Chapter 5 Data Handling (Additional Questions)

The correct answer is (B) Frequency.

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In statistics, the frequency of an observation is the number of times the observation occurs in the data set. It is a fundamental concept used in summarizing and analyzing data.

The correct answer is (B) Primary data.

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Primary data refers to data that is collected directly by the researcher or investigator for the first time for a specific research purpose. This contrasts with secondary data, which is collected by someone else and used by the researcher.

The correct answer is (C) Bar graph.

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A bar graph is the most suitable type of graph for comparing the sales of different products in a shop.

This is because bar graphs use separate bars to represent the value (sales) for each distinct category (product), making direct comparison of the heights of the bars easy and clear.

The correct answer is (D) Class size.

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In a frequency distribution, the class size (or class width) is the difference between the upper limit and the lower limit of a class interval.

The correct answer is (C) Class mark.

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The Class mark is the mid-point of a class interval. It is calculated by adding the upper limit and the lower limit of the class interval and dividing the sum by 2.

Class Mark $=$ $\frac{\text{Upper Limit} + \text{Lower Limit}}{2}$.

The correct answer is (C) Histogram.

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A histogram is a graphical representation of grouped data, specifically frequency distributions with continuous class intervals.

In a histogram, the bars are adjacent to each other with no gaps, representing consecutive class intervals.

A bar graph, on the other hand, is used for categorical data and typically has gaps between the bars.

The correct answer is (D) Pie chart.

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A pie chart is a circular statistical graphic which is divided into slices to illustrate numerical proportion.

In a pie chart, the arc length of each slice, is proportional to the quantity it represents, showing the relationship between the part (each category) and the whole (total). This makes it ideal for visualizing proportions or percentages of a whole.

The correct answer is (B) $360^\circ$.

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A pie chart represents a whole, which is a full circle.

The total angle at the center of a circle is always $360^\circ$.

Each sector in a pie chart corresponds to a portion of the whole, and its central angle is proportional to that portion.

Therefore, the sum of all the central angles representing the different categories in a pie chart must equal the total angle of the circle, which is $360^\circ$.

The correct answer is (B) Event.

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In probability, an event is a set of outcomes of an experiment to which a probability is assigned.

An event can be a single outcome or a collection of several outcomes from the sample space.

The sample space is the set of all possible outcomes, and a trial is a single performance of the experiment.

The correct answer is (B) $0$.

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An impossible event is an event that has no chance of occurring.

In probability theory, an impossible event is represented by the empty set ($\emptyset$).

The probability of an impossible event is always $0$.

The correct answer is (B) $\frac{1}{2}$.

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When a standard six-sided die is rolled, the possible outcomes are $1, 2, 3, 4, 5, 6$.

The total number of possible outcomes is $6$.

We want to find the probability of getting a prime number. The prime numbers in the set of outcomes $\{1, 2, 3, 4, 5, 6\}$ are $2, 3, 5$.

The number of favorable outcomes (getting a prime number) is $3$.

The probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

In this case, the event is getting a prime number.

$P(\text{Prime Number}) = \frac{3}{6}$

Simplifying the fraction:

$P(\text{Prime Number}) = \frac{1}{2}$

The correct answer is (A) $\frac{3}{10}$.

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First, we find the total number of balls in the bag.

Total number of balls $=$ Number of red balls $+$ Number of blue balls $+$ Number of green balls

Total number of balls $=$ $5 + 3 + 2 = 10$

The number of favorable outcomes (drawing a blue ball) is the number of blue balls, which is $3$.

The probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

For the event of drawing a blue ball:

$P(\text{Drawing a blue ball}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}}$

$P(\text{Drawing a blue ball}) = \frac{3}{10}$

The correct answer is (B) $1 - P(E)$.

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In probability theory, the sum of the probability of an event happening and the probability of the event not happening (also called the complement of the event) is always equal to $1$.

Let $P(E)$ be the probability of event E happening.

Let $P(\text{not } E)$ be the probability of event E not happening, often denoted as $P(E')$ or $P(\overline{E})$.

The fundamental relationship is:

$P(E) + P(\text{not } E) = 1$

To find the probability of the event not happening, we rearrange the equation:

$P(\text{not } E) = 1 - P(E)$

The correct answer is (A) $17$.

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The range of a data set is the difference between the highest value and the lowest value in the set.

Given data set: 15, 20, 8, 12, 25, 10.

First, identify the maximum value in the data set.

Maximum value $=$ $25$

Next, identify the minimum value in the data set.

Minimum value $=$ $8$

Now, calculate the range by subtracting the minimum value from the maximum value.

Range $=$ Maximum value $-$ Minimum value

Range $=$ $25 - 8$

Range $=$ $17$

The correct answer is (B) 20-30.

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In frequency distributions with continuous class intervals like 10-20, 20-30, 30-40, etc., a standard convention is followed.

The convention is that the lower limit of a class interval is included in the class, but the upper limit is excluded. This prevents any data value from falling into two classes or being left out.

So, the interval 10-20 represents values from 10 up to (but not including) 20. Mathematically, this is often written as $[10, 20)$.

The interval 20-30 represents values from 20 up to (but not including) 30. Mathematically, this is often written as $[20, 30)$.

Therefore, the data value 20 is included in the class interval where it is the lower limit, which is 20-30.

The correct answers are (A) It has more than one possible outcome. and (B) It is not possible to predict the outcome in advance.

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A random experiment is an experiment or a process for which the outcome cannot be determined in advance, but the set of all possible outcomes is known.

Let's analyze the options:

(A) It has more than one possible outcome. This is a characteristic of a random experiment. If there were only one possible outcome, the result would be certain, not random.

(B) It is not possible to predict the outcome in advance. This is a key characteristic of a random experiment. Before the experiment is conducted, the specific outcome cannot be known with certainty.

(C) The outcome is always the same every time it is performed. This is incorrect. If the outcome were always the same, it would be a deterministic experiment, not a random one.

(D) All outcomes are equally likely. This is a characteristic of some random experiments (like flipping a fair coin or rolling a fair die), but it is not a requirement for all random experiments. For example, drawing a card from a deck where certain cards have been removed is a random experiment, but the outcomes might not be equally likely.

Therefore, the essential characteristics of a random experiment among the given options are having more than one possible outcome and the inability to predict the outcome in advance.

The correct answer is (C) A is true, but R is false.

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Assertion (A): Histograms are used to represent continuous grouped data.

This statement is True. Histograms are graphical representations of the frequency distribution of continuous numerical data, where the data is grouped into class intervals.

Reason (R): The bars in a histogram have gaps between them to show distinct categories.

This statement is False. The bars in a histogram are drawn adjacent to each other with no gaps. This adjacency visually represents the continuous nature of the data and the class intervals. Bars with gaps are a characteristic of bar graphs, which are used for discrete or categorical data.

Since Assertion (A) is true and Reason (R) is false, option (C) is the correct choice.

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

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Let's match each term with its correct definition:

(i) Experiment: A process that results in some well-defined outcomes. This matches definition (c).

(ii) Outcome: A single result of an experiment. This matches definition (d).

(iii) Event: A collection of one or more outcomes. This matches definition (b).

(iv) Sample Space: The set of all possible outcomes of an experiment. This matches definition (a).

Combining these matches, we get: (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a), which corresponds to option (A).

The correct answer is (B) $-0.2$.

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The probability of any event must satisfy the following conditions:

1. The probability of any event E is a number between $0$ and $1$, inclusive. That is, $0 \leq P(E) \leq 1$.

2. The probability of an impossible event is $0$.

3. The probability of a certain event is $1$.

Let's examine the given options:

(A) $0.7$: This value is between $0$ and $1$ ($0 \leq 0.7 \leq 1$). So, it can be a probability.

(B) $-0.2$: This value is less than $0$. Probabilities cannot be negative.

(C) $1$: This value is equal to $1$. This is the probability of a certain event.

(D) $\frac{3}{4}$: This value is $0.75$, which is between $0$ and $1$ ($0 \leq 0.75 \leq 1$). So, it can be a probability.

Therefore, $-0.2$ cannot be a probability of an event because probabilities must be non-negative.

Solution:

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The total number of students surveyed is given as 50.

The number of students who like Banana is 12.

In a pie chart, the central angle for a sector is proportional to the value it represents.

The sum of all central angles in a pie chart is $360^\circ$.

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The formula to calculate the central angle for a sector is:

Central angle = $\left(\frac{\text{Value of the component}}{\text{Total value}}\right) \times 360^\circ$

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For the 'Banana' sector, the value of the component is the number of students who like Banana, which is 12.

The total value is the total number of students, which is 50.

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Central angle for 'Banana' sector = $\left(\frac{\text{Number of students who like Banana}}{\text{Total number of students}}\right) \times 360^\circ$

Central angle for 'Banana' sector = $\left(\frac{12}{50}\right) \times 360^\circ$

Central angle for 'Banana' sector = $\frac{12}{50} \times 360^\circ$

Central angle for 'Banana' sector = $\frac{12}{\cancel{50}_{5} \times \cancel{360}^{36}^\circ$

Central angle for 'Banana' sector = $\frac{12}{5} \times 36^\circ$

Central angle for 'Banana' sector = $12 \times \frac{36}{5}^\circ$

Central angle for 'Banana' sector = $12 \times 7.2^\circ$

Central angle for 'Banana' sector = $86.4^\circ$

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Comparing the calculated angle with the given options:

(A) $86.4^\circ$

(B) $72^\circ$

(C) $90^\circ$

(D) $60^\circ$

The calculated angle matches option (A).

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Final Answer:

The central angle for the 'Banana' sector is $86.4^\circ$.

(A) $86.4^\circ$

Solution:

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The question asks for the term that defines the difference between the highest and lowest observation in a data set.

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Let the given data set be $X = \{x_1, x_2, ..., x_n\}$.

Let the highest observation be $x_{max}$ and the lowest observation be $x_{min}$.

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The Range of a data set is defined as the difference between the highest and the lowest observations.

Range = Highest observation - Lowest observation

Range = $x_{max} - x_{min}$

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Let's look at the options:

(A) Class interval: A range of values used to group data, typically in frequency distributions.

(B) Frequency: The number of times a particular observation or value occurs in a data set.

(C) Range: The difference between the highest and lowest values in a data set.

(D) Class size: The difference between the upper and lower limits of a class interval.

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Based on the definition, the term that describes the difference between the highest and lowest observation is the Range.

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Final Answer:

The difference between the highest and lowest observation in a data set is called the Range.

(C) Range

Solution:

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The given data represents the daily wages (in $\textsf{₹}$) of 25 employees. The data points are:

200, 250, 220, 280, 300, 250, 220, 280, 300, 200,

250, 280, 300, 220, 250, 200, 280, 300, 220, 250,

280, 300, 220, 250, 200

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We are asked to find the frequency of the class interval 225-250 when creating a frequency distribution table with class intervals 200-225, 225-250, etc., described as "exclusive".

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The common convention for class intervals like $[L, U)$ means including the lower limit $L$ and excluding the upper limit $U$. In this case, the interval 225-250 would mean wages $\geq 225$ and $< 250$. If we apply this strictly to the given discrete data, no values fall in this range (220 is less than 225, and 250 is not less than 250).

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However, considering the discrete nature of the data points (all are multiples of 10 or 50) and the provided options, it is likely that the class intervals are intended to group the specific wage values present in the data.

Given the options, the most plausible interpretation is that the interval "225-250" specifically refers to the data point $250$, or perhaps values $\geq 225$ and $\leq 250$. Given the standard practice of non-overlapping intervals and the wording "200-225, 225-250", it is most probable that 250 is included in the second interval (225-250), while 225 would be included in the first (200-225) if it were present in the data.

Let's assume the class interval 225-250 includes the value 250 and excludes values less than 225.

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We need to count how many times the value 250 appears in the given data set.

• Data row 1: 250 appears 2 times.
• Data row 2: 250 appears 3 times.
• Data row 3: 250 appears 2 times.

Total frequency of 250 = $2 + 3 + 2 = 7$.

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Under the interpretation that the class interval 225-250 includes the value 250, the frequency of this class interval is 7.

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Comparing this frequency with the given options:

(A) 5

(B) 6

(C) 7

(D) 8

The calculated frequency matches option (C).

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Final Answer:

The frequency of the class interval 225-250 (interpreting it as including the wage $\textsf{₹}250$) is 7.

(C) 7

Solution:

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The class mark of a class interval is the midpoint of the interval. It is calculated as the average of the lower and upper class limits.

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The given class interval is 275-300.

Lower limit of the class interval ($L$) = 275

Upper limit of the class interval ($U$) = 300

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The formula for the class mark is:

Class Mark = $\frac{\text{Lower Limit} + \text{Upper Limit}}{2}$

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Substituting the values:

Class Mark = $\frac{275 + 300}{2}$

Class Mark = $\frac{575}{2}$

Class Mark = $287.5$

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Comparing the calculated class mark with the given options:

(A) 280

(B) 287.5

(C) 290

(D) 275.5

The calculated class mark is 287.5, which matches option (B).

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Final Answer:

The class mark for the class interval 275-300 is 287.5.

(B) 287.5

Solution:

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A histogram is a graphical representation used for displaying the frequency distribution of continuous data.

Unlike bar graphs which are used for categorical or discrete data and have spaces between the bars, the bars in a histogram are adjacent, representing continuous intervals or ranges of data.

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Histograms require data to be divided into class intervals or bins. This grouping is necessary to visualize the distribution pattern of continuous variables.

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Let's analyze the given options:

(A) Number of students in each class: This is discrete data, where the values (number of students) correspond to distinct categories (classes). A bar graph is appropriate here.

(B) Favourite colours of students: This is categorical or qualitative data. A bar graph or a pie chart is suitable for this type of data.

(C) Heights of students in a class: Height is a continuous variable. The heights can take any value within a given range. To represent the distribution of heights using a histogram, the data must be grouped into class intervals (e.g., 150 cm - 155 cm, 155 cm - 160 cm, etc.).

(D) Number of cars of different brands sold: This is discrete data, where the values (number of cars) correspond to distinct categories (brands). A bar graph is appropriate here.

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Therefore, data that is continuous, such as the heights of students, requires grouping into class intervals before it can be effectively represented using a histogram.

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Final Answer:

The type of data that requires grouping before representing it using a histogram is continuous data.

Among the given options, Heights of students in a class represents continuous data.

(C) Heights of students in a class

Solution:

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In probability, an experiment is a process that results in one of several possible outcomes.

The set of all possible outcomes of an experiment is called the sample space. It is usually denoted by $S$.

An event is any subset of the sample space.

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In this problem, the experiment is rolling a die.

The possible outcomes are the numbers {1, 2, 3, 4, 5, 6}.

So, the sample space is $S = \{1, 2, 3, 4, 5, 6\}$.

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Now let's examine each option based on the definition of an event:

(A) Rolling a 7: The outcome 7 is not in the sample space $S = \{1, 2, 3, 4, 5, 6\}$. This describes an impossible event, which corresponds to the empty set, $\emptyset$. The empty set is a subset of any set, so $\emptyset \subset S$. Thus, 'Rolling a 7' represents the event $\emptyset$. While technically an event, this option describes an outcome not in the sample space.

(B) Getting an even number ({2, 4, 6}): This option describes the collection of outcomes {2, 4, 6}. This set is a subset of the sample space $S = \{1, 2, 3, 4, 5, 6\}$. Since $\{2, 4, 6\} \subset S$, this is an event.

(C) The number of dots on the top face: This is a description of the variable or the outcome itself, not a specific subset of outcomes. It refers to any element in the sample space (1, 2, 3, 4, 5, or 6), not a particular event defined as a collection of these outcomes.

(D) Rolling the die: This describes the experiment being performed, not a specific result or set of results from the experiment (which is what an event is).

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Comparing the options, option (B) is a set of outcomes that is a subset of the sample space, fitting the definition of an event directly.

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Final Answer:

The option that represents an event is getting an even number, which corresponds to the subset $\{2, 4, 6\}$ of the sample space.

(B) Getting an even number ({2, 4, 6})

Solution:

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When a fair coin is tossed, there are two possible outcomes: Head (H) or Tail (T).

The set of all possible outcomes is called the Sample Space, denoted by $S$.

$S = \{H, T\}$

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The total number of possible outcomes is the number of elements in the sample space, which is $|S| = 2$.

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The event we are interested in is getting a tail.

Let $E$ be the event of getting a tail.

$E = \{T\}$

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The number of favourable outcomes for the event $E$ is the number of elements in $E$, which is $|E| = 1$.

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The Probability of an event $E$ is calculated as the ratio of the number of favourable outcomes to the total number of possible outcomes, assuming all outcomes are equally likely.

Probability $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{Getting a tail}) = \frac{|E|}{|S|} = \frac{1}{2}$

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Comparing the calculated probability with the given options:

(A) $0$

(B) $1$

(C) $\frac{1}{2}$

(D) $2$

The calculated probability is $\frac{1}{2}$, which matches option (C).

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Final Answer:

The probability of getting a tail when a coin is tossed is $\frac{1}{2}$.

(C) $\frac{1}{2}$

Solution:

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Equally likely outcomes are outcomes of a random experiment that have the same probability of occurring.

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Let's examine each option:

(A) Getting a head or a tail when tossing a fair coin.

When a fair coin is tossed, there are two possible outcomes: Head (H) and Tail (T). Assuming the coin is fair, the probability of getting a head is $P(H) = \frac{1}{2}$, and the probability of getting a tail is $P(T) = \frac{1}{2}$. Since $P(H) = P(T)$, the outcomes Head and Tail are equally likely.

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(B) Getting a '1' or a '6' when rolling a standard die.

When a standard (fair) six-sided die is rolled, there are six possible outcomes: {1, 2, 3, 4, 5, 6}. Assuming the die is fair, the probability of getting any specific face is $\frac{1}{6}$. Thus, the probability of getting a '1' is $P(\text{'1'}) = \frac{1}{6}$, and the probability of getting a '6' is $P(\text{'6'}) = \frac{1}{6}$. Since $P(\text{'1'}) = P(\text{'6'})$, the outcomes '1' and '6' are equally likely.

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(C) Getting a boy or a girl child.

The outcomes are having a boy or having a girl. While in reality the probabilities are slightly different, in many introductory probability contexts, it is assumed that the probability of having a boy is approximately $0.5$ and the probability of having a girl is approximately $0.5$. Under this common simplifying assumption, the outcomes Boy and Girl are treated as equally likely.

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(D) Picking any specific card from a well-shuffled deck.

A standard deck of cards has 52 unique cards. If the deck is well-shuffled, each of the 52 cards has an equal chance of being selected. The probability of picking any specific card is $\frac{1}{52}$. Therefore, picking any specific card (e.g., Ace of Spades, 7 of Hearts, etc.) are all equally likely outcomes.

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Considering the standard assumptions made in probability problems of this type (fairness, randomness), all the described scenarios involve outcomes that are considered equally likely.

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Final Answer:

All the options represent examples where the listed outcomes are equally likely under standard probability assumptions.

(A), (B), (C), (D)

Solution:

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We are given the number of students absent for 6 days. The data points are 2, 0, 1, 3, 5, and 4.

The total number of days for which the data is recorded is 6.

To find the average number of absent students per day, we need to calculate the mean of the given data.

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The formula for the mean (average) is:

Mean = $\frac{\text{Sum of observations}}{\text{Number of observations}}$

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

Sum = $2 + 0 + 1 + 3 + 5 + 4$

Sum = $3 + 3 + 5 + 4$

Sum = $6 + 5 + 4$

Sum = $11 + 4$

Sum = $15$

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Now, we calculate the average:

Average = $\frac{15}{6}$

Average = $\frac{5 \times \cancel{3}}{2 \times \cancel{3}}$

Average = $\frac{5}{2}$

Average = $2.5$

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Comparing the calculated average with the given options:

(A) 2.5

(B) 3

(C) 2.25

(D) 3.5

The calculated average is 2.5, which matches option (A).

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Final Answer:

The average number of absent students per day is 2.5.

(A) 2.5

Solution:

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The question asks for the number of students whose weight is between 35 kg (inclusive) and 45 kg (exclusive). This means we are interested in the students whose weight $W$ satisfies $35 \leq W < 45$.

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The given frequency distribution table provides the number of students in different weight categories (class intervals).

The class intervals and their corresponding frequencies (derived from the tally marks) are:

• Class Interval 30-35: Frequency = $\bcancel{||||}$ = 5 students. (Assuming this interval is $30 \leq W < 35$)
• Class Interval 35-40: Frequency = $\bcancel{||||}\ ||$ = 7 students. (Assuming this interval is $35 \leq W < 40$)
• Class Interval 40-45: Frequency = $\bcancel{||||}\ |||$ = 8 students. (Assuming this interval is $40 \leq W < 45$)
• Class Interval 45-50: Frequency = $||||$ = 4 students. (Assuming this interval is $45 \leq W < 50$)

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We need to find the number of students whose weight is in the range $35 \leq W < 45$. This range is covered by the class intervals where the lower limit is 35 or more, and the upper limit is less than 45.

Looking at the intervals and our assumptions based on standard practice:

• The interval 35-40 covers weights $35 \leq W < 40$. These students satisfy the condition $35 \leq W < 45$. The frequency is 7.
• The interval 40-45 covers weights $40 \leq W < 45$. These students also satisfy the condition $35 \leq W < 45$. The frequency is 8.

The other intervals (30-35 and 45-50) do not fall within the range $35 \leq W < 45$.

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To find the total number of students with weight between 35 kg (inclusive) and 45 kg (exclusive), we sum the frequencies of the intervals 35-40 and 40-45.

Total number of students = Frequency of (35-40) + Frequency of (40-45)

Total number of students = $7 + 8$

Total number of students = $15$

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Comparing the calculated number of students with the given options:

(A) 7

(B) 8

(C) 10

(D) 15

The calculated number of students is 15, which matches option (D).

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Final Answer:

The number of students who have a weight between 35 kg (inclusive) and 45 kg (exclusive) is 15.

(D) 15

Solution:

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A pie chart represents the whole data as a circle, where the total angle at the center is $360^\circ$.

Each sector of the pie chart represents a part of the data, and its central angle is proportional to the value of that part.

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We are given that the central angle for the rent sector is $90^\circ$.

The total angle of the pie chart is $360^\circ$.

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To find the percentage of the total expenditure spent on rent, we can use the formula:

Percentage of expenditure = $\left(\frac{\text{Central angle of the sector}}{\text{Total central angle}}\right) \times 100\%$

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In this case:

Central angle of rent sector = $90^\circ$

Total central angle = $360^\circ$

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Percentage of expenditure on rent = $\left(\frac{90^\circ}{360^\circ}\right) \times 100\%$

Percentage of expenditure on rent = $\frac{90}{360} \times 100\%$

Percentage of expenditure on rent = $\frac{\cancel{90}^1}{\cancel{360}_4} \times 100\%$

Percentage of expenditure on rent = $\frac{1}{4} \times 100\%$

Percentage of expenditure on rent = $\frac{100}{4}\%$

Percentage of expenditure on rent = $25\%$

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Comparing the calculated percentage with the given options:

(A) $25\%$

(B) $50\%$

(C) $75\%$

(D) $90\%$

The calculated percentage is $25\%$, which matches option (A).

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Final Answer:

The percentage of the total expenditure spent on rent is $25\%$.

(A) $25\%$

Solution:

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

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Assertion (A): The sum of the probabilities of all elementary events of an experiment is always 1.

An elementary event is an event that consists of a single outcome of the random experiment. For any random experiment, the sample space is the set of all possible outcomes. The elementary events are essentially the individual outcomes in the sample space considered as single-element sets.

By the definition of probability, the sum of the probabilities of all possible outcomes (elementary events) in the sample space must equal the probability of the sample space itself, which represents the certainty of some outcome occurring. The probability of the sample space is always 1.

Thus, the sum of the probabilities of all elementary events of an experiment is indeed always 1.

So, Assertion (A) is true.

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Reason (R): Each elementary event is an outcome with a single result.

This is the standard definition of an elementary event. An elementary event is the most basic outcome of a random experiment; it cannot be broken down further into simpler events. It corresponds to a single point in the sample space.

So, Reason (R) is true.

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

Assertion (A) states that the sum of probabilities of elementary events is 1. Reason (R) defines what an elementary event is.

The fact that an elementary event is a single outcome (R) does not directly explain *why* the sum of their probabilities equals 1 (A). The property that the sum of probabilities of all elementary events is 1 is a fundamental axiom of probability theory, stemming from the fact that the union of all elementary events constitutes the entire sample space, and the probability of the sample space is defined as 1. Reason (R) provides a correct definition, but it doesn't explain the summation property stated in A.

Therefore, both A and R are true, but R is not the correct explanation of A.

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Comparing this conclusion with the given options:

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

(B) Both A and R are true, but R is not the correct explanation of A.

(C) A is true, but R is false.

(D) A is false, but R is true.

Our conclusion matches option (B).

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Final Answer:

Both Assertion (A) and Reason (R) are true statements, but Reason (R) does not explain why Assertion (A) is true.

(B) Both A and R are true, but R is not the correct explanation of A.

Solution:

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Given:

Total number of students surveyed = 100.

Number of students who prefer Bus = 40.

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To Find:

The probability that a randomly chosen student prefers going to school by Bus.

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The probability of an event is defined as:

Probability of an event = $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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In this case:

The total number of possible outcomes is the total number of students surveyed, which is 100.

The number of favourable outcomes is the number of students who prefer going to school by Bus, which is 40.

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Probability (student prefers Bus) = $\frac{\text{Number of students who prefer Bus}}{\text{Total number of students}}$

Probability (student prefers Bus) = $\frac{40}{100}$

Probability (student prefers Bus) = $0.4$

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Comparing the calculated probability with the given options:

(A) $0.4$

(B) $0.25$

(C) $0.2$

(D) $0.15$

The calculated probability is $0.4$, which matches option (A).

---

Final Answer:

The probability that the student prefers going to school by Bus is $0.4$.

(A) $0.4$

Solution:

---

A bar graph is used to represent categorical data. The length or height of each bar in a bar graph is proportional to the frequency or value it represents.

The given data shows the number of students preferring different modes of transport:

• Bus: 40 students
• Car: 25 students
• Bicycle: 20 students
• Walking: 15 students

---

When this data is represented by a bar graph, each mode of transport will have a bar, and the height of the bar will correspond to the number of students who prefer that mode.

The mode of transport with the lowest number of students will have the shortest bar.

---

Let's compare the number of students for each mode:

• Bus: 40
• Car: 25
• Bicycle: 20
• Walking: 15

The minimum number of students is 15, which corresponds to the 'Walking' mode of transport.

---

Therefore, the bar representing 'Walking' will be the shortest in the bar graph.

---

Comparing this with the given options:

(A) Bus (40 students)

(B) Car (25 students)

(C) Bicycle (20 students)

(D) Walking (15 students)

The mode with the fewest students is Walking.

---

Final Answer:

Walking would have the shortest bar.

(D) Walking

Solution:

---

A histogram is a type of graphical representation used to display the frequency distribution of continuous data. It is constructed using rectangular bars.

---

Let's examine the characteristics of a histogram:

• Histograms are used for representing grouped continuous data. The horizontal axis represents the class intervals of the data. (Option D is a necessary feature).
• The width of each bar corresponds to the class width or size of the respective class interval. The widths can be equal or unequal, but they represent the range of values in the interval. Thus, the width of each bar is proportional to the class size. (Option A is a necessary feature).
• The height of each bar represents the frequency density ($\frac{\text{Frequency}}{\text{Class Width}}$) of the class interval. For class intervals of equal width, the height of the bar is directly proportional to the frequency. The area of the bar is always proportional to the frequency. While height proportionality holds for equal widths, it is a very common scenario. (Option B is generally considered a feature, especially for equal class widths).
• Crucially, in a histogram, the bars are drawn adjacent to each other without any gaps between them. This adjacency visually emphasizes the continuous nature of the data and the flow from one class interval to the next.

---

Based on these characteristics, the statement "There are gaps between consecutive bars" is the opposite of a necessary feature of a histogram. Gaps between bars are a characteristic of bar graphs, which are typically used for discrete or categorical data.

---

Therefore, the statement that is NOT a necessary feature of a histogram is (C).

---

Final Answer:

The property that is not a necessary feature of a histogram is the presence of gaps between consecutive bars.

(C) There are gaps between consecutive bars.

Solution:

---

The total number of cards in the bag is 10, numbered from 1 to 10.

The sample space $S$ is the set of all possible outcomes when drawing a card.

$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

The total number of possible outcomes is $|S| = 10$.

---

We are interested in the event of drawing a card with an odd number.

Let $E$ be the event of drawing a card with an odd number.

The odd numbers from 1 to 10 are 1, 3, 5, 7, and 9.

$E = \{1, 3, 5, 7, 9\}$

The number of favourable outcomes for the event $E$ is $|E| = 5$.

---

The probability of event $E$ is given by:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{Drawing an odd number}) = \frac{|E|}{|S|} = \frac{5}{10}$

---

Simplifying the fraction:

$P(\text{Drawing an odd number}) = \frac{\cancel{5}^1}{\cancel{10}_2} = \frac{1}{2}$

---

Comparing the calculated probability with the given options:

(A) $\frac{1}{10}$

(B) $\frac{1}{2}$

(C) $\frac{3}{10}$

(D) $\frac{1}{5}$

The calculated probability is $\frac{1}{2}$, which matches option (B).

---

Final Answer:

The probability of drawing a card with an odd number is $\frac{1}{2}$.

(B) $\frac{1}{2}$

Solution:

---

The given data set represents the marks obtained by 15 students in a test. The marks are:

12, 15, 18, 12, 20, 15, 18, 12, 15, 20, 18, 12, 15, 18, 20

---

To find the mark with the highest frequency, we need to count how many times each unique mark appears in the data set. This is the frequency of each mark.

---

Let's count the frequency for each mark present in the data:

• Mark 12: Appears at positions 1, 4, 8, 12. Frequency of 12 is 4.
• Mark 15: Appears at positions 2, 6, 9, 13. Frequency of 15 is 4.
• Mark 18: Appears at positions 3, 7, 11, 14. Frequency of 18 is 4.
• Mark 20: Appears at positions 5, 10, 15. Frequency of 20 is 3.

---

The frequencies of the unique marks are:

• Frequency of 12 = 4
• Frequency of 15 = 4
• Frequency of 18 = 4
• Frequency of 20 = 3

---

The highest frequency observed is 4.

The marks that have a frequency of 4 are 12, 15, and 18.

The question asks "Which mark has the highest frequency?" and provides options among these marks.

Options (A) 12, (B) 15, and (C) 18 all have the highest frequency (4).

In a multiple-choice question, if multiple options are technically correct based on the data, there might be an issue with the question design or the options provided.

However, assuming a single answer is expected from the options, and based on standard practices when multiple options share the highest frequency, any of 12, 15, or 18 could be considered a valid answer. Given that 12 is the first option listed that has the highest frequency, we will select (A).

---

Final Answer:

The marks with the highest frequency are 12, 15, and 18, each having a frequency of 4.

Among the given options, 12 is listed.

(A) 12

Solution:

---

Let's consider how each type of graphical representation is constructed:

(A) Bar graph: Uses rectangular bars of equal width, with heights proportional to the values or frequencies they represent. The bars are typically separated by uniform gaps. Central angles are not used.

(B) Double bar graph: A variation of the bar graph used to compare two sets of data simultaneously, often for the same categories. It uses pairs of adjacent bars for each category. Central angles are not used.

(C) Histogram: Used for grouped continuous data. It uses adjacent rectangular bars where the width of the bar represents the class interval and the area (or height, for equal class widths) is proportional to the frequency. Central angles are not used.

(D) Pie chart: Represents data as sectors of a circle. The entire circle represents the sum of all values. Each sector corresponds to a category or component, and its size is determined by the central angle. The central angle for each component is calculated as a proportion of the total angle ($360^\circ$).

The formula for the central angle of a sector in a pie chart is:

Central Angle = $\left(\frac{\text{Value of the component}}{\text{Total value}}\right) \times 360^\circ$

---

Therefore, a pie chart is the type of graph that requires calculating central angles to represent the data.

---

Comparing this with the given options, the Pie chart (D) is the correct answer.

---

Final Answer:

Calculating central angles is required for representing data in a pie chart.

(D) Pie chart

Given: Marks obtained by 5 students in a test are $15, 18, 20, 12, 15$.

---

To Find: The mean and mode of these marks.

---

Solution:

To find the Mean, we use the formula for the mean of ungrouped data:

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Sum of observations (marks) = $15 + 18 + 20 + 12 + 15$

Sum of marks = $80$

Number of observations (students) = $5$

Mean = $\frac{80}{5}$

Mean = $16$

---

To find the Mode, we identify the observation that occurs most frequently in the given data set.

The given marks are $15, 18, 20, 12, 15$.

Let's list the frequencies of each mark:

Mark 12: occurs 1 time

Mark 15: occurs 2 times

Mark 18: occurs 1 time

Mark 20: occurs 1 time

The mark with the highest frequency is $15$, which occurs $2$ times.

Therefore, the Mode of the marks is $15$.

---

The Mean of the marks is $16$.

The Mode of the marks is $15$.

Given: The weights (in kg) of 7 students are $45, 50, 48, 52, 45, 55, 49$.

---

To Find: The median weight of the students.

---

Solution:

To find the Median of a data set, we must first arrange the observations in ascending or descending order.

Let's arrange the given weights in ascending order:

$45, 45, 48, 49, 50, 52, 55$

The number of observations is $n = 7$.

Since the number of observations ($n$) is odd, the median is the value of the $\left(\frac{n+1}{2}\right)^{\text{th}}$ observation.

Position of the median = $\left(\frac{7+1}{2}\right)^{\text{th}}$ observation

Position of the median = $\left(\frac{8}{2}\right)^{\text{th}}$ observation

Position of the median = $4^{\text{th}}$ observation.

Looking at the sorted list, the $4^{\text{th}}$ observation is $49$.

Therefore, the Median weight is $49$ kg.

---

The Median weight of the 7 students is $49$ kg.

Given: The data set is $11, 15, 9, 20, 13, 18, 10$.

---

To Find: The range of the given data.

---

Solution:

The Range of a data set is the difference between the highest and the lowest observation in the data set.

Range = Highest value - Lowest value

Let's identify the highest and lowest values from the given data: $11, 15, 9, 20, 13, 18, 10$.

The highest value in the data set is $20$.

The lowest value in the data set is $9$.

Now, we calculate the range:

Range = $20 - 9$

Range = $11$

---

The Range of the given data is $11$.

Given: A standard six-sided die is thrown.

---

To Find: The probability of getting (a) an even number and (b) a prime number.

---

Solution:

When a standard six-sided die is thrown, the possible outcomes are the numbers $1, 2, 3, 4, 5, 6$.

The sample space is $S = \{1, 2, 3, 4, 5, 6\}$.

The Total number of outcomes is $n(S) = 6$.

---

(a) Probability of getting an even number:

The even numbers in the sample space are $2, 4, 6$.

Let $E$ be the event of getting an even number. The favorable outcomes are $E = \{2, 4, 6\}$.

The Number of favorable outcomes is $n(E) = 3$.

The Probability of an event is given by:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

$P(E) = \frac{n(E)}{n(S)} = \frac{3}{6}$

$P(E) = \frac{1}{2}$

---

(b) Probability of getting a prime number:

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

The prime numbers in the sample space $\{1, 2, 3, 4, 5, 6\}$ are $2, 3, 5$. (Note: 1 is not a prime number).

Let $P$ be the event of getting a prime number. The favorable outcomes are $P = \{2, 3, 5\}$.

The Number of favorable outcomes is $n(P) = 3$.

Using the probability formula:

$P(P) = \frac{n(P)}{n(S)} = \frac{3}{6}$

$P(P) = \frac{1}{2}$

---

The probability of getting an even number is $\frac{1}{2}$.

The probability of getting a prime number is $\frac{1}{2}$.

Given:

Number of red balls in the bag = $5$

Number of blue balls in the bag = $3$

---

To Find: The probability of drawing a blue ball.

---

Solution:

First, we find the total number of balls in the bag.

Total number of balls = Number of red balls + Number of blue balls

Total number of balls = $5 + 3 = 8$

So, the Total number of outcomes when drawing a ball at random is $8$.

The event we are interested in is drawing a blue ball.

The Number of favorable outcomes (drawing a blue ball) is equal to the number of blue balls, which is $3$.

The Probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Probability of drawing a blue ball = $\frac{\text{Number of blue balls}}{\text{Total number of balls}}$

Probability of drawing a blue ball = $\frac{3}{8}$

---

The Probability of drawing a blue ball from the bag is $\frac{3}{8}$.

Given: The number of absentees for 6 days are $5, 3, 0, 7, 4, 5$.

---

To Find: The mean number of absentees per day.

---

Solution:

To find the Mean, we use the formula for the mean of ungrouped data:

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

The observations are the number of absentees on each day.

Sum of observations (absentees) = $5 + 3 + 0 + 7 + 4 + 5$

Sum of absentees = $24$

Number of observations (days) = $6$

Mean = $\frac{24}{6}$

Mean = $4$

---

The Mean number of absentees per day is $4$.

Definition of Frequency Distribution:

A frequency distribution is a table that summarizes a set of data by showing the frequency (or count) of each distinct value or range of values (class interval) within the data set. It organises raw data into a more structured and understandable format.

---

Usefulness in Handling Large Data Sets:

Frequency distributions are extremely useful when dealing with large data sets for several reasons:

1. Simplification and Organisation: Large raw data sets are often unmanageable and difficult to interpret directly. A frequency distribution condenses the data, making it easier to see the pattern, spread, and central tendency.

2. Understanding Data Characteristics: It provides a clear picture of how often different values occur. This helps in understanding the distribution of the data, identifying the most frequent values (mode), and spotting unusual observations.

3. Basis for Further Analysis: Frequency distributions serve as the basis for calculating various statistical measures like mean, median, mode, range, variance, and standard deviation more easily for grouped data. They are also used to create visual representations like histograms, frequency polygons, and bar graphs.

4. Comparison: It facilitates the comparison of different data sets or subgroups by presenting their distributions side-by-side.

In essence, a frequency distribution transforms chaotic raw data into a concise summary, allowing for quicker analysis and better insights into the characteristics of the data.

In a grouped frequency distribution, data is grouped into ranges or categories to make it more manageable and understandable.

---

Class Interval:

A class interval is one of the groups or ranges into which the entire range of data is divided. Each class interval has a lower limit and an upper limit. For example, in a grouped frequency distribution of test scores, $20-30$ or $30-40$ would be class intervals, representing scores falling within those ranges.

---

Class Width (or Size):

The class width is the difference between the upper limit and the lower limit of a class interval. It represents the size or spread of each interval. Ideally, all class intervals in a frequency distribution should have the same width.

---

For the class interval $20-30$:

In the class interval $20-30$, the lower limit is $20$ and the upper limit is $30$.

The class width is calculated as: Upper Limit - Lower Limit

Class Width = $30 - 20 = 10$

The upper limit of the class interval $20-30$ is $30$.

What is a Histogram?

A histogram is a graphical representation of the distribution of numerical data. It is similar in appearance to a bar graph but is used for continuous data that has been divided into class intervals. In a histogram, the data is plotted as a series of rectangles, where:

1. The base of each rectangle represents a class interval.

2. The width of the base is equal to the class width.

3. The height of each rectangle represents the frequency of the observations falling within that class interval.

4. There are no gaps between consecutive bars (unless a class interval has a frequency of zero).

---

How is it different from a Bar Graph?

A bar graph is a graphical representation used to compare the values of categorical or discrete data. It also uses rectangles (bars), but the key differences from a histogram are:

1. Bar graphs are used for categorical or discrete variables (e.g., types of fruits, favourite colors, number of siblings), whereas histograms are used for continuous variables (e.g., height, weight, age, test scores).

2. In a bar graph, each bar represents a specific category or discrete value, and there are typically gaps between the bars to show that the categories are distinct and not continuous.

3. The order of bars in a bar graph can often be rearranged (unless the categories have a natural order like months), while the bars in a histogram are always plotted in the order of the class intervals along the numerical scale.

4. In a histogram, the area of each bar is proportional to the frequency (or frequency density), especially if class widths are unequal. In a bar graph, the height of the bar is proportional to the value or frequency.

---

In summary, the main distinction lies in the type of data represented (continuous vs. categorical/discrete) and the presence or absence of gaps between bars.

What does a 'pie chart' represent?

A pie chart is a circular statistical graphic that is divided into slices to illustrate numerical proportion. In a pie chart, each slice represents a category, and the size of the slice (both in area and arc length) is proportional to the quantity it represents compared to the whole. Essentially, it shows the relative contribution of different components to a whole.

---

Total angle at the center of a pie chart:

A pie chart is a complete circle. The total angle around the center of a circle is always $360^\circ$. Each slice's angle at the center is calculated as a fraction of $360^\circ$, proportional to the slice's share of the total data.

The total angle at the center of a pie chart is $360^\circ$.

Given:

Percentage of students who are girls = $40\%$

The distribution is to be shown in a pie chart.

---

To Find: The central angle for the sector representing girls in the pie chart.

---

Solution:

In a pie chart, the entire circle represents $100\%$ of the data.

The total angle at the center of a circle is $360^\circ$.

Therefore, $100\%$ of the data corresponds to a central angle of $360^\circ$.

To find the central angle for a specific percentage, we can use the formula:

Central Angle = $\left(\text{Percentage} \times \frac{360^\circ}{100}\right)$

We are given that $40\%$ of the students are girls. So, the percentage is $40\%$.

Central angle for girls = $\left(40\% \times \frac{360^\circ}{100}\right)$

Central angle for girls = $\left(\frac{40}{100} \times 360^\circ\right)$

Central angle for girls = $\left(\frac{4}{10} \times 360^\circ\right)$

Central angle for girls = $(0.4 \times 360^\circ)$

Central angle for girls = $144^\circ$

---

The central angle for the sector representing girls in the pie chart is $144^\circ$.

In the context of probability:

---

Experiment:

An experiment is a process or action that has a well-defined set of possible results, and the result of the experiment is uncertain before it is performed. It is a repeatable procedure that produces a set of possible outcomes.

---

Outcome:

An outcome is any single, possible result of an experiment.

---

Example: Drawing a card from a well-shuffled deck of 52 cards

In this scenario:

The Experiment is the act of drawing one card at random from a well-shuffled deck of 52 cards.

The Outcomes are the specific cards that could potentially be drawn. Since there are 52 distinct cards in a standard deck (e.g., Ace of Spades, King of Hearts, 3 of Clubs, etc.), there are 52 possible individual outcomes for this experiment. Each specific card drawn represents one outcome.

In the context of probability:

---

Event:

An event is a specific set of one or more possible outcomes of an experiment. It is a subset of the sample space (the set of all possible outcomes).

---

Example: Throwing a die

Consider the experiment of throwing a standard six-sided die.

The possible outcomes are $\{1, 2, 3, 4, 5, 6\}$. This is the sample space.

An event could be, for example, "getting an even number". The outcomes that satisfy this condition are $\{2, 4, 6\}$. This set of outcomes constitutes the event "getting an even number".

Other examples of events in this experiment could be:

- Getting a prime number: $\{2, 3, 5\}$

- Getting a number greater than 4: $\{5, 6\}$

- Getting a number less than or equal to 1: $\{1\}$

- Getting a number 7: $\emptyset$ (This is an impossible event)

So, an event is essentially a collection of one or more desired outcomes from the total set of possible outcomes.

Given:

A box contains slips numbered from 1 to 10.

The numbers on the slips are $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.

---

To Find: The probability of drawing a slip with a number greater than 7.

---

Solution:

The Sample Space ($S$) for this experiment is the set of all possible outcomes when drawing one slip.

$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

The Total number of outcomes is the number of elements in the sample space, which is $n(S) = 10$.

Let $E$ be the event of drawing a slip with a number greater than 7.

The numbers greater than 7 in the sample space are $\{8, 9, 10\}$.

The Favorable outcomes for event $E$ are $\{8, 9, 10\}$.

The Number of favorable outcomes is the number of elements in event $E$, which is $n(E) = 3$.

The Probability of an event $E$ is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

$P(E) = \frac{n(E)}{n(S)}$

$P(\text{number > 7}) = \frac{3}{10}$

---

The Probability of drawing a slip with a number greater than 7 is $\frac{3}{10}$.

Sure Event:

In probability, a sure event (or certain event) is an event that contains all the possible outcomes of an experiment. It is an event that is guaranteed to happen every time the experiment is performed.

---

Probability of a Sure Event:

Since a sure event includes all possible outcomes, the number of favorable outcomes is equal to the total number of outcomes. Using the probability formula:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

For a sure event, Number of favorable outcomes = Total number of outcomes.

Therefore, the probability of a sure event is $\frac{\text{Total number of outcomes}}{\text{Total number of outcomes}} = 1$.

The probability of a sure event is always $1$.

---

Example:

Consider the experiment of throwing a standard six-sided die.

The possible outcomes are $\{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is $6$.

Consider the event $E$: "getting a number less than 7".

The outcomes that satisfy this event are $\{1, 2, 3, 4, 5, 6\}$.

The number of favorable outcomes for event $E$ is $6$.

Since the set of favorable outcomes $\{1, 2, 3, 4, 5, 6\}$ is the same as the set of all possible outcomes, this is a sure event.

The probability of this event is $P(E) = \frac{6}{6} = 1$.

Thus, getting a number less than 7 when throwing a standard die is a sure event.

Impossible Event:

In probability, an impossible event is an event that contains none of the possible outcomes of an experiment. It is an event that can never happen when the experiment is performed.

---

Probability of an Impossible Event:

Since an impossible event has no outcomes that satisfy it, the number of favorable outcomes for an impossible event is $0$. Using the probability formula:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

For an impossible event, Number of favorable outcomes = $0$.

Therefore, the probability of an impossible event is $\frac{0}{\text{Total number of outcomes}} = 0$ (assuming the total number of outcomes is not zero, which it is for any meaningful experiment).

The probability of an impossible event is always $0$.

---

Example:

Consider the experiment of throwing a standard six-sided die.

The possible outcomes are $\{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is $6$.

Consider the event $E$: "getting a number greater than 7".

When you throw a standard six-sided die, it is impossible to get a number greater than 7, as the maximum possible outcome is 6.

The set of outcomes that satisfy event $E$ is the empty set, denoted by $\emptyset$. The number of favorable outcomes is $n(E) = 0$.

The probability of this event is $P(E) = \frac{0}{6} = 0$.

Thus, getting a number greater than 7 when throwing a standard die is an impossible event.

Given:

A spinner has 5 equal sectors with colours: Red, Blue, Green, Yellow, and Red.

---

To Find: The probability of landing on the Red sector.

---

Solution:

The total number of equal sectors on the spinner represents the Total number of outcomes.

Total number of sectors = $5$.

So, the total number of outcomes is $5$.

The event we are interested in is landing on the Red sector.

We need to count how many sectors are Red.

Looking at the colours listed: Red, Blue, Green, Yellow, and Red.

The number of Red sectors is $2$.

The Number of favorable outcomes (landing on Red) is $2$.

The Probability of an event is calculated as:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Probability of landing on Red = $\frac{\text{Number of Red sectors}}{\text{Total number of sectors}}$

Probability of landing on Red = $\frac{2}{5}$

---

The Probability of landing on the Red sector is $\frac{2}{5}$.

Given:

The mean of 4 numbers is $25$.

Three of the numbers are $20, 22, 28$.

---

To Find: The fourth number.

---

Solution:

The formula for the Mean of a set of numbers is:

Mean = $\frac{\text{Sum of all numbers}}{\text{Number of numbers}}$

We are given the Mean ($25$) and the Number of numbers ($4$).

We can rearrange the formula to find the Sum of all numbers:

Sum of all numbers = Mean $\times$ Number of numbers

Sum of the 4 numbers = $25 \times 4$

Sum of the 4 numbers = $100$

Let the fourth number be $x$.

The sum of the four numbers is also the sum of the three given numbers plus the fourth number:

Sum of the 4 numbers = $20 + 22 + 28 + x$

We know the sum is $100$, so we can set up an equation:

$20 + 22 + 28 + x = 100$

First, sum the three known numbers:

$20 + 22 + 28 = 70$

Now, the equation becomes:

$70 + x = 100$

To find $x$, subtract $70$ from both sides of the equation:

$x = 100 - 70$

$x = 30$

---

The fourth number is $30$.

Given: The ages (in years) of 6 teachers are $28, 35, 42, 28, 45, 35$.

---

To Find: The mode and median of their ages.

---

Solution:

To find the Mode, we need to identify the age(s) that appear most frequently in the data set.

The given ages are $28, 35, 42, 28, 45, 35$.

Let's count the frequency of each age:

Age 28: occurs 2 times

Age 35: occurs 2 times

Age 42: occurs 1 time

Age 45: occurs 1 time

Both ages $28$ and $35$ have the highest frequency (2). When a data set has two modes, it is called bimodal.

Therefore, the Mode of the ages is $28$ and $35$.

---

To find the Median, we must first arrange the ages in ascending or descending order.

Let's arrange the given ages in ascending order:

$28, 28, 35, 35, 42, 45$

The number of observations is $n = 6$.

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

The $\left(\frac{6}{2}\right)^{\text{th}}$ observation is the $3^{\text{rd}}$ observation.

The $\left(\frac{6}{2}+1\right)^{\text{th}}$ observation is the $(3+1)^{\text{th}} = 4^{\text{th}}$ observation.

From the sorted list ($28, 28, 35, 35, 42, 45$), the $3^{\text{rd}}$ observation is $35$, and the $4^{\text{th}}$ observation is $35$.

Median = $\frac{\text{3}^{\text{rd}}\text{ observation} + \text{4}^{\text{th}}\text{ observation}}{2}$

Median = $\frac{35 + 35}{2}$

Median = $\frac{70}{2}$

Median = $35$

---

The Mode of the ages is $28$ and $35$.

The Median of the ages is $35$ years.

Given: The word is 'MATHEMATICS'.

---

To Find: The probability of getting a vowel when a letter is chosen at random from the word.

---

Solution:

First, let's list all the letters in the word 'MATHEMATICS'.

M, A, T, H, E, M, A, T, I, C, S

The Total number of letters in the word 'MATHEMATICS' is $11$.

This is the total number of possible outcomes when choosing a letter at random.

Total number of outcomes = $11$.

Next, we need to identify the vowels in the English alphabet and count how many of these vowels are present in the word 'MATHEMATICS'.

The vowels are A, E, I, O, U.

Looking at the letters in 'MATHEMATICS' (M, A, T, H, E, M, A, T, I, C, S), the vowels are:

A (occurs 2 times)

E (occurs 1 time)

I (occurs 1 time)

The number of vowels in the word is the sum of the occurrences of A, E, and I.

Number of vowels = $2 + 1 + 1 = 4$

These are the favorable outcomes for the event of choosing a vowel.

Number of favorable outcomes = $4$.

The Probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Probability of getting a vowel = $\frac{\text{Number of vowels}}{\text{Total number of letters}}$

Probability of getting a vowel = $\frac{4}{11}$

---

The Probability of getting a vowel when a letter is chosen at random from the word 'MATHEMATICS' is $\frac{4}{11}$.

A double bar graph is used to compare two sets of data simultaneously, often across different categories or periods.

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In a double bar graph, the key (also called the legend) is an essential component that helps the reader understand which set of data is represented by each bar within a pair. Each pair of bars corresponds to a category, and the two bars in the pair represent the values for that category from the two different data sets being compared.

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The key typically consists of coloured or patterned boxes, each associated with a label. For example, if a double bar graph is comparing the sales of Product A and Product B over several months, the key might show a blue box labelled "Product A Sales" and a red box labelled "Product B Sales". This indicates that all blue bars in the graph represent the sales of Product A, and all red bars represent the sales of Product B.

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Therefore, the key indicates which specific data set corresponds to which visual representation (colour or pattern) used for the bars in the double bar graph, making the graph readable and allowing for accurate comparison between the two data sets.

A double bar graph is particularly useful for comparing two different sets of data side-by-side for the same categories.

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Situation:

Comparing the marks obtained by students in two different subjects (e.g., Mathematics and Science) across several students or classes.

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Why a Double Bar Graph is More Appropriate:

If you wanted to visualise and compare the performance of, say, 5 students in both Mathematics and Science, a single bar graph would require two separate graphs (one for Math marks and one for Science marks). This would make it difficult to directly compare a single student's performance between the two subjects, or to quickly see which subject generally had higher or lower marks across the group.

Using a double bar graph, you can represent each student's marks for both subjects using a pair of bars next to each other, where each bar in the pair represents one subject (distinguished by colour/pattern as shown in the key). This allows for a clear and immediate comparison of a single student's performance in Math versus Science, and also enables easy visual comparison of performance across different students for each subject.

Thus, when the primary goal is to compare values from two different categories or groups across the same set of items, a double bar graph is significantly more effective and informative than a single bar graph.

In a histogram, the bars represent the frequencies of observations falling within consecutive class intervals of continuous data.

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The x-axis of a histogram represents a continuous numerical scale, and each bar's base covers a specific range (a class interval) on this scale.

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Because the class intervals in a histogram typically cover a continuous range of values without any breaks (e.g., $0-10$, $10-20$, $20-30$), the end point of one interval is the starting point of the next (when using inclusive or exclusive boundaries consistently).

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The absence of gaps between the bars visually represents this continuity of the data distribution across the entire range covered by the histogram. It shows that the data is continuous and that observations exist across the scale, grouped into adjacent intervals.

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This is a key difference from a bar graph, where gaps between bars are used to emphasize that the categories being represented are discrete and separate.

In a frequency distribution table, tally marks are used to count the occurrences of each observation or item as you go through the raw data. A block of five is represented by four vertical lines crossed by a diagonal line.

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To show the tally marks for a frequency of 13, we make groups of five and then add the remaining count.

$13 = 5 + 5 + 3$

First group of 5: $\bcancel{||||}$

Second group of 5: $\bcancel{||||}$

Remaining 3: $|||$

Combining these, the tally marks for a frequency of 13 are:

$\bcancel{||||}\ \bcancel{||||}\ |||$

Given: The number of bicycles manufactured in a factory from 2018 to 2022 is provided in the table:

Year	2018	2019	2020	2021	2022
Number of Bicycles	800	1000	900	1200	1100

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To Find:

1. Draw a bar graph to represent the data.

2. Find the average number of bicycles manufactured per year.

---

Solution:

Part 1: Drawing the Bar Graph

To draw the bar graph, we will follow these steps:

1. Draw two perpendicular axes, the horizontal axis (x-axis) and the vertical axis (y-axis).

2. On the horizontal axis, represent the Years (2018, 2019, 2020, 2021, 2022). Leave equal space between each bar.

3. On the vertical axis, represent the Number of Bicycles manufactured. Since the number of bicycles ranges from 800 to 1200, an appropriate scale would be to let 1 unit length represent 100 bicycles. Mark the axis with values like 0, 100, 200, ..., 1300.

4. For each year, draw a rectangular bar of equal width. The height of the bar should correspond to the number of bicycles manufactured in that year, according to the chosen scale on the vertical axis.

5. The height of the bars will be:

- For 2018: 800 units (or 8 units if scale is 100 per unit)

- For 2019: 1000 units (or 10 units)

- For 2020: 900 units (or 9 units)

- For 2021: 1200 units (or 12 units)

- For 2022: 1100 units (or 11 units)

6. Label both axes clearly: "Year" on the x-axis and "Number of Bicycles Manufactured" on the y-axis. Give the bar graph a title, such as "Bicycle Manufacturing (2018-2022)".

The bars should be separated by equal gaps.

---

Part 2: Finding the Average Number of Bicycles

To find the average number of bicycles manufactured per year, we need to calculate the mean of the number of bicycles manufactured each year.

The numbers of bicycles manufactured each year are: $800, 1000, 900, 1200, 1100$.

The number of years (observations) is $5$.

The formula for the mean is:

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Sum of bicycles manufactured = $800 + 1000 + 900 + 1200 + 1100$

Sum of bicycles manufactured = $5000$

Number of years = $5$

Average number of bicycles = $\frac{5000}{5}$

Average number of bicycles = $1000$

---

The average number of bicycles manufactured per year during the period 2018 to 2022 is $1000$.

Given: The marks obtained by a student in half-yearly and annual examinations for five subjects are provided in the table:

Subject	Hindi	English	Maths	Science	Social Science
Half-Yearly Marks (out of 100)	75	60	90	80	70
Annual Marks (out of 100)	80	65	95	85	75

---

To Find:

1. Draw a double bar graph to represent this data.

2. Draw a conclusion about the student's performance from the half-yearly to the annual examination.

---

Solution:

Part 1: Drawing the Double Bar Graph

To draw the double bar graph, we will follow these steps:

1. Draw two perpendicular axes, the horizontal axis (x-axis) and the vertical axis (y-axis).

2. On the horizontal axis, represent the Subjects (Hindi, English, Maths, Science, Social Science). For each subject, we will draw a pair of bars.

3. On the vertical axis, represent the Marks Obtained. Since the marks are out of 100, an appropriate scale would be to let 1 unit length represent 10 marks. Mark the axis with values like 0, 10, 20, ..., 100.

4. For each subject, draw two adjacent rectangular bars of equal width. One bar in the pair will represent the Half-Yearly marks, and the other will represent the Annual marks. Use different colours or patterns for the Half-Yearly and Annual bars to distinguish them.

5. The height of the bars will correspond to the marks obtained in each examination for each subject, according to the chosen scale on the vertical axis.

- Hindi: Half-Yearly 75, Annual 80

- English: Half-Yearly 60, Annual 65

- Maths: Half-Yearly 90, Annual 95

- Science: Half-Yearly 80, Annual 85

- Social Science: Half-Yearly 70, Annual 75

6. Label both axes clearly: "Subject" on the x-axis and "Marks Obtained (out of 100)" on the y-axis. Give the graph a title, such as "Student Performance in Examinations".

7. Include a Key to indicate which colour/pattern represents the Half-Yearly marks and which represents the Annual marks.

The pairs of bars for different subjects should be separated by equal gaps.

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Part 2: Conclusion about Student's Performance

By comparing the heights of the Half-Yearly bars with the corresponding Annual bars for each subject, we can observe the change in performance.

- In Hindi, marks increased from 75 to 80 (a gain of 5 marks).

- In English, marks increased from 60 to 65 (a gain of 5 marks).

- In Maths, marks increased from 90 to 95 (a gain of 5 marks).

- In Science, marks increased from 80 to 85 (a gain of 5 marks).

- In Social Science, marks increased from 70 to 75 (a gain of 5 marks).

In every subject, the marks in the Annual examination are higher than the marks in the Half-Yearly examination.

---

Conclusion:

From the double bar graph and the data, it is clear that the student's performance has improved in all subjects from the half-yearly examination to the annual examination. The student scored exactly 5 more marks in the annual exam compared to the half-yearly exam for every subject listed.

Given: The heights (in cm) of 20 students in a class are: $150, 152, 155, 151, 160, 158, 155, 152, 150, 156, 160, 158, 152, 155, 151, 150, 158, 155, 152, 156$.

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To Find: Organize the data into a grouped frequency distribution table using class intervals $150-155, 155-160$, etc.

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Solution:

We will create a grouped frequency distribution table using the given class intervals. Following the convention for continuous data where the upper limit of one class is the lower limit of the next, the interval $150-155$ includes heights from 150 up to (but not including) 155 cm. Similarly, $155-160$ includes heights from 155 up to (but not including) 160 cm, and $160-165$ includes heights from 160 up to (but not including) 165 cm.

Let's count the frequency of heights falling into each interval using tally marks:

• 150-155: (150, 152, 151, 152, 150, 152, 151, 150, 152) - Counts: 9
• 155-160: (155, 158, 155, 156, 158, 155, 158, 155, 156) - Counts: 9
• 160-165: (160, 160) - Counts: 2

Now we can construct the frequency distribution table:

Class Interval (Height in cm)	Tally Marks	Frequency
150-155	$\bcancel{||||}\ ||||$	9
155-160	$\bcancel{||||}\ ||||$	9
160-165	$||$	2
Total		20

The total frequency (9 + 9 + 2 = 20) matches the total number of students.

Given: The grouped frequency distribution table from Question 3:

Class Interval (Height in cm)	Frequency
150-155	9
155-160	9
160-165	2
Total	20

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To Draw: A histogram for the given frequency distribution.

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Solution:

To draw the histogram, we will follow these steps:

1. Draw two perpendicular axes. The horizontal axis (x-axis) will represent the Class Intervals (Height in cm), and the vertical axis (y-axis) will represent the Frequency (Number of Students).

2. Mark the class boundaries on the horizontal axis. The class intervals are 150-155, 155-160, and 160-165. So, mark points at 150, 155, 160, and 165 cm on the x-axis.

3. Choose an appropriate scale for the vertical axis (Frequency). The frequencies are 9, 9, and 2. A scale of 1 unit length representing 1 student is suitable. Mark the y-axis with values from 0 up to a value slightly greater than the maximum frequency (e.g., 10).

4. Draw rectangular bars for each class interval. The base of each bar should cover the corresponding class interval on the x-axis, and the height of the bar should be equal to the frequency of that class.

- For the class interval 150-155, draw a bar from 150 to 155 on the x-axis with a height of 9 units.

- For the class interval 155-160, draw a bar from 155 to 160 on the x-axis with a height of 9 units.

- For the class interval 160-165, draw a bar from 160 to 165 on the x-axis with a height of 2 units.

5. Ensure there are no gaps between consecutive bars, as the class intervals are contiguous (the upper limit of one interval is the lower limit of the next).

6. Label the axes: "Height (in cm)" on the x-axis and "Number of Students (Frequency)" on the y-axis. Give the histogram a title, such as "Histogram of Student Heights".

(Note: As I am unable to directly generate an image of the graph, the above steps describe how to construct it.)

Given: The daily wages (in $\textsf{₹}$) of 25 workers are: $200, 250, 220, 200, 250, 280, 200, 220, 250, 280, 200, 220, 250, 280, 200, 220, 250, 280, 200, 220, 250, 280, 200, 220, 250$.

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To Find:

1. Organize the data into an ungrouped frequency distribution table using tally marks.

2. Find the mean and mode of the daily wages.

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Solution:

Part 1: Ungrouped Frequency Distribution Table

We identify the distinct wage values present in the data and count their occurrences using tally marks. The distinct wages are $\textsf{₹}200, \textsf{₹}220, \textsf{₹}250, \textsf{₹}280$.

• Wage $\textsf{₹}200$: Occurrences - $7$ ($\bcancel{||||}\ ||$)
• Wage $\textsf{₹}220$: Occurrences - $6$ ($\bcancel{||||}\ |$)
• Wage $\textsf{₹}250$: Occurrences - $8$ ($\bcancel{||||}\ |||$)
• Wage $\textsf{₹}280$: Occurrences - $4$ ($||||$)

Total frequency = $7 + 6 + 8 + 4 = 25$, which matches the number of workers.

The ungrouped frequency distribution table is:

Daily Wage ($\textsf{₹}$)	Tally Marks	Frequency
200	$\bcancel{||||}\ ||$	7
220	$\bcancel{||||}\ |$	6
250	$\bcancel{||||}\ |||$	8
280	$||||$	4
Total		25

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Part 2: Mean of Daily Wages

To find the mean of the daily wages from the frequency table, we use the formula:

Mean = $\frac{\sum (x_i \times f_i)}{\sum f_i}$

where $x_i$ is the daily wage and $f_i$ is its frequency.

Sum of $(x_i \times f_i) = (200 \times 7) + (220 \times 6) + (250 \times 8) + (280 \times 4)$

Sum of products = $1400 + 1320 + 2000 + 1120$

Sum of products = $5840$

Total frequency ($\sum f_i$) = $25$

Mean = $\frac{5840}{25}$

Mean = $233.6$

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Part 3: Mode of Daily Wages

The mode is the value that appears most frequently in the data set. From the frequency distribution table, the highest frequency is 8, which corresponds to the daily wage of $\textsf{₹}250$.

Therefore, the Mode of the daily wages is $\textsf{₹}250$.

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The Mean daily wage is $\textsf{₹}233.6$.

The Mode daily wage is $\textsf{₹}250$.

Explanation of Pie Chart Construction:

A pie chart is a circular graph divided into sectors, where each sector represents a part of the whole. The size of each sector is proportional to the quantity it represents. To construct a pie chart, we divide the total quantity into $360^\circ$ (the total angle at the center of a circle) and allocate a proportional angle to each category.

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Given Data:

The table shows the favourite colours of 45 students:

Colour	Number of Students (Frequency)
Red	10
Blue	15
Green	12
Yellow	8
Total	45

Total number of students = $45$.

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Calculating Central Angles:

The central angle for each sector is calculated using the formula:

Central Angle = $\frac{\text{Frequency of the Category}}{\text{Total Frequency}} \times 360^\circ$

Let's calculate the central angle for each colour:

1. Red:

Central Angle (Red) = $\frac{10}{45} \times 360^\circ$

Central Angle (Red) = $\frac{2}{9} \times 360^\circ$

Central Angle (Red) = $2 \times \frac{360^\circ}{9} = 2 \times 40^\circ = 80^\circ$

2. Blue:

Central Angle (Blue) = $\frac{15}{45} \times 360^\circ$

Central Angle (Blue) = $\frac{1}{3} \times 360^\circ$

Central Angle (Blue) = $1 \times \frac{360^\circ}{3} = 1 \times 120^\circ = 120^\circ$

3. Green:

Central Angle (Green) = $\frac{12}{45} \times 360^\circ$

Central Angle (Green) = $\frac{4}{15} \times 360^\circ$

Central Angle (Green) = $4 \times \frac{360^\circ}{15} = 4 \times 24^\circ = 96^\circ$

4. Yellow:

Central Angle (Yellow) = $\frac{8}{45} \times 360^\circ$

Central Angle (Yellow) = $\frac{8}{45} \times 360^\circ = 8 \times \frac{360^\circ}{45}$

Central Angle (Yellow) = $8 \times 8^\circ = 64^\circ$

Let's verify the sum of the angles: $80^\circ + 120^\circ + 96^\circ + 64^\circ = 360^\circ$. The angles sum up correctly.

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Steps to Draw the Pie Chart:

1. Draw a circle of any convenient radius.

2. Draw a radius from the center of the circle to any point on the circumference. This will be the starting line for the first sector.

3. Using a protractor, draw the first sector corresponding to the angle calculated for the first category (e.g., Red, $80^\circ$). Place the protractor's center at the circle's center and its base line along the radius drawn in step 2. Mark the angle $80^\circ$ and draw a new radius to that mark.

4. For the second sector (e.g., Blue, $120^\circ$), place the protractor's base line along the newly drawn radius (the boundary of the first sector). Measure the angle $120^\circ$ from this line and draw another radius.

5. Repeat this process for the remaining categories (Green, $96^\circ$, and Yellow, $64^\circ$), using the previously drawn radius as the base line for the next sector.

6. The last sector's angle should automatically match the calculated angle for the final category (Yellow, $64^\circ$), and its boundary should align with the initial radius drawn in step 2.

7. Label each sector clearly with the name of the category it represents (Colour) and optionally the number of students or percentage. Use different colours or patterns to fill each sector for better visual distinction.

8. Give the pie chart an appropriate title (e.g., "Favourite Colours of Students").

Given:

Number of red marbles = $10$

Number of blue marbles = $8$

Number of green marbles = $6$

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To Find: The probability of drawing (a) a red marble, (b) a green marble, (c) a marble that is not blue, and (d) a yellow marble.

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Solution:

First, calculate the total number of marbles in the bag. This represents the Total number of outcomes.

Total number of marbles = Number of red marbles + Number of blue marbles + Number of green marbles

Total number of marbles = $10 + 8 + 6 = 24$

So, the total number of outcomes is $24$.

The formula for probability is $P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.

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(a) Probability of drawing a red marble:

The event is drawing a red marble.

Number of favorable outcomes (red marbles) = $10$.

Probability of drawing a red marble = $\frac{\text{Number of red marbles}}{\text{Total number of marbles}}$

Probability (Red) = $\frac{10}{24}$

Simplify the fraction:

Probability (Red) = $\frac{\cancel{10}^5}{\cancel{24}_{12}} = \frac{5}{12}$

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(b) Probability of drawing a green marble:

The event is drawing a green marble.

Number of favorable outcomes (green marbles) = $6$.

Probability of drawing a green marble = $\frac{\text{Number of green marbles}}{\text{Total number of marbles}}$

Probability (Green) = $\frac{6}{24}$

Simplify the fraction:

Probability (Green) = $\frac{\cancel{6}^1}{\cancel{24}_4} = \frac{1}{4}$

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(c) Probability of drawing a marble that is not blue:

A marble that is not blue must be either red or green.

Number of marbles that are not blue = Number of red marbles + Number of green marbles

Number of marbles that are not blue = $10 + 6 = 16$.

The event is drawing a marble that is not blue.

Number of favorable outcomes (not blue marbles) = $16$.

Probability of drawing a marble that is not blue = $\frac{\text{Number of marbles not blue}}{\text{Total number of marbles}}$

Probability (Not Blue) = $\frac{16}{24}$

Simplify the fraction:

Probability (Not Blue) = $\frac{\cancel{16}^2}{\cancel{24}_3} = \frac{2}{3}$

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(d) Probability of drawing a yellow marble:

The event is drawing a yellow marble.

Looking at the given information, there are no yellow marbles in the bag.

Number of favorable outcomes (yellow marbles) = $0$.

Probability of drawing a yellow marble = $\frac{\text{Number of yellow marbles}}{\text{Total number of marbles}}$

Probability (Yellow) = $\frac{0}{24}$

Probability (Yellow) = $0$

This is an impossible event.

---

The probabilities are:

(a) Probability of drawing a red marble = $\frac{5}{12}$.

(b) Probability of drawing a green marble = $\frac{1}{4}$.

(c) Probability of drawing a marble that is not blue = $\frac{2}{3}$.

(d) Probability of drawing a yellow marble = $0$.

Given: Two coins are tossed simultaneously.

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To Find:

1. The possible outcomes.

2. The probability of getting: (a) Exactly one Head, (b) At least one Tail, (c) No Heads.

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Solution:

When two coins are tossed simultaneously, the possible outcomes are the combinations of results for each coin. Let H represent getting a Head and T represent getting a Tail.

The possible outcomes are:

• Head on the first coin and Head on the second coin (HH)
• Head on the first coin and Tail on the second coin (HT)
• Tail on the first coin and Head on the second coin (TH)
• Tail on the first coin and Tail on the second coin (TT)

The set of all possible outcomes, also known as the Sample Space ($S$), is:

$S = \{HH, HT, TH, TT\}$

The Total number of outcomes is $n(S) = 4$.

The probability of an event $E$ is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

---

(a) Probability of getting Exactly one Head:

The event is getting exactly one Head. The outcomes in the sample space with exactly one Head are $HT$ and $TH$.

Let $E_a$ be the event of getting exactly one Head.

Favorable outcomes for $E_a = \{HT, TH\}$.

Number of favorable outcomes, $n(E_a) = 2$.

Probability of getting exactly one Head = $P(E_a) = \frac{n(E_a)}{n(S)} = \frac{2}{4}$

$P(\text{Exactly one Head}) = \frac{\cancel{2}^1}{\cancel{4}_2} = \frac{1}{2}$

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(b) Probability of getting At least one Tail:

The event is getting at least one Tail. This means getting one Tail or two Tails. The outcomes in the sample space with at least one Tail are $HT, TH$, and $TT$.

Let $E_b$ be the event of getting at least one Tail.

Favorable outcomes for $E_b = \{HT, TH, TT\}$.

Number of favorable outcomes, $n(E_b) = 3$.

Probability of getting at least one Tail = $P(E_b) = \frac{n(E_b)}{n(S)} = \frac{3}{4}$

$P(\text{At least one Tail}) = \frac{3}{4}$

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(c) Probability of getting No Heads:

The event is getting no Heads. This means both coins must be Tails. The outcome in the sample space with no Heads is $TT$.

Let $E_c$ be the event of getting no Heads.

Favorable outcomes for $E_c = \{TT\}$.

Number of favorable outcomes, $n(E_c) = 1$.

Probability of getting no Heads = $P(E_c) = \frac{n(E_c)}{n(S)} = \frac{1}{4}$

$P(\text{No Heads}) = \frac{1}{4}$

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The possible outcomes are $\{HH, HT, TH, TT\}$.

(a) The probability of getting exactly one Head is $\frac{1}{2}$.

(b) The probability of getting at least one Tail is $\frac{3}{4}$.

(c) The probability of getting no Heads is $\frac{1}{4}$.

Given: A well-shuffled deck of 52 playing cards.

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To Find: The probability of drawing (a) a King of Red colour, (b) a Face card, (c) the Ace of Spades, and (d) a Red card.

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Solution:

A standard deck of 52 playing cards consists of 4 suits (Hearts, Diamonds, Clubs, Spades), with 13 cards in each suit.

Hearts and Diamonds are red suits (26 cards total).

Clubs and Spades are black suits (26 cards total).

Each suit has cards Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.

Face cards are Jack, Queen, and King.

The Total number of outcomes when drawing one card from the deck is the total number of cards, which is $52$.

The probability of an event $E$ is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

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(a) Probability of getting a King of Red colour:

The red suits are Hearts and Diamonds. Each red suit has one King.

The Kings of red colour are the King of Hearts and the King of Diamonds.

Number of favorable outcomes (Kings of red colour) = $2$.

Probability of getting a King of Red colour = $\frac{\text{Number of Red Kings}}{\text{Total number of cards}}$

$P(\text{Red King}) = \frac{2}{52}$

Simplify the fraction:

$P(\text{Red King}) = \frac{\cancel{2}^1}{\cancel{52}_{26}} = \frac{1}{26}$

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(b) Probability of getting a Face card:

The face cards are Jack, Queen, and King. There are 3 face cards in each of the 4 suits.

Number of face cards = $3 \times 4 = 12$.

Number of favorable outcomes (Face cards) = $12$.

Probability of getting a Face card = $\frac{\text{Number of Face cards}}{\text{Total number of cards}}$

$P(\text{Face card}) = \frac{12}{52}$

Simplify the fraction:

$P(\text{Face card}) = \frac{\cancel{12}^3}{\cancel{52}_{13}} = \frac{3}{13}$

---

(c) Probability of getting the Ace of Spades:

There is only one Ace of Spades in a deck of 52 cards.

Number of favorable outcomes (Ace of Spades) = $1$.

Probability of getting the Ace of Spades = $\frac{\text{Number of Ace of Spades}}{\text{Total number of cards}}$

$P(\text{Ace of Spades}) = \frac{1}{52}$

---

(d) Probability of getting a Red card:

The red suits are Hearts and Diamonds. Each red suit has 13 cards.

Number of red cards = $13 (\text{Hearts}) + 13 (\text{Diamonds}) = 26$.

Number of favorable outcomes (Red cards) = $26$.

Probability of getting a Red card = $\frac{\text{Number of Red cards}}{\text{Total number of cards}}$

$P(\text{Red card}) = \frac{26}{52}$

Simplify the fraction:

$P(\text{Red card}) = \frac{\cancel{26}^1}{\cancel{52}_2} = \frac{1}{2}$

---

The probabilities are:

(a) Probability of getting a King of Red colour is $\frac{1}{26}$.

(b) Probability of getting a Face card is $\frac{3}{13}$.

(c) Probability of getting the Ace of Spades is $\frac{1}{52}$.

(d) Probability of getting a Red card is $\frac{1}{2}$.

Given: The number of students speaking different languages in a hostel are provided in the table:

Language	Hindi	English	Marathi	Bengali	Tamil	Total
Number of Students	40	12	9	7	4	72

Total number of students = $72$.

---

To Find:

1. Calculate the central angle for each language.

2. Describe the steps to draw a pie chart for this data.

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Solution:

To represent the data in a pie chart, we need to find the central angle for each language group. The total number of students (72) corresponds to the total angle of the circle, which is $360^\circ$. The central angle for each language will be a fraction of $360^\circ$, proportional to the number of students speaking that language compared to the total number of students.

The formula for the central angle of a sector is:

Central Angle = $\frac{\text{Number of students for the language}}{\text{Total number of students}} \times 360^\circ$

---

Calculating Central Angles for each language:

1. Hindi:

Number of students = $40$

Central Angle (Hindi) = $\frac{40}{72} \times 360^\circ$

Central Angle (Hindi) = $\frac{\cancel{40}^{\,\,5}}{\cancel{72}_{\,\,9}} \times 360^\circ$

Central Angle (Hindi) = $\frac{5}{9} \times 360^\circ = 5 \times \frac{360^\circ}{9} = 5 \times 40^\circ = 200^\circ$

2. English:

Number of students = $12$

Central Angle (English) = $\frac{12}{72} \times 360^\circ$

Central Angle (English) = $\frac{\cancel{12}^{\,\,1}}{\cancel{72}_{\,\,6}} \times 360^\circ$

Central Angle (English) = $\frac{1}{6} \times 360^\circ = \frac{360^\circ}{6} = 60^\circ$

3. Marathi:

Number of students = $9$

Central Angle (Marathi) = $\frac{9}{72} \times 360^\circ$

Central Angle (Marathi) = $\frac{\cancel{9}^{\,\,1}}{\cancel{72}_{\,\,8}} \times 360^\circ$

Central Angle (Marathi) = $\frac{1}{8} \times 360^\circ = \frac{360^\circ}{8} = 45^\circ$

4. Bengali:

Number of students = $7$

Central Angle (Bengali) = $\frac{7}{72} \times 360^\circ$

Central Angle (Bengali) = $7 \times \frac{360^\circ}{72}$

Central Angle (Bengali) = $7 \times 5^\circ = 35^\circ$

5. Tamil:

Number of students = $4$

Central Angle (Tamil) = $\frac{4}{72} \times 360^\circ$

Central Angle (Tamil) = $\frac{\cancel{4}^{\,\,1}}{\cancel{72}_{\,\,18}} \times 360^\circ$

Central Angle (Tamil) = $\frac{1}{18} \times 360^\circ = \frac{360^\circ}{18} = 20^\circ$

Let's verify the sum of the angles: $200^\circ + 60^\circ + 45^\circ + 35^\circ + 20^\circ = 360^\circ$. The total angle is correct.

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Steps to Draw the Pie Chart:

1. Draw a circle of a suitable radius using a compass.

2. Draw a radius from the center of the circle to any point on the circumference. This will serve as the starting line for the first sector.

3. Using a protractor, measure and draw the central angle for the first language (Hindi, $200^\circ$) from the starting radius. Draw a new radius to mark the end of this sector.

4. From the end of the previous sector (the newly drawn radius), measure and draw the central angle for the second language (English, $60^\circ$) using the protractor. Draw another radius.

5. Continue this process for the remaining languages (Marathi $45^\circ$, Bengali $35^\circ$, Tamil $20^\circ$), using the end of the previously drawn sector as the starting line for the next angle.

6. The final sector drawn should automatically have the correct angle for the last language (Tamil, $20^\circ$) and should meet the initial radius drawn in step 2.

7. Label each sector with the language name it represents and optionally the number of students or the percentage. Fill each sector with different colours or patterns to distinguish them clearly.

8. Add a title to the pie chart (e.g., "Languages Spoken by Hostel Students").

Given: Maximum temperature ($^\circ\text{C}$) for the first 7 days of June and July are provided in the table:

Day	1	2	3	4	5	6	7
June Temp ($^\circ\text{C}$)	38	40	39	38	41	40	42
July Temp ($^\circ\text{C}$)	35	37	36	35	38	37	39

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To Find:

1. Draw a double bar graph to represent this data (describe steps).

2. Calculate the mean temperature for the first 7 days of June.

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Solution:

Part 1: Drawing the Double Bar Graph

To draw the double bar graph, we will follow these steps:

1. Draw two perpendicular axes, the horizontal axis (x-axis) and the vertical axis (y-axis).

2. On the horizontal axis, represent the Days (1, 2, 3, 4, 5, 6, 7). For each day, we will draw a pair of bars.

3. On the vertical axis, represent the Temperature ($^\circ\text{C}$). The temperatures range roughly from 35 to 42$^\circ\text{C}$. An appropriate scale would be to start the axis from a value below the minimum temperature (e.g., $30^\circ\text{C}$) and use a unit length for each degree Celsius or every 2 degrees. Mark the axis with values like $30, 32, 34, \dots, 44$. Use a zig-zag mark ($\mathbf{\sim}$) on the y-axis near the origin if starting from a value greater than 0.

4. For each day, draw two adjacent rectangular bars of equal width. One bar in the pair will represent the June temperature, and the other will represent the July temperature. Use different colours or patterns for the June and July bars to distinguish them.

5. The height of the bars will correspond to the temperature on each day for each month, according to the chosen scale on the vertical axis.

6. Label both axes clearly: "Day" on the x-axis and "Temperature ($^\circ\text{C}$)" on the y-axis. Give the graph a title, such as "Maximum Temperature Comparison: June vs July (First 7 Days)".

7. Include a Key to indicate which colour/pattern represents the June temperatures and which represents the July temperatures.

The pairs of bars for different days should be separated by equal gaps.

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Part 2: Calculating the Mean Temperature for June

To find the mean temperature for the first 7 days of June, we will sum the temperatures for these days and divide by the number of days.

The temperatures for the first 7 days of June are: $38, 40, 39, 38, 41, 40, 42$.

Number of days = $7$.

Sum of temperatures = $38 + 40 + 39 + 38 + 41 + 40 + 42$

Sum of temperatures = $278^\circ\text{C}$.

Mean temperature = $\frac{\text{Sum of temperatures}}{\text{Number of days}}$

Mean temperature = $\frac{278}{7}^\circ\text{C}$

To calculate the division:

$\begin{array}{r} 39.71\dots \\ 7{\overline{\smash{\big)}\,278.00\dots\phantom{)}}} \\ \underline{-~\phantom{(}(21)\phantom{...)}}\\ 68\phantom{...}\\ \underline{-~\phantom{()}(63)\phantom{..)}}\\ 50\phantom{..}\\ \underline{-~\phantom{()}(49)\phantom{.}}\\ 10\phantom{.}\\ \underline{-~\phantom{()}(7)}\\ 3 \end{array}$

Mean temperature $\approx 39.71^\circ\text{C}$ (rounded to two decimal places).

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The mean temperature for the first 7 days of June is approximately $39.71^\circ\text{C}$.

Role of Representative Values:

Representative values, also known as measures of central tendency (like mean, median, and mode), play a crucial role in summarising data. Instead of looking at every single observation in a data set, these values provide a single number that attempts to describe the "center" or typical value of the data. They help in getting a quick overview of the data's characteristics, comparing different data sets, and making inferences.

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Mean:

Role: The Mean (or average) is calculated by summing all the values in a data set and dividing by the number of values. It represents the arithmetic average and is sensitive to every value in the data set.

Suitability: The mean is most suitable for numerical data that is relatively symmetrical and does not have extreme outliers (values that are significantly different from the rest of the data). It uses all the data points in its calculation.

Example: To find the mean height of students in a class, you add up all their heights and divide by the number of students. For the data set $\{150, 155, 160, 152, 158\}$, the mean is $\frac{150+155+160+152+158}{5} = \frac{775}{5} = 155$.

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Median:

Role: The Median is the middle value in a data set that has been ordered from least to greatest. It divides the data into two equal halves.

Suitability: The median is most suitable for numerical data, especially when the data contains outliers or is significantly skewed (not symmetrical). Unlike the mean, the median is not affected by extreme values, making it a better indicator of the typical value in such cases.

Example: Consider the salaries (in $\textsf{₹}$) of 5 employees: $20,000, 25,000, 30,000, 35,000, 500,000$. Sorted data: $20,000, 25,000, 30,000, 35,000, 500,000$. The median is the middle value, $\textsf{₹}30,000$. The mean would be $\textsf{₹}122,000$, which is heavily influenced by the high salary and doesn't represent the typical salary as well as the median does.

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Mode:

Role: The Mode is the value that appears most frequently in a data set. A data set can have one mode (unimodal), multiple modes (multimodal), or no mode.

Suitability: The mode is suitable for all types of data, including categorical (non-numerical) data and numerical data. It is the only measure of central tendency that can be used for categorical data.

Example: In a survey of favourite colours among students, the results are: Red, Blue, Green, Red, Yellow, Blue, Red. The frequencies are: Red (3 times), Blue (2 times), Green (1 time), Yellow (1 time). The colour that appears most frequently is Red. Therefore, the mode is Red.

For numerical data $\{15, 18, 20, 12, 15, 15\}$, the value 15 appears most often. The mode is 15.

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In summary, these representative values provide different perspectives on the center of a data set, and the choice of which measure to use depends on the type of data and the presence of outliers or skewness.