ($\bar{x}$) and the standard deviation ($\sigma$) of the given data.

The formula for the coefficient of variation is: CV = $\frac{\sigma}{\bar{x}} \times 100\%$.

Step 1: Calculate the Mean ($\bar{x}$).

We need to find the class marks ($x_i$) and then use the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

Mean ($\bar{x}$) = $\frac{\sum f_i x_i}{\sum f_i} = \frac{2260}{50}$

Step 2: Calculate the Standard Deviation ($\sigma$).

We use the formula $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$.

Variance ($\sigma^2$) = $\frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} = \frac{6298}{50}$

Standard Deviation ($\sigma$) = $\sqrt{125.96}$

Step 3: Calculate the Coefficient of Variation (CV).

CV = $\frac{\sigma}{\bar{x}} \times 100\%$

CV = $\frac{11.223}{45.2} \times 100\%$

Therefore, the coefficient of variation for the given data is approximately 24.83%.

Given:

Frequency distribution:

To Find:

Variance ($\sigma^2$) and Standard Deviation ($\sigma$) using the Step-deviation method.

Solution:

The Step-deviation method is a simplified way to calculate the mean, variance, and standard deviation when the class intervals are equal.

Steps involved:

1. Find the class mark ($x_i$) for each class interval.

2. Choose a convenient assumed mean ($a$). It's usually the class mark of the middle class.

3. Calculate the step deviation ($d_i = \frac{x_i - a}{h}$), where $h$ is the class width.

4. Calculate $f_i d_i$ and $f_i d_i^2$.

5. Calculate the mean using $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \times h$.

6. Calculate the variance using $\sigma^2 = h^2 \left[ \frac{\sum f_i d_i^2}{\sum f_i} - \left( \frac{\sum f_i d_i}{\sum f_i} \right)^2 \right]$.

7. Calculate the standard deviation ($\sigma$) by taking the square root of the variance.

Let's construct a table for these calculations:

The class width ($h$) for all intervals is 10.

The middle class is 20-30, so let's assume the mean ($a$) is the class mark of this class, which is 25.

Now, let's fill in the $d_i$, $f_i d_i$, and $f_i d_i^2$ columns:

Calculate the Mean ($\bar{x}$):

$\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \times h$

$\bar{x} = 25 + \frac{10}{50} \times 10$

$\bar{x} = 25 + \frac{1}{5} \times 10$

$\bar{x} = 25 + 2$

Calculate the Variance ($\sigma^2$):

$\sigma^2 = h^2 \left[ \frac{\sum f_i d_i^2}{\sum f_i} - \left( \frac{\sum f_i d_i}{\sum f_i} \right)^2 \right]$

$\sigma^2 = 10^2 \left[ \frac{68}{50} - \left( \frac{10}{50} \right)^2 \right]$

$\sigma^2 = 100 \left[ \frac{68}{50} - \left( \frac{1}{5} \right)^2 \right]$

$\sigma^2 = 100 \left[ \frac{68}{50} - \frac{1}{25} \right]$

To subtract the fractions, find a common denominator, which is 50:

$\sigma^2 = 100 \left[ \frac{68}{50} - \frac{2}{50} \right]$

$\sigma^2 = 100 \left[ \frac{66}{50} \right]$

$\sigma^2 = 100 \times \frac{33}{25}$

$\sigma^2 = 4 \times 33$

Calculate the Standard Deviation ($\sigma$):

$\sigma = \sqrt{\sigma^2} = \sqrt{132}$

Therefore:

Mean ($\bar{x}$) = 27

Variance ($\sigma^2$) = 132

Standard Deviation ($\sigma$) $\approx$ 11.489