Chapter 3 Data Handling (Additional Questions)

Raw data refers to data that has been collected but has not yet been processed, organized, or analyzed. It is in its original form directly as it was gathered from the source.

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Let's examine the given options:

(A) Data already organised in a table: This is data that has been processed and structured. It is not raw data.

(B) Information collected from newspapers: Information in newspapers is typically processed, interpreted, and presented in a summarized or narrative format. It is not the raw data collected directly from the source.

(C) Figures collected in their original form: This definition perfectly matches the definition of raw data. It is the data as it is initially gathered.

(D) Data presented in a bar graph: A bar graph is a visual representation of data that has been organized and summarized. It is not raw data.

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Therefore, the option that represents raw data is the one describing figures collected in their original form.

The correct option is (C).

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

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Given the marks obtained by 10 students:

55, 36, 95, 73, 60, 42, 25, 78, 75, 62

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To Find:

The range of this data.

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Solution:

First, let's identify the highest value in the given data set.

The values are: 55, 36, 95, 73, 60, 42, 25, 78, 75, 62.

Comparing these values, the highest value is 95.

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Next, let's identify the lowest value in the given data set.

The values are: 55, 36, 95, 73, 60, 42, 25, 78, 75, 62.

Comparing these values, the lowest value is 25.

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Now, we calculate the range using the formula:

Performing the subtraction:

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The range of the given data is 70.

Comparing this result with the given options:

(A) 70

(B) 95

(C) 25

(D) 73

The calculated range matches option (A).

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The correct option is (A).

Tally marks are used for counting things. A single vertical line ($|$) represents 1 unit.

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A group of four vertical lines crossed by a diagonal line ($\bcancel{||||}$) represents a group of 5 units. This is a common way to make counting in fives easier.

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The given tally marks are: $\bcancel{||||}\ \bcancel{||||}\ |||$.

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Let's break down the given tally marks:

The first group is $\bcancel{||||}$, which represents 5.

The second group is $\bcancel{||||}$, which represents 5.

The third group is $|||$, which represents 3 (since there are three vertical lines).

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To find the total number represented by these tally marks, we add the values represented by each group:

Performing the addition:

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The number represented by the tally marks $\bcancel{||||}\ \bcancel{||||}\ |||$ is 13.

Comparing this result with the given options:

(A) 10

(B) 13

(C) 15

(D) 8

The calculated number matches option (B).

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The correct option is (B).

Given is a table showing the number of absent students in a class for each day of a week from Monday to Friday.

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To Find:

The total number of absent students during the week.

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Solution:

From the table, the number of absent students on each day is:

• Monday: 3
• Tuesday: 1
• Wednesday: 0
• Thursday: 2
• Friday: 4

To find the total number of absent students during the week, we need to add the number of absent students from each day.

Let's perform the addition:

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The total number of absent students during the week is 10.

Comparing this result with the given options:

(A) 5

(B) 8

(C) 10

(D) 12

The calculated total matches option (C).

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The correct option is (C).

A bar graph is a graphical representation of data using rectangular bars of equal width, drawn either horizontally or vertically with equal spacing between them. The length (or height) of each bar is proportional to the value it represents.

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One of the key characteristics of a bar graph is the uniformity in the representation of different categories.

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In a bar graph:

• The bars have uniform width.
• The distance between consecutive bars is equal.
• The height (or length) of each bar is proportional to the frequency or value of the category it represents.

Keeping the width of the bars the same is essential so that the visual comparison between categories is based solely on the height (or length) of the bars, which represents the data value. If the widths were different, the area of the bars might influence the perception of the data, which is not the purpose of a standard bar graph.

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Therefore, the width of each bar in a bar graph is usually the same.

Comparing this with the given options:

(A) Different

(B) Same

(C) Not important

(D) Depends on the data

The correct statement is that the width of each bar is usually the same.

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The correct option is (B).

Given is the number of books read by five students.

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Given:

Number of books read by Student C = 7

Number of books read by Student E = 5

Number of books read by Student A = 6

Number of books read by Student B = 8

Number of books read by Student D = 4

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To Find:

Which student read the maximum number of books.

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Solution:

We are given the number of books read by each of the five students: A, B, C, D, and E.

Student A read 6 books.

Student B read 8 books.

Student C read 7 books.

Student D read 4 books.

Student E read 5 books.

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To find which student read the maximum number of books, we need to compare the number of books read by each student and find the largest value.

Comparing the values: 6, 8, 7, 4, 5.

The values in ascending order are: 4, 5, 6, 7, 8.

The maximum value is 8.

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Now, we identify the student who read 8 books.

From the given information, Student B read 8 books.

Therefore, Student B read the maximum number of books.

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Comparing this result with the given options:

(A) Student A

(B) Student B

(C) Student C

(D) Student D

Our finding that Student B read the maximum number of books matches option (B).

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The correct option is (B).

A bar graph is a graphical representation of data using bars. A simple bar graph is used to represent a single set of data.

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A double bar graph is a visual tool used to compare two sets of data simultaneously, especially when the data is related to the same categories but measured under different conditions or at different times.

In a double bar graph, for each category on the x-axis, there are two adjacent bars, each representing a value from one of the two data sets being compared. Different colors or patterns are typically used for the bars representing the two different data sets, and a key (legend) is provided to identify which bar represents which data set.

The primary purpose of using two bars side-by-side for each category is to make direct comparisons between the two data sets easy and visually clear for each category.

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Let's evaluate the given options:

(A) Show a single set of data: This is the purpose of a simple bar graph, not a double bar graph.

(B) Compare two sets of data simultaneously: This is precisely the purpose of a double bar graph, allowing for a visual comparison of corresponding values from two data sets.

(C) Show the frequency distribution of data: While bar graphs can show frequency, the term "double" specifically indicates a comparison between two sets, which is a more specific use case than just showing frequency distribution.

(D) Represent data using pictures: This describes a pictograph, where symbols or pictures are used to represent data, not a bar graph.

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Based on the definition and usage of a double bar graph, its main purpose is to compare two sets of data simultaneously.

The correct option is (B).

In statistics, the term "average" is most commonly used to refer to a measure of central tendency that gives a single value representing the center of a set of data.

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There are several measures of central tendency, including the mean, median, and mode. Each represents a different way of calculating the "average" or central value of a data set.

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Let's consider the given options:

• Mode: The mode is the value that appears most frequently in a data set. It is useful for categorical data but does not represent a mathematical average in the typical sense.
• Median: The median is the middle value in a data set when it is arranged in order from least to greatest. It divides the data set into two equal halves.
• Mean: The mean is calculated by summing all the values in a data set and dividing by the number of values. This is the most widely used definition of the "average" of a set of observations. It is often denoted by $\overline{x}$ (for a sample) or $\mu$ (for a population). The formula for the mean of $n$ observations $x_1, x_2, ..., x_n$ is $\overline{x} = \frac{x_1 + x_2 + ... + x_n}{n}$ or $\overline{x} = \frac{\sum_{i=1}^{n} x_i}{n}$.
• Range: The range is the difference between the highest and lowest values in a data set. It is a measure of spread or dispersion, not a measure of central tendency or average.

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Based on the standard definition in statistics, the average of a set of observations is referred to as the Mean.

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Comparing this with the given options:

(A) Mode

(B) Median

(C) Mean

(D) Range

The term "average" corresponds to the definition of the Mean.

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The correct option is (C).

Given is a set of data: 10, 20, 30, 40, 50.

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To Find:

The mean of the given data.

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Solution:

The mean of a set of observations is calculated by dividing the sum of all observations by the number of observations.

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The formula for the mean ($\overline{x}$) of $n$ observations is:

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

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Next, let's count the number of observations.

The given observations are 10, 20, 30, 40, 50.

There are 5 observations.

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Now, substitute the sum and the number of observations into the mean formula:

Perform the division:

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The mean of the given data is 30.

Comparing this result with the given options:

(A) 25

(B) 30

(C) 35

(D) 150

The calculated mean matches option (B).

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The correct option is (B).

The median of a set of data is the middle value when the data is arranged in order from least to greatest or greatest to least.

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Given is the data set: 3, 5, 7, 9, 11.

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To Find:

The median of the given data.

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Solution:

First, we need to arrange the data in ascending or descending order. The given data is already arranged in ascending order: 3, 5, 7, 9, 11.

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Next, we need to find the middle value.

There are 5 observations in the data set. The number of observations ($n$) is 5.

Since $n$ is an odd number, the median is the value at the $\left(\frac{n+1}{2}\right)^{\text{th}}$ position in the ordered data.

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Now, we find the value at the $3^{\text{rd}}$ position in the ordered data (3, 5, 7, 9, 11).

The value at the $1^{\text{st}}$ position is 3.

The value at the $2^{\text{nd}}$ position is 5.

The value at the $3^{\text{rd}}$ position is 7.

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Therefore, the median of the given data is 7.

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Comparing this result with the given options:

(A) 5

(B) 7

(C) 9

(D) 11

The calculated median matches option (B).

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The correct option is (B).

The median of a set of data is the middle value when the data is arranged in order.

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Given is the data set: 2, 4, 6, 8, 10, 12.

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To Find:

The median of the given data.

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Solution:

First, we need to arrange the data in ascending or descending order. The given data is already arranged in ascending order: 2, 4, 6, 8, 10, 12.

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Next, we need to find the middle value(s).

There are 6 observations in the data set. The number of observations ($n$) is 6.

Since $n$ is an even number, the median is the average of the two middle values. The positions of the two middle values are $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2} + 1\right)^{\text{th}}$ in the ordered data.

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Now, we find the values at the $3^{\text{rd}}$ and $4^{\text{th}}$ positions in the ordered data (2, 4, 6, 8, 10, 12).

The value at the $3^{\text{rd}}$ position is 6.

The value at the $4^{\text{th}}$ position is 8.

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The median is the average of these two middle values:

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The median of the given data is 7.

Comparing this result with the given options:

(A) 6

(B) 8

(C) 7

(D) 9

The calculated median matches option (C).

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The correct option is (C).

The mode of a data set is the value that appears most frequently in the set.

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Given is the data set: 2, 3, 4, 2, 5, 2, 6.

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To Find:

The mode of the given data.

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Solution:

To find the mode, we need to count the frequency of each value in the data set.

• The value 2 appears 3 times.
• The value 3 appears 1 time.
• The value 4 appears 1 time.
• The value 5 appears 1 time.
• The value 6 appears 1 time.

The frequencies of the values are:

• Frequency of 2: 3
• Frequency of 3: 1
• Frequency of 4: 1
• Frequency of 5: 1
• Frequency of 6: 1

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The value with the highest frequency is the mode.

Comparing the frequencies, the highest frequency is 3, which corresponds to the value 2.

Therefore, the mode of the data set is 2.

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Comparing this result with the given options:

(A) 3

(B) 4

(C) 5

(D) 2

The calculated mode matches option (D).

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The correct option is (D).

The mode of a data set is the value that appears most frequently in the set.

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A data set has no mode if all values in the set appear with the same frequency.

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Let's examine the frequency of each value in the given data sets:

(A) 1, 2, 3, 4, 5

• Frequency of 1: 1
• Frequency of 2: 1
• Frequency of 3: 1
• Frequency of 4: 1
• Frequency of 5: 1

In this data set, every value appears exactly once. Therefore, all values have the same frequency (1). There is no value that appears more frequently than others. Thus, this data set has no mode.

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(B) 1, 2, 2, 3, 3

• Frequency of 1: 1
• Frequency of 2: 2
• Frequency of 3: 2

In this data set, the values 2 and 3 both appear 2 times, which is more frequent than the value 1. This data set has two modes: 2 and 3.

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(C) 1, 1, 2, 2, 3

• Frequency of 1: 2
• Frequency of 2: 2
• Frequency of 3: 1

In this data set, the values 1 and 2 both appear 2 times, which is more frequent than the value 3. This data set has two modes: 1 and 2.

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(D) 1, 1, 1, 2, 2

• Frequency of 1: 3
• Frequency of 2: 2

In this data set, the value 1 appears 3 times, which is more frequent than the value 2. This data set has one mode: 1.

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Based on the analysis, the data set where all values appear with the same frequency is 1, 2, 3, 4, 5. This data set has no mode.

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The correct option is (A).

Given is the experiment of tossing a fair coin.

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To Find:

The probability of getting a head.

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Solution:

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

The set of all possible outcomes is called the sample space. For a single coin toss, the sample space is {H, T}.

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We are interested in the event of getting a head.

The favourable outcome is getting a Head (H).

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The probability of an event is calculated using the formula:

In this case, the event is getting a head. Let's denote this event as E.

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

Comparing this result with the given options:

(A) 0

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

(C) 1

(D) $\frac{1}{4}$

The calculated probability matches option (B).

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The correct option is (B).

The probability of an event is the ratio of the number of favourable outcomes to the total number of possible outcomes, assuming all outcomes are equally likely.

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Given:

Number of red balls in the bag = 3

Number of blue balls in the bag = 5

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To Find:

The probability of drawing a red ball from the bag.

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Solution:

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

These are the total number of possible outcomes when drawing one ball from the bag.

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Next, we identify the number of favourable outcomes for the event of drawing a red ball.

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Now, we apply the probability formula:

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The probability of drawing a red ball is $\frac{3}{8}$.

Comparing this result with the given options:

(A) $\frac{3}{5}$

(B) $\frac{5}{8}$

(C) $\frac{3}{8}$

(D) $\frac{5}{3}$

The calculated probability matches option (C).

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The correct option is (C).

An event is considered impossible if it has no chance of happening.

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In probability theory, the probability of an event is defined as the ratio of the number of favourable outcomes to the total number of possible outcomes, assuming all outcomes are equally likely.

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For an impossible event:

The number of outcomes that result in the event occurring is 0.

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The total number of possible outcomes in any experiment is typically a positive number (unless the experiment has no outcomes, which is not a standard scenario). Let the total number of possible outcomes be $N$, where $N > 0$.

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Now, we can calculate the probability of the impossible event using the formula:

Since $N > 0$, any fraction with a numerator of 0 and a non-zero denominator is equal to 0.

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Thus, the probability of an impossible event is 0.

Comparing this result with the given options:

(A) 1

(B) 0

(C) 0.5

(D) Cannot be determined

The calculated probability matches option (B).

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The correct option is (B).

In probability theory, the set of all possible outcomes of an experiment is called the sample space.

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A fundamental rule of probability states that the sum of the probabilities of all mutually exclusive and exhaustive outcomes in a sample space must equal 1.

Since the set of all possible outcomes in an experiment constitutes the entire sample space, the sum of their individual probabilities must be 1.

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Let the possible outcomes of an experiment be $O_1, O_2, ..., O_n$. The sum of their probabilities is $P(O_1) + P(O_2) + ... + P(O_n)$.

According to the axioms of probability, this sum is always equal to 1.

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Therefore, the sum of all probabilities of all possible outcomes of an experiment is 1.

Comparing this with the given options:

(A) 0

(B) 1

(C) Less than 1

(D) Greater than 1

The correct value is 1.

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The correct option is (B).

The probability of any event must be a value between 0 and 1, inclusive.

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That is, for any event E, the probability $P(E)$ satisfies the inequality:

where 0 represents the probability of an impossible event, and 1 represents the probability of a sure event.

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Let's examine each of the given options:

(A) $\frac{1}{3}$

To check if this value is between 0 and 1, we can convert the fraction to a decimal:

Since $0 \leq 0.333... \leq 1$, this value can be a probability.

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(B) 0.75

This is already in decimal form. We check if it's between 0 and 1:

Since the inequality holds, this value can be a probability.

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(C) -0.2

This is a negative value.

Since probabilities cannot be negative, this value cannot be a probability.

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

To check if this value is between 0 and 1, we simplify the fraction:

Since $0 \leq 1 \leq 1$, this value (which represents the probability of a sure event) can be a probability.

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Based on the property that probability must be between 0 and 1 inclusive, the value -0.2 is the only option that falls outside this range.

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The correct option is (C).

When a standard six-sided die is rolled, the possible outcomes are the numbers on its faces.

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The sample space (set of all possible outcomes) for rolling a standard die is {1, 2, 3, 4, 5, 6}.

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To Find:

The probability of getting a number greater than 6.

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Solution:

We are interested in the event of getting a number greater than 6 from the sample space {1, 2, 3, 4, 5, 6}.

Let's check which outcomes in the sample space are greater than 6.

The numbers in the sample space are 1, 2, 3, 4, 5, 6. None of these numbers are greater than 6.

Therefore, there are no favourable outcomes for the event "getting a number greater than 6".

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The probability of an event is calculated using the formula:

Let E be the event of getting a number greater than 6.

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The probability of getting a number greater than 6 when a standard die is rolled is 0.

This is an example of an impossible event.

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Comparing this result with the given options:

(A) $\frac{1}{6}$

(B) $\frac{5}{6}$

(C) 0

(D) 1

The calculated probability matches option (C).

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The correct option is (C).

In probability, specific terms are used to describe the components of an experiment and its outcomes.

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Let's define the given options:

• Event: An event is a subset of the sample space. It is a specific outcome or a collection of specific outcomes of an experiment. For example, when rolling a die, getting an even number (which includes outcomes 2, 4, and 6) is an event.
• Sample space: The sample space, denoted by $S$ or $\Omega$, is the set of all possible outcomes of a random experiment. It lists every single result that could occur. For example, when rolling a standard die, the sample space is $\{1, 2, 3, 4, 5, 6\}$.
• Trial: A trial is a single performance of a random experiment. For example, rolling a die once is a trial.
• Probability: Probability is a numerical measure of the likelihood that a specific event will occur. It is a value between 0 and 1, inclusive.

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The question asks for the term that describes the collection of all possible outcomes of an experiment.

Based on the definitions, this corresponds to the definition of the Sample space.

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Comparing this with the given options:

(A) Event

(B) Sample space

(C) Trial

(D) Probability

The term that fits the description is Sample space.

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The correct option is (B).

The mode of a data set is the value that appears most frequently.

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Let's analyze each statement to determine if it is true about the mode:

(A) It is always one of the observations in the data.

The mode is defined as the value that occurs most often in the data set. Therefore, the mode must be one of the values that are present in the data set. For example, if the data is {2, 3, 3, 4, 5}, the mode is 3, which is one of the observations.

This statement is TRUE.

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(B) It is affected by extreme values.

Extreme values (very high or very low values compared to the rest of the data) can significantly affect the mean. However, the mode is determined solely by the frequency of values. The presence or absence of extreme values does not change the value that occurs most frequently, unless those extreme values become the most frequent. But a few extreme values usually do not become the most frequent. Consider the data {1, 2, 3, 4, 100}. The mode is not clearly defined (or we could say there's no mode if all appear once), but adding 100 doesn't change the frequency of 1, 2, 3, or 4. If we had {1, 1, 2, 3, 1000}, the mode is 1. The extreme value 1000 does not affect the mode. The mode is not generally affected by extreme values.

This statement is FALSE.

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(C) It is the middle value of the data.

The middle value of the data when arranged in order is called the median, not the mode. The mode is based on frequency, not position in the ordered data.

This statement is FALSE.

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(D) It is the average of the data.

The term "average" usually refers to the mean. The mode is a measure of central tendency, but it is not calculated by summing the data and dividing by the number of observations, which is how the mean is calculated. The mode is based on the frequency of values.

This statement is FALSE.

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Based on the analysis, the only true statement about the mode among the given options is that it is always one of the observations in the data.

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The correct option is (A).

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

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Assertion (A): The mean of the data 2, 4, 6, 8 is 5.

To check if the assertion is true, we need to calculate the mean of the data set {2, 4, 6, 8}.

The data has 4 observations: 2, 4, 6, 8.

The sum of the observations is $2 + 4 + 6 + 8 = 20$.

The number of observations is 4.

The mean is calculated as:

The calculated mean is 5. This matches the value given in the assertion.

Therefore, Assertion (A) is TRUE.

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Reason (R): Mean is calculated by summing all observations and dividing by the number of observations.

This statement provides the correct definition and formula for calculating the mean of a data set.

Therefore, 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 the mean of a specific data set. Reason (R) provides the method by which the mean is calculated, which is exactly the method used to verify Assertion (A). The reason explains *why* the mean of the data is 5 (because you sum them and divide by the count).

Thus, Reason (R) is the correct explanation of Assertion (A).

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Based on the analysis:

• Assertion (A) is true.
• Reason (R) is true.
• Reason (R) correctly explains Assertion (A).

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Comparing this 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.

Option (A) accurately describes the relationship between A and R.

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The correct option is (A).

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

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Assertion (A): The mode is the most appropriate measure of central tendency for data like shoe sizes.

Data like shoe sizes are typically discrete values. While they are numerical, the differences between sizes are fixed steps (e.g., size 8, size 8.5, size 9). Calculating the mean shoe size might result in a value that is not an actual available shoe size (e.g., a mean of 7.3). The median would give the middle size in a sorted list, but the most relevant information for someone like a shoe retailer is often the size that is purchased most frequently, as this impacts inventory and stock management.

The mode identifies the most frequent value in a data set. For shoe sizes, the mode would tell you the size that occurs most often. This information is highly practical and meaningful in this context.

Therefore, the mode is indeed often considered the most appropriate measure of central tendency for data like shoe sizes, especially from a practical standpoint (like determining which size to stock the most).

Assertion (A) is TRUE.

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Reason (R): The mode represents the most frequent value in a data set.

This statement is the definition of the mode in statistics. The mode is the value that appears with the highest frequency in a collection of data.

Reason (R) is TRUE.

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

Assertion (A) claims that the mode is appropriate for shoe sizes. Reason (R) defines the mode as the most frequent value. The fact that the mode identifies the most frequent value (i.e., the most commonly occurring shoe size) is precisely why it is considered an appropriate measure for shoe sizes. Retailers want to know the size that is most popular or most in demand, which is given by the mode.

Therefore, Reason (R) provides the underlying justification for why Assertion (A) is true.

Reason (R) is the correct explanation of Assertion (A).

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Based on the analysis:

• Assertion (A) is true.
• Reason (R) is true.
• Reason (R) correctly explains Assertion (A).

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Comparing this 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.

Option (A) accurately reflects our findings.

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The correct option is (A).

Given is the data collected from a survey in Anand Nagar regarding the number of vehicles per household.

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Given Data for Anand Nagar:

The number of vehicles per household for 10 households in Anand Nagar is:

1, 2, 1, 3, 2, 1, 1, 2, 3, 1

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To Find:

The mean number of vehicles per household in Anand Nagar.

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Solution:

The mean of a data set is calculated as the sum of all observations divided by the number of observations.

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First, let's find the sum of the observations for Anand Nagar:

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Next, let's count the number of observations for Anand Nagar.

There are 10 values in the data set for Anand Nagar.

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Now, we calculate the mean:

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The mean number of vehicles per household in Anand Nagar is 1.7.

Comparing this result with the given options:

(A) 1.5

(B) 1.7

(C) 2.0

(D) 2.5

The calculated mean matches option (B).

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The correct option is (B).

Given is the data collected from a survey in Shanti Park regarding the number of vehicles per household.

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Given Data for Shanti Park:

The number of vehicles per household for 10 households in Shanti Park is:

2, 2, 3, 1, 4, 2, 2, 1, 3, 2

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To Find:

The mode of the number of vehicles per household in Shanti Park.

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Solution:

The mode of a data set is the value that appears most frequently in the set.

To find the mode, we need to count the frequency of each unique value in the data set for Shanti Park.

The unique values in the data set {2, 2, 3, 1, 4, 2, 2, 1, 3, 2} are 1, 2, 3, and 4.

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Let's count the frequency of each value:

• Value 1 appears 2 times.
• Value 2 appears 5 times.
• Value 3 appears 2 times.
• Value 4 appears 1 time.

The frequencies are:

• Frequency of 1: 2
• Frequency of 2: 5
• Frequency of 3: 2
• Frequency of 4: 1

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The value with the highest frequency is the mode.

Comparing the frequencies, the highest frequency is 5, which corresponds to the value 2.

Therefore, the mode of the number of vehicles per household in Shanti Park is 2.

---

Comparing this result with the given options:

(A) 1

(B) 2

(C) 3

(D) 4

The calculated mode matches option (B).

---

The correct option is (B).

To determine which colony has a higher mean number of vehicles per household, we need to compare the mean calculated for each colony based on the data provided in Question 24.

---

Given Data (from Question 24):

Anand Nagar: 1, 2, 1, 3, 2, 1, 1, 2, 3, 1

Shanti Park: 2, 2, 3, 1, 4, 2, 2, 1, 3, 2

---

To Find:

Which colony has a higher mean number of vehicles per household.

---

Solution:

We need to calculate the mean for Anand Nagar and Shanti Park separately.

The mean is calculated using the formula:

---

Mean for Anand Nagar:

Data: 1, 2, 1, 3, 2, 1, 1, 2, 3, 1

---

Mean for Shanti Park:

Data: 2, 2, 3, 1, 4, 2, 2, 1, 3, 2

---

Comparison:

Mean (Anand Nagar) = 1.7

Mean (Shanti Park) = 2.2

Comparing the two mean values, we see that $2.2 > 1.7$.

Therefore, the mean number of vehicles per household is higher in Shanti Park than in Anand Nagar.

---

Comparing our result with the given options:

(A) Anand Nagar

(B) Shanti Park

(C) Both have the same mean

(D) Cannot be determined from the data

Our finding matches option (B).

---

The correct option is (B).

Measures of central tendency are statistical values that represent the center or typical value of a data set. The most common measures of central tendency are the mean, median, and mode.

---

Let's define each of the given options:

• Mean: The mean is the arithmetic average of a data set, calculated by summing all the values and dividing by the number of values. It represents the balance point of the data.
• Median: The median is the middle value in a data set that has been ordered from least to greatest. It divides the data set into two equal halves.
• Mode: The mode is the value that appears most frequently in a data set. It represents the most common value.
• Range: The range of a data set is the difference between the highest value and the lowest value in the set. It is a measure of the spread or dispersion of the data, indicating how much the data varies, rather than representing a central value.

---

The question asks which of the following is NOT a measure of central tendency.

Based on the definitions, the Mean, Median, and Mode are all measures of central tendency as they represent a central or typical value of the data.

The Range measures the spread, not the center, of the data.

---

Therefore, the Range is not a measure of central tendency.

Comparing this with the given options:

(A) Mean

(B) Median

(C) Mode

(D) Range

The option that is not a measure of central tendency is (D).

---

The correct option is (D).

The median is a measure of central tendency that represents the middle value of a data set when it is arranged in order.

---

Let's analyze each statement about finding the median of a data set:

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(A) The data must be arranged in ascending or descending order.

This is a fundamental step in finding the median. The median is defined based on the position of values in an ordered list. If the data is not ordered, finding the middle value is meaningless in the context of the median.

This statement is TRUE.

---

(B) If the number of observations is odd, the median is the middle observation.

If a data set has an odd number of observations, say $n$, then after arranging the data in order, there is exactly one middle observation at the $\left(\frac{n+1}{2}\right)^{\text{th}}$ position. This middle observation is the median.

For example, in the ordered data set {3, 5, 7, 9, 11} ($n=5$), the middle observation is at the $\left(\frac{5+1}{2}\right)^{\text{th}} = 3^{\text{rd}}$ position, which is 7. The median is 7.

This statement is TRUE.

---

(C) If the number of observations is even, the median is the average of the two middle observations.

If a data set has an even number of observations, say $n$, then after arranging the data in order, there are two middle observations. These are at the $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2} + 1\right)^{\text{th}}$ positions. The median is defined as the arithmetic mean (average) of these two middle observations.

For example, in the ordered data set {2, 4, 6, 8} ($n=4$), the middle observations are at the $\left(\frac{4}{2}\right)^{\text{th}} = 2^{\text{nd}}$ position (which is 4) and the $\left(\frac{4}{2} + 1\right)^{\text{th}} = 3^{\text{rd}}$ position (which is 6). The median is $\frac{4+6}{2} = \frac{10}{2} = 5$.

This statement is TRUE.

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(D) The median is always one of the observations in the data set.

As shown in the example for an even number of observations in statement (C), the median (5) for the data set {2, 4, 6, 8} is not present in the original data set. The median is only guaranteed to be one of the observations if the number of observations is odd.

Because the statement says "always", and it is not true for data sets with an even number of observations, this statement is INCORRECT.

---

We are looking for the incorrect statement. Based on our analysis, statement (D) is incorrect.

---

The correct option is (D).

Let's review the definitions of the terms provided in the options:

---

• Mean: The arithmetic average of a data set.
• Median: The middle value of a data set when arranged in order.
• Mode: The value that appears most often or with the highest frequency in a data set.
• Frequency: The number of times a particular observation occurs in a data set.

---

The statement asks for the term that describes the observation (value) that occurs most frequently.

Based on the definitions, the term that fits this description is the Mode.

---

The completed statement is: The observation that occurs most frequently in a data set is called the Mode.

---

Comparing this with the given options:

(A) Mean

(B) Median

(C) Mode

(D) Frequency

The correct term is Mode, which is option (C).

---

The correct option is (C).

Let's review the definitions of the terms provided in the options related to statistical measures:

---

• Mean: This is the average of a data set. It is calculated by summing all the values and dividing by the total number of values.
• Range: This is a measure of the spread or dispersion of a data set. It is calculated as the difference between the maximum (highest) value and the minimum (lowest) value in the data set.
• Median: This is the middle value of a data set when the observations are arranged in ascending or descending order.
• Mode: This is the value that appears most frequently in a data set.

---

The statement asks for the term that represents "The difference between the highest and the lowest observation in a data set".

Based on the definitions above, this description perfectly matches the definition of the Range.

---

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

---

Comparing this with the given options:

(A) Mean

(B) Range

(C) Median

(D) Mode

The correct term is Range, which corresponds to option (B).

---

The correct option is (B).

We need to match each statistical term with its correct definition.

---

Let's consider each term and find its corresponding definition from the given list:

(i) Mean: The mean is the arithmetic average of a set of observations.

Looking at the definitions:

• (a) The difference between the highest and lowest values. (This is the Range)
• (b) The middle value of a sorted data set. (This is the Median)
• (c) The average of the observations. (This matches the definition of Mean)
• (d) The most frequent observation. (This is the Mode)

So, (i) matches with (c).

---

(ii) Median: The median is the middle value of a data set when it is arranged in ascending or descending order.

Looking at the remaining definitions:

• (b) The middle value of a sorted data set. (This matches the definition of Median)

So, (ii) matches with (b).

---

(iii) Mode: The mode is the value that occurs most frequently in a data set.

Looking at the remaining definitions:

• (d) The most frequent observation. (This matches the definition of Mode)

So, (iii) matches with (d).

---

(iv) Range: The range of a data set is the difference between the highest and the lowest values.

Looking at the remaining definitions:

• (a) The difference between the highest and lowest values. (This matches the definition of Range)

So, (iv) matches with (a).

---

The correct matching is:

• (i) Mean - (c) The average of the observations.
• (ii) Median - (b) The middle value of a sorted data set.
• (iii) Mode - (d) The most frequent observation.
• (iv) Range - (a) The difference between the highest and lowest values.

This can be written as: (i)-(c), (ii)-(b), (iii)-(d), (iv)-(a).

---

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

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

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

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

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

Option (A) matches our correct matching.

---

The correct option is (A).

Given that a bag contains slips with numbers from 1 to 10 written on them.

---

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

The sample space (set of all possible outcomes when drawing a slip) is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

---

To Find:

The probability of drawing a slip with an even number.

---

Solution:

We are interested in the event of drawing a slip with an even number.

The even numbers in the sample space (from 1 to 10) are 2, 4, 6, 8, 10.

The set of favourable outcomes is {2, 4, 6, 8, 10}.

---

The probability of an event is calculated using the formula:

Let E be the event of drawing a slip with an even number.

---

The fraction $\frac{5}{10}$ can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 5.

---

The probability of drawing a slip with an even number is $\frac{5}{10}$ or its simplified form $\frac{1}{2}$.

Comparing this result with the given options:

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

(B) $\frac{5}{10}$

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

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

Options (B) and (D) represent the same probability. Option (D) is the simplified form of option (B).

Both $\frac{5}{10}$ and $\frac{1}{2}$ are correct probabilities. However, typically the simplified fraction is preferred.

Both (B) and (D) are mathematically equivalent and correct answers. In a multiple-choice question, if both simplified and unsimplified correct answers are provided, either one is technically correct unless a specific format (like simplest form) is requested. Let's check if one is a better fit based on common practice or if there's a most simplified form provided.

Option (D) $\frac{1}{2}$ is the simplified form of $\frac{5}{10}$. Both are listed as options.

Often, multiple-choice questions with fractional answers include the simplest form. Let's assume the intended answer is the most simplified correct option.

---

The probability is $\frac{5}{10}$, which simplifies to $\frac{1}{2}$. Both $\frac{5}{10}$ (Option B) and $\frac{1}{2}$ (Option D) are among the choices.

Since $\frac{1}{2}$ is the simplified form, let's select option (D) as the most likely intended correct answer.

---

The correct option is (D).

Given is a set of 5 observations and their mean.

---

Given:

Observations: $x, x+2, x+4, x+6, x+8$

Number of observations ($n$) = 5

Mean of the observations = 11

---

To Find:

The value of $x$.

---

Solution:

The mean of a set of observations is calculated using the formula:

---

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

Combine the $x$ terms and the constant terms:

---

Now, substitute the sum, the number of observations, and the given mean into the mean formula:

To solve for $x$, multiply both sides of the equation by 5:

Subtract 20 from both sides of the equation:

Divide both sides by 5:

---

The value of $x$ is 7.

---

Checking the options:

(A) 5

(B) 7

(C) 9

(D) 11

The calculated value of $x$ matches option (B).

---

The correct option is (B).

A double bar graph is a type of bar graph used to compare two related sets of data. It displays two bars side-by-side for each category being compared, allowing for easy visual comparison between the two data sets within each category.

---

Let's examine each option to see if the data described would be suitable for a double bar graph:

(A) Marks of students in one subject over different tests.

This could mean the marks of a single student over several tests (best represented by a line graph or simple bar graph) or the marks of multiple students on a single test (simple bar graph). If it meant comparing, for instance, the average marks of two different classes on a series of tests, a double bar graph would be suitable. However, the phrasing is general ("students" plural). A standard double bar graph is usually for comparing two *sets* of measures across several items/categories, not necessarily comparing multiple individual students simultaneously in this specific format.

---

(B) Height of students in a class.

This is a single set of data (heights). It would typically be represented by a simple bar graph (for individual heights) or a histogram (if heights are grouped into intervals). A double bar graph would only be appropriate if you were comparing, for example, the heights of students in Class A versus Class B.

---

(C) Comparison of performance of two teams in different matches.

This is a classic example where a double bar graph is very useful. You can list the different matches on the x-axis and the performance metric (like score, points, etc.) on the y-axis. For each match, you would have two bars, one representing Team 1's performance and the other representing Team 2's performance in that specific match. This allows for a direct comparison of how the two teams performed against each other match by match.

---

(D) Favourite colours of students.

This data typically involves counting the frequency of each colour preference among a group of students. It is best represented by a simple bar graph or a pie chart. A double bar graph would be used if you were comparing, for example, the favourite colours of boys versus girls.

---

Out of the given options, the scenario that most directly and commonly utilizes a double bar graph for visualization is the comparison of two sets of related data (the performance of two teams) across multiple categories (different matches).

---

The correct option is (C).

The mode of a data set is the value that appears most frequently in the set.

---

Given is the data set representing the number of siblings students have:

1, 2, 0, 1, 3, 2, 1, 0, 1, 2

---

To Find:

The mode of this data.

---

Solution:

To find the mode, we need to determine the frequency of each unique value in the data set.

The unique values present in the data are 0, 1, 2, and 3.

---

Let's count how many times each value appears:

• The value 0 appears 2 times.
• The value 1 appears 4 times.
• The value 2 appears 3 times.
• The value 3 appears 1 time.

The frequencies of the values are:

• Frequency of 0: 2
• Frequency of 1: 4
• Frequency of 2: 3
• Frequency of 3: 1

---

The value with the highest frequency is the mode.

Comparing the frequencies (2, 4, 3, 1), the highest frequency is 4. This frequency corresponds to the value 1.

Therefore, the mode of the data set is 1.

---

Comparing this result with the given options:

(A) 0

(B) 1

(C) 2

(D) 3

The calculated mode matches option (B).

---

The correct option is (B).

The median of a data set is the middle value when the data is arranged in order.

---

Given is the data set: 12, 15, 11, 13, 16, 14.

---

To Find:

The median of the given data.

---

Solution:

First, we need to arrange the data in ascending or descending order. Let's arrange it in ascending order:

11, 12, 13, 14, 15, 16

---

Next, we need to find the middle value(s).

There are 6 observations in the data set. The number of observations ($n$) is 6.

Since $n$ is an even number, the median is the average of the two middle values. The positions of the two middle values are at the $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2} + 1\right)^{\text{th}}$ positions in the ordered data.

---

Now, we find the values at the $3^{\text{rd}}$ and $4^{\text{th}}$ positions in the ordered data (11, 12, 13, 14, 15, 16).

The value at the $3^{\text{rd}}$ position is 13.

The value at the $4^{\text{th}}$ position is 14.

---

The median is the average of these two middle values:

---

The median of the given data set is 13.5.

Comparing this result with the given options:

(A) 13

(B) 14

(C) 13.5

(D) 14.5

The calculated median matches option (C).

---

The correct option is (C).

Given is the result of an experiment involving coin tosses.

---

Given:

Total number of times the coin was tossed = 50

Number of times a Head appeared = 28

---

To Find:

The experimental probability of getting a Tail.

---

Solution:

In a coin toss, the outcomes are either Head or Tail. The total number of tosses is the sum of the number of times a Head appeared and the number of times a Tail appeared.

We know the Total Tosses and the Number of Heads. We can find the Number of Tails:

Subtract 28 from both sides:

---

The experimental probability of an event is given by the formula:

Here, the event is getting a Tail. The number of times a Tail occurred is 22, and the total number of trials is 50.

---

The probability of getting a tail is $\frac{22}{50}$. This fraction can be simplified by dividing the numerator and the denominator by their greatest common divisor, which is 2:

However, the simplified form $\frac{11}{25}$ is not one of the options. The unsimplified fraction $\frac{22}{50}$ is provided as an option.

---

Comparing our result with the given options:

(A) $\frac{28}{50}$ (This is the probability of getting a Head)

(B) $\frac{22}{50}$ (This is the probability of getting a Tail, matching our calculation)

(C) $\frac{1}{2}$ (This is the theoretical probability for a fair coin, not based on the experiment)

(D) $\frac{28}{22}$ (This is the ratio of Heads to Tails, not a probability)

The calculated probability of getting a tail based on the experiment is $\frac{22}{50}$, which is option (B).

---

The correct option is (B).

Definition of Data:

Data refers to raw facts, figures, observations, or measurements collected for a specific purpose. Data can be in the form of numbers, text, images, sounds, or other types of information. It is the basic input for analysis, interpretation, and decision-making.

---

Example of Data from Surroundings:

An example of data you can collect from your surroundings is the number of different types of vehicles (like cars, bikes, buses) that pass by your window in a specific time interval (e.g., 15 minutes). This would give you quantitative data.

Alternatively, you could collect data on the colour of the flowers in a park (e.g., counting how many are red, yellow, blue, etc.). This would be qualitative data that can be counted.

Raw Data:

Raw data is the data collected in its original, unprocessed form. It is a collection of facts and figures that have not been sorted, classified, or analyzed in any way. Raw data is typically scattered, unarranged, and may contain errors or inconsistencies.

---

Organized Data:

Organized data is data that has been arranged, sorted, classified, and presented in a systematic manner. This process makes the data understandable, easy to analyze, and suitable for drawing conclusions. Organizing data involves steps like grouping data, creating tables, charts, or graphs.

---

In essence, raw data is the initial, unstructured information, while organized data is the refined and structured version of raw data, made ready for interpretation and use.

Purpose of a Bar Graph:

A bar graph (or bar chart) is used to visually represent and compare data using rectangular bars of different heights or lengths. It is particularly useful for comparing different categories of data, showing trends over time, or illustrating the distribution of a single variable across different groups.

---

Key Features of a Bar Graph:

Here are the key features of a bar graph:

1. Axes: A bar graph has two axes, usually a horizontal axis (x-axis) and a vertical axis (y-axis).

2. Labels: Both axes are labeled clearly. One axis (often the horizontal) represents the categories being compared (e.g., months, items, groups), while the other axis (often the vertical) represents the values or frequency of each category (e.g., sales figures, counts, percentages).

3. Bars: Rectangular bars are drawn for each category. The height (or length) of each bar corresponds to the value it represents on the value axis.

4. Uniform Width: All bars should have the same width to avoid misleading comparisons.

5. Uniform Spacing: There should be uniform space between consecutive bars. This spacing is usually less than the width of the bars.

6. Scale: The value axis has a scale marked with appropriate intervals, starting from zero, to accurately represent the magnitudes.

7. Title: A clear and concise title is usually provided to explain what the graph represents.

Definition of a Double Bar Graph:

A double bar graph is a variation of a bar graph that uses pairs of bars to compare two sets of data side-by-side for each category. For each category on the horizontal axis, there are two adjacent bars, each representing a value from one of the two data sets being compared. Different colors or patterns are typically used for the bars of the two different data sets to distinguish them, and a legend is provided to identify which bar represents which data set.

---

When is it Useful?

A double bar graph is particularly useful when you need to:

1. Compare two related data sets: For example, comparing the sales figures of two different products over several months, or comparing the performance of two groups of students on the same test.

2. Show changes over time for two variables: Such as comparing the population of males and females in a town over different census years.

3. Make direct visual comparisons: It allows for easy visual comparison of the values of the two data sets for each category, making it simple to see differences, trends, or relationships between the two sets of information.

Representative values (also known as measures of central tendency) are single values that are used to describe or summarize an entire set of data. They aim to find a "typical" or "central" value within the data set, giving us a concise way to understand the overall distribution of the data without looking at every single data point.

---

These values help in comparing different data sets or understanding the general characteristics of a single data set. Instead of listing all the individual observations, a representative value provides a snapshot of where the data tends to cluster.

---

The most common representative values are:

1. Mean (Average): Calculated by summing all the data points and dividing by the number of data points. It represents the arithmetic average.

2. Median: The middle value in a data set that has been arranged in ascending or descending order. It divides the data into two equal halves.

3. Mode: The value that appears most frequently in the data set.

---

Choosing the appropriate representative value depends on the nature of the data and the purpose of the analysis. For example, the mean is sensitive to extreme values, while the median is not.

Definition of Mean (Average):

The mean, or average, of a data set is a measure of central tendency. It is calculated by dividing the sum of all the observations or values in the data set by the total number of observations.

---

Formula for the Mean of Ungrouped Data:

For a data set with $n$ observations: $x_1, x_2, x_3, ..., x_n$, the mean ($\overline{x}$) is calculated using the formula:

$\overline{x} = \frac{\text{Sum of all observations}}{\text{Total number of observations}}$

In mathematical notation, this can be written as:

$\overline{x} = \frac{x_1 + x_2 + x_3 + ... + x_n}{n}$

This can also be represented using summation notation:

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

where $\sum$ represents the sum of the terms, $x_i$ represents the $i$-th observation, and $n$ is the total number of observations.

Given:

The data set is: $5, 8, 10, 12, 15$.

---

To Find:

The mean (average) of the given data.

---

Solution:

The mean ($\overline{x}$) of ungrouped data is calculated using the formula:

$\overline{x} = \frac{\text{Sum of all observations}}{\text{Total number of observations}}$

The observations are $5, 8, 10, 12, 15$.

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

Sum of observations = $5 + 8 + 10 + 12 + 15$

Sum of observations = $50$

Now, applying the formula for the mean:

$\overline{x} = \frac{50}{5}$

$\overline{x} = 10$

---

The mean of the given data is $\mathbf{10}$.

Definition of Median:

The median is a measure of central tendency that represents the middle value of a data set when the data points are arranged in either ascending or descending order. It is the value that divides the data set into two halves, with an equal number of observations above and below it.

---

Finding the Median with an Odd Number of Observations:

To find the median of a data set with an odd number of observations, follow these steps:

1. Arrange the data: Arrange all the observations in either ascending (smallest to largest) or descending (largest to smallest) order.

2. Identify the middle position: If there are $n$ observations in total, and $n$ is an odd number, the median will be the observation at the $\frac{n+1}{2}$-th position in the ordered list.

3. The middle value is the median: The value of the observation located at the $\frac{n+1}{2}$-th position is the median of the data set.

For example, if you have 7 observations ($n=7$), the median will be the observation at the $\frac{7+1}{2} = \frac{8}{2} = 4$-th position after arranging the data.

Given:

The data set is: $11, 15, 18, 20, 22, 25, 28$.

---

To Find:

The median of the given data.

---

Solution:

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

The given data is already arranged in ascending order: $11, 15, 18, 20, 22, 25, 28$.

Next, we count the number of observations ($n$) in the data set.

Number of observations, $n = 7$.

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

Median position = $\frac{7+1}{2}$ = $\frac{8}{2}$ = $4$-th position.

Now, we find the value of the observation at the 4th position in the ordered data set:

1st position: 11

2nd position: 15

3rd position: 18

4th position: 20

5th position: 22

6th position: 25

7th position: 28

The value at the 4th position is 20.

---

Therefore, the median of the data is $\mathbf{20}$.

To find the median of a data set with an even number of observations, follow these steps:

---

1. Arrange the data: Arrange all the observations in either ascending (smallest to largest) or descending (largest to smallest) order.

---

2. Identify the two middle positions: If there are $n$ observations in total, and $n$ is an even number, the median will be the average of the two middle observations. These two middle observations are located at the $\frac{n}{2}$-th position and the $(\frac{n}{2} + 1)$-th position in the ordered list.

---

3. Calculate the average of the middle values: Find the values of the observations at the $\frac{n}{2}$-th position and the $(\frac{n}{2} + 1)$-th position. The median is the mean (average) of these two values.

Median = $\frac{\text{Value at } (\frac{n}{2})\text{-th position} + \text{Value at } (\frac{n}{2} + 1)\text{-th position}}{2}$

For example, if you have 8 observations ($n=8$), the two middle positions are the $\frac{8}{2} = 4$-th position and the $(\frac{8}{2} + 1) = 5$-th position. The median would be the average of the values at the 4th and 5th positions.

Given:

The data set is: $4, 7, 9, 11, 13, 16$.

---

To Find:

The median of the given data.

---

Solution:

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

The given data is already arranged in ascending order: $4, 7, 9, 11, 13, 16$.

Next, we count the number of observations ($n$) in the data set.

Number of observations, $n = 6$.

Since the number of observations ($n=6$) is an even number, the median is the average of the two middle values.

The positions of the two middle values are the $\frac{n}{2}$-th position and the $(\frac{n}{2} + 1)$-th position.

$\frac{n}{2} = \frac{6}{2} = 3$-rd position.

$\frac{n}{2} + 1 = 3 + 1 = 4$-th position.

The values at these positions in the ordered list are:

3rd position: 9

4th position: 11

The median is the average of these two values:

Median = $\frac{9 + 11}{2}$

Median = $\frac{20}{2}$

Median = 10

---

Therefore, the median of the data is $\mathbf{10}$.

Definition of Mode:

The mode is a measure of central tendency that represents the value or observation that appears most frequently in a data set. It is the value with the highest frequency.

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Can a Data Set Have More Than One Mode?

Yes, a data set can have more than one mode. This occurs when two or more values have the same highest frequency.

1. If a data set has one mode, it is called unimodal.

2. If a data set has two modes with the same highest frequency, it is called bimodal.

3. If a data set has more than two modes with the same highest frequency, it is called multimodal.

4. If all values in the data set appear with the same frequency, then the data set has no mode (or sometimes considered that every value is a mode, but the standard convention is often 'no mode' in this case).

Given:

The data set is: $2, 3, 5, 2, 4, 2, 3, 5, 2, 1, 5$.

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To Find:

The mode of the given data.

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Solution:

To find the mode, we need to identify the observation that occurs most frequently in the data set.

Let's count the frequency of each distinct value in the data set:

• Value 1 appears 1 time.
• Value 2 appears 4 times.
• Value 3 appears 2 times.
• Value 4 appears 1 time.
• Value 5 appears 3 times.

Comparing the frequencies, the value '2' has the highest frequency, which is 4.

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Therefore, the mode of the data set is $\mathbf{2}$.

Definition of Range:

The range of a data set is a measure of dispersion or spread. It is the difference between the highest value and the lowest value in the data set. It gives a simple indication of how spread out the data is.

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Given:

The data set is: $25, 32, 18, 45, 20, 30$.

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To Find:

The range of the given data.

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Solution:

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

The given data points are $25, 32, 18, 45, 20, 30$.

The highest value in the data set is $\mathbf{45}$.

The lowest value in the data set is $\mathbf{18}$.

The formula for the range is:

Range = Highest value - Lowest value

Range = $45 - 18$

Range = $27$

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Therefore, the range of the data set is $\mathbf{27}$.

Definition of Probability:

Probability is a branch of mathematics that deals with the study of random events. It is a numerical measure of the likelihood or chance that a specific event will occur. Probability is expressed as a number between 0 and 1, inclusive, where 0 indicates that the event is impossible, and 1 indicates that the event is certain to occur. A higher probability value means a greater chance of the event happening.

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How is it measured?

For simple events in a situation where all possible outcomes are equally likely, probability is typically measured using the following formula:

Probability of an Event = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

where:

Favorable outcomes are the outcomes where the specific event you are interested in occurs.

Total number of possible outcomes is the total number of distinct outcomes that can occur in the given random experiment.

For example, if you flip a fair coin, there are two possible outcomes: Heads or Tails (Total outcomes = 2). The probability of getting Heads is $\frac{1}{2}$ because there is 1 favorable outcome (Heads) out of 2 total possible outcomes.

When a single fair coin is tossed, there are two possible outcomes.

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The possible outcomes are:

1. Heads (H)

2. Tails (T)

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These two outcomes are typically considered equally likely for a fair coin.

When a standard six-sided die is thrown, the total possible outcomes are the numbers on its faces.

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Total possible outcomes = $\{1, 2, 3, 4, 5, 6\}$.

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We are interested in the outcomes that are odd numbers.

The odd numbers within the set of possible outcomes are 1, 3, and 5.

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Therefore, the possible outcomes of getting an odd number when a die is thrown are:

Outcomes for getting an odd number = $\{1, 3, 5\}$.

Given:

A fair coin is tossed.

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To Find:

The probability of getting a Head.

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Solution:

When a fair coin is tossed, there are two possible outcomes:

Total possible outcomes = $\{$Head, Tail$\}$.

The total number of possible outcomes is $2$.

The favorable outcome (the event we are interested in) is getting a Head.

Favorable outcome = $\{$Head$\}$.

The number of favorable outcomes is $1$.

The probability of an event is calculated using 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 Head.

$P(\text{Getting a Head}) = \frac{\text{Number of Heads}}{\text{Total number of outcomes}}$

$P(\text{Getting a Head}) = \frac{1}{2}$

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Therefore, the probability of getting a Head when a coin is tossed is $\mathbf{\frac{1}{2}}$.

Given:

A standard six-sided die is thrown.

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To Find:

The probability of getting an even number.

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Solution:

When a standard six-sided die is thrown, the total possible outcomes are the numbers on its faces.

Total possible outcomes = $\{1, 2, 3, 4, 5, 6\}$.

The total number of possible outcomes is $6$.

We are interested in the event of getting an even number. The even numbers in the set of possible outcomes are 2, 4, and 6.

Favorable outcomes (getting an even number) = $\{2, 4, 6\}$.

The number of favorable outcomes is $3$.

The probability of an event is calculated using the formula:

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

In this case, the event is getting an even number.

$P(\text{Getting an even number}) = \frac{\text{Number of even numbers}}{\text{Total number of outcomes}}$

$P(\text{Getting an even number}) = \frac{3}{6}$

This fraction can be simplified:

$P(\text{Getting an even number}) = \frac{\cancel{3}^{1}}{\cancel{6}_{2}}$

$P(\text{Getting an even number}) = \frac{1}{2}$

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Therefore, the probability of getting an even number when a die is thrown is $\mathbf{\frac{1}{2}}$.

Given:

Number of red balls in the bag = $3$.

Number of blue balls in the bag = $5$.

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To Find:

The probability of drawing a red ball from the bag.

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Solution:

First, we need to find the total number of balls in the bag. This is the total number of possible outcomes when drawing a ball at random.

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

Total number of possible outcomes = $3 + 5 = 8$.

The event we are interested in is drawing a red ball. The number of favorable outcomes for this event is the number of red balls in the bag.

Number of favorable outcomes (drawing a red ball) = Number of red balls = $3$.

The probability of an event is calculated using the formula:

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

In this case, the event is drawing a red ball.

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

$P(\text{Drawing a red ball}) = \frac{3}{8}$

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Therefore, the probability of drawing a red ball from the bag is $\mathbf{\frac{3}{8}}$.

No, the probability of an event cannot be greater than $1$.

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Explanation:

Probability measures the likelihood of an event occurring relative to the total number of possible outcomes. The formula for calculating the probability of a simple event is:

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

Consider the numerator, "Number of favorable outcomes". This represents the count of outcomes where the specific event we are interested in happens.

Consider the denominator, "Total number of possible outcomes". This represents the count of all distinct outcomes that can possibly occur in the random experiment.

The number of favorable outcomes for any event must always be less than or equal to the total number of possible outcomes. It is impossible for the number of ways a specific event can happen to exceed the total number of different things that could possibly happen.

So, mathematically, for any event:

When you divide a number by another number that is greater than or equal to it (and not zero), the result will always be less than or equal to $1$.

For example, if there are 10 possible outcomes and 3 are favorable, the probability is $\frac{3}{10} = 0.3$. If all 10 outcomes are favorable (the event is certain), the probability is $\frac{10}{10} = 1$. It's not possible to have more than 10 favorable outcomes out of 10 total outcomes.

Thus, the probability of any event must lie in the range from $0$ to $1$, inclusive:

$0 \leq P(\text{Event}) \leq 1$

A probability of $0$ means the event is impossible, and a probability of $1$ means the event is certain.

Definition of a Sure Event:

A sure event (or certain event) is an event that is guaranteed to happen in a random experiment. It is an event that includes all possible outcomes of the experiment.

The probability of a sure event is always $\mathbf{1}$ (or 100%).

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Example:

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

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

An example of a sure event is "getting a number less than 7" when rolling the die. The outcomes for this event are $\{1, 2, 3, 4, 5, 6\}$, which include all the possible outcomes of the experiment.

Since all possible outcomes are part of this event, it is certain to occur.

The probability of this sure event is:

$P(\text{Number less than 7}) = \frac{\text{Number of outcomes less than 7}}{\text{Total number of outcomes}} = \frac{6}{6} = 1$

Definition of an Impossible Event:

An impossible event is an event that cannot happen in a random experiment. There are no outcomes in the sample space that correspond to this event.

The probability of an impossible event is always $\mathbf{0}$.

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Example:

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

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

An example of an impossible event is "getting a number greater than 6" when rolling the die. The outcomes for this event are an empty set, as no face of a standard die shows a number greater than 6.

Since there are no favorable outcomes for this event, it cannot occur.

The probability of this impossible event is:

$P(\text{Number greater than 6}) = \frac{\text{Number of outcomes greater than 6}}{\text{Total number of outcomes}} = \frac{0}{6} = 0$

In probability, the terms 'experiment' and 'outcome' are fundamental concepts.

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Experiment:

An experiment is an action or process that can be repeated and that produces a set of well-defined possible results. The result of any single performance of the experiment is not known beforehand with certainty.

In the context of drawing a card from a well-shuffled deck of $52$ cards, the experiment is the act of drawing one card from the deck.

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Outcome:

An outcome is one of the possible results of a random experiment. Each outcome is equally likely if the experiment is fair.

In the context of drawing a card from a deck of $52$ cards, an outcome is the specific card that is drawn. Since there are $52$ different cards in a standard deck (Ace of Spades, 3 of Hearts, King of Diamonds, etc.), there are $52$ possible outcomes for this experiment.

Given:

A spinner with 4 equal sectors colored Red, Blue, Green, and Yellow.

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To Find:

The probability of landing on the Green sector.

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Solution:

When the spinner is spun, there are 4 possible outcomes, corresponding to each of the sectors.

Total number of possible outcomes = 4 (Red, Blue, Green, Yellow).

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

Number of favorable outcomes (landing on Green) = 1 (only the Green sector).

The probability of an event is calculated using the formula:

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

In this case, the event is landing on the Green sector.

$P(\text{Landing on Green}) = \frac{\text{Number of Green sectors}}{\text{Total number of sectors}}$

$P(\text{Landing on Green}) = \frac{1}{4}$

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Therefore, the probability of landing on the Green sector is $\mathbf{\frac{1}{4}}$.

Given:

The heights (in cm) of 10 students are recorded as: $140, 142, 135, 148, 140, 145, 138, 140, 146, 136$.

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To Find:

Mean, Median, and Mode of the given data. Which measure best represents the typical height?

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Solution:

We will calculate the Mean, Median, and Mode step-by-step.

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1. Mean

The mean ($\overline{x}$) of ungrouped data is calculated by the formula:

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

Given observations ($x_i$): $140, 142, 135, 148, 140, 145, 138, 140, 146, 136$

Sum of observations ($ \sum x $):

$140 + 142 + 135 + 148 + 140 + 145 + 138 + 140 + 146 + 136 = 1410$

Number of observations ($ n $):

$n = 10$

Mean ($\overline{x}$):

$\overline{x} = \frac{1410}{10} = 141$

The mean height is $141$ cm.

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2. Median

To find the median, we first arrange the data in ascending order.

Sorted data:

Student Index (Sorted)	Height (cm)
1	135
2	136
3	138
4	140
5	140
6	140
7	142
8	145
9	146
10	148

The number of observations is $ n = 10 $, which is an even number.

For an even number of observations, the median is the average of the $(\frac{n}{2})$-th term and the $(\frac{n}{2} + 1)$-th term.

Here, $\frac{n}{2} = \frac{10}{2} = 5$-th term and $\frac{n}{2} + 1 = 5 + 1 = 6$-th term.

From the sorted data, the 5th term is $140$ and the 6th term is $140$.

Median = $\frac{\text{5th term} + \text{6th term}}{2}$

Median = $\frac{140 + 140}{2} = \frac{280}{2} = 140$

The median height is $140$ cm.

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3. Mode

The mode is the observation that occurs most frequently in the data.

Let's count the frequency of each height:

135: 1

136: 1

138: 1

140: 3

142: 1

145: 1

146: 1

148: 1

The height $140$ cm occurs most frequently (3 times).

The mode height is $140$ cm.

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Which measure of central tendency best represents the typical height?

We have found the following measures:

Mean = $141$ cm

Median = $140$ cm

Mode = $140$ cm

All three measures are very close to each other. The dataset does not contain extreme outliers, suggesting that the mean is a suitable measure of the average height.

However, the term "typical" can also refer to the value that is most common (mode) or the value that lies exactly in the middle of the data (median).

Since the mode ($140$ cm) and the median ($140$ cm) are identical and $140$ cm is the most frequent height observed among the students, the Median and Mode are arguably the best measures to represent the typical height of the students in this specific group, as $140$ cm is both the most common height and the middle value in the sorted data.

Given:

Data showing the number of ice cream cones of different flavours sold on two consecutive days.

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To Plot:

A double bar graph to represent the given data.

Draw conclusions from the graph.

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Solution (Steps to draw the Double Bar Graph):

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

2. Label the x-axis as "Flavour" and mark points for each flavour (Vanilla, Chocolate, Strawberry, Mango).

3. Label the y-axis as "Number of Ice Cream Cones Sold".

4. Choose an appropriate scale for the y-axis. Since the maximum number of cones sold for any flavour on any day is 45, a scale of 1 unit = 5 cones would be suitable. Mark values from 0 up to 50 or 60 on the y-axis with intervals of 5 or 10.

5. For each flavour on the x-axis, draw two adjacent bars. One bar will represent the sales on Day 1, and the other will represent the sales on Day 2.

6. Use different colours or patterns to distinguish the bars for Day 1 and Day 2.

7. The height of the bar for Day 1 of a particular flavour should correspond to the number of cones sold on Day 1 for that flavour, according to the chosen scale on the y-axis.

8. Similarly, the height of the adjacent bar for Day 2 of the same flavour should correspond to the number of cones sold on Day 2.

9. Leave a uniform gap between different pairs of bars (representing different flavours).

10. Provide a key or legend to indicate which colour/pattern represents Day 1 and which represents Day 2.

Based on the data:

• For Vanilla, draw bars up to 25 (Day 1) and 30 (Day 2).
• For Chocolate, draw bars up to 40 (Day 1) and 35 (Day 2).
• For Strawberry, draw bars up to 30 (Day 1) and 45 (Day 2).
• For Mango, draw bars up to 20 (Day 1) and 25 (Day 2).

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Conclusions from the Graph:

By observing the double bar graph, we can draw the following conclusions:

1. Strawberry flavour saw a significant increase in sales from Day 1 (30) to Day 2 (45), which is the largest increase among all flavours.

2. Chocolate was the most popular flavour on Day 1 (40 cones sold).

3. Strawberry became the most popular flavour on Day 2 (45 cones sold).

4. Vanilla and Mango flavours also saw an increase in sales from Day 1 to Day 2.

5. Only Chocolate flavour experienced a decrease in sales from Day 1 to Day 2.

6. The total sales on Day 2 (30+35+45+25 = 135) were higher than the total sales on Day 1 (25+40+30+20 = 115).

Given:

A die is thrown $100$ times. The frequencies of the outcomes are provided in the table.

Total number of trials = $100$.

Frequency of outcome 1 = 15

Frequency of outcome 2 = 18

Frequency of outcome 3 = 17

Frequency of outcome 4 = 16

Frequency of outcome 5 = 14

Frequency of outcome 6 = 20

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To Find:

The experimental probability of getting:

(a) The number $6$

(b) An even number

(c) A number less than $3$

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Solution:

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

$P(E) = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

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(a) Probability of getting the number 6

Let E be the event of getting the number 6.

Number of times the outcome 6 occurred (Frequency of 6) = 20.

Total number of trials = 100.

$P(\text{getting 6}) = \frac{\text{Frequency of 6}}{\text{Total number of trials}}$

$P(\text{getting 6}) = \frac{20}{100}$

$P(\text{getting 6}) = \frac{1}{5} = 0.2$

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(b) Probability of getting an even number

The even numbers that can be obtained when rolling a die are 2, 4, and 6.

Let F be the event of getting an even number.

The outcomes corresponding to event F are 2, 4, and 6.

Number of times an even number occurred = (Frequency of 2) + (Frequency of 4) + (Frequency of 6)

Number of times an even number occurred = $18 + 16 + 20 = 54$

Total number of trials = 100.

$P(\text{getting an even number}) = \frac{\text{Number of times an even number occurred}}{\text{Total number of trials}}$

$P(\text{getting an even number}) = \frac{54}{100}$

$P(\text{getting an even number}) = \frac{27}{50} = 0.54$

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(c) Probability of getting a number less than 3

The numbers less than 3 that can be obtained when rolling a die are 1 and 2.

Let G be the event of getting a number less than 3.

The outcomes corresponding to event G are 1 and 2.

Number of times a number less than 3 occurred = (Frequency of 1) + (Frequency of 2)

Number of times a number less than 3 occurred = $15 + 18 = 33$

Total number of trials = 100.

$P(\text{getting a number less than 3}) = \frac{\text{Number of times a number less than 3 occurred}}{\text{Total number of trials}}$

$P(\text{getting a number less than 3}) = \frac{33}{100} = 0.33$

Explanation of Mean, Median, and Mode:

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1. Mean:

The mean (or arithmetic average) is calculated by summing all the values in a dataset and dividing by the number of values. It represents the 'balance point' of the dataset.

Formula: $\overline{x} = \frac{\sum x_i}{n}$, where $\sum x_i$ is the sum of all observations and $n$ is the number of observations.

Example: Consider the test scores of 5 students: 85, 90, 78, 92, 88.

Mean = $\frac{85 + 90 + 78 + 92 + 88}{5} = \frac{433}{5} = 86.6$

The mean test score is 86.6.

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2. Median:

The median is the middle value in a dataset that has been ordered from least to greatest. It divides the data into two equal halves.

To find the median:

• Arrange the data in ascending or descending order.
• If the number of observations ($n$) is odd, the median is the middle value, i.e., the $(\frac{n+1}{2})$-th term.
• If the number of observations ($n$) is even, the median is the average of the two middle values, i.e., the average of the $(\frac{n}{2})$-th term and the $(\frac{n}{2}+1)$-th term.

Example: Using the same test scores: 85, 90, 78, 92, 88.

First, arrange them in ascending order: 78, 85, 88, 90, 92.

The number of observations is $n=5$ (odd).

The median is the $(\frac{5+1}{2})$-th term = 3rd term.

The 3rd term is 88.

The median test score is 88.

Example (Even $n$): Consider the ages of 6 people: 25, 30, 22, 35, 28, 40.

Arrange in ascending order: 22, 25, 28, 30, 35, 40.

The number of observations is $n=6$ (even).

The median is the average of the $(\frac{6}{2})$-th term (3rd term) and the $(\frac{6}{2}+1)$-th term (4th term).

The 3rd term is 28, and the 4th term is 30.

Median = $\frac{28 + 30}{2} = \frac{58}{2} = 29$.

The median age is 29.

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3. Mode:

The mode is the value that appears most frequently in a dataset. A dataset can have one mode (unimodal), more than one mode (multimodal), or no mode at all if all values appear with the same frequency.

Example: Using the test scores: 85, 90, 78, 92, 88.

Each score appears only once. This dataset has no mode.

Example (with mode): Consider the shoe sizes of 10 students: 7, 8, 7, 9, 8, 7, 10, 8, 7, 9.

Count the frequency of each size:

• Size 7: 4 times
• Size 8: 3 times
• Size 9: 2 times
• Size 10: 1 time

The size 7 appears most frequently.

The mode shoe size is 7.

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Differences:

• Calculation: Mean uses the sum of all values, Median uses the position of values after sorting, and Mode uses the frequency of values.
• Sensitivity to Outliers: The mean is heavily affected by extreme values (outliers). The median is less affected by outliers. The mode is not affected by outliers at all, as it only depends on the frequency of values.
• Data Type: Mean and Median are typically used for numerical data. Mode can be used for both numerical and categorical data.
• Uniqueness: A dataset always has exactly one mean and one median. A dataset can have no mode, one mode, or multiple modes.

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Situation where Mode is more appropriate:

The mode is the most appropriate measure of central tendency when dealing with categorical data or when you want to know the most common category or value in a dataset. It is particularly useful when calculating a mean or median is not possible or meaningful.

Example Situation:

Suppose a survey asks 100 people about their favourite colour, and the responses are: Red, Blue, Green, Red, Yellow, Blue, Red, Green, Red, ...

In this case, the data is categorical (Red, Blue, Green, Yellow). You cannot calculate a mean or a median for these colours because they are not numerical values that can be summed or ordered meaningfully.

However, you can easily find the frequency of each colour:

Red: 45 times

Blue: 25 times

Green: 15 times

Yellow: 15 times

The colour 'Red' occurs most frequently (45 times). The mode favourite colour is Red.

Here, the mode (Red) is the only measure of central tendency that makes sense and tells us the most popular choice among the surveyed group. The mean or median would be completely inappropriate.

Given:

The marks obtained by $11$ students in a class test are: $15, 12, 18, 10, 15, 16, 12, 15, 17, 13, 15$.

Number of students ($n$) = $11$.

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To Find:

The mean, median, and mode of the given marks.

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Solution:

We will calculate the Mean, Median, and Mode for the given set of marks.

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1. Mean

The mean ($\overline{x}$) is the sum of all observations divided by the number of observations.

Formula: $\overline{x} = \frac{\text{Sum of all observations}}{\text{Number of observations}}$

Sum of marks ($ \sum x $):

$15 + 12 + 18 + 10 + 15 + 16 + 12 + 15 + 17 + 13 + 15 = 158$

Number of observations ($n$) = $11$

Mean ($\overline{x}$):

$\overline{x} = \frac{158}{11}$

$\overline{x} \approx 14.36$

The mean mark is approximately $14.36$.

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2. Median

To find the median, we first arrange the marks in ascending order.

Sorted marks:

$10, 12, 12, 13, 15, 15, 15, 15, 16, 17, 18$

The number of observations is $n = 11$, which is an odd number.

For an odd number of observations, the median is the $(\frac{n+1}{2})$-th term.

Median position = $(\frac{11+1}{2})$-th term = $\frac{12}{2}$-th term = 6th term.

From the sorted list, the 6th term is $15$.

Median = $15$

The median mark is $15$.

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3. Mode

The mode is the observation that occurs most frequently in the data.

Let's count the frequency of each mark:

$10$: 1 time

$12$: 2 times

$13$: 1 time

$15$: 4 times

$16$: 1 time

$17$: 1 time

$18$: 1 time

The mark $15$ appears most frequently (4 times).

The mode mark is $15$.

Given:

Number of red marbles = 5

Number of blue marbles = 7

Number of green marbles = 3

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To Find:

The probability of drawing:

(a) A red marble

(b) A blue marble

(c) A green marble

(d) A marble that is not blue

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Solution:

First, calculate the total number of marbles in the bag.

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

Total number of marbles = $5 + 7 + 3 = 15$

The total number of possible outcomes when drawing a marble is 15.

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

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

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(a) Probability of drawing a red marble

Let R be the event of drawing a red marble.

Number of favourable outcomes (drawing a red marble) = Number of red marbles = 5.

Total number of possible outcomes = 15.

$P(\text{Red}) = \frac{5}{15}$

$P(\text{Red}) = \frac{\cancel{5}^{1}}{\cancel{15}_{3}} = \frac{1}{3}$

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(b) Probability of drawing a blue marble

Let B be the event of drawing a blue marble.

Number of favourable outcomes (drawing a blue marble) = Number of blue marbles = 7.

Total number of possible outcomes = 15.

$P(\text{Blue}) = \frac{7}{15}$

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(c) Probability of drawing a green marble

Let G be the event of drawing a green marble.

Number of favourable outcomes (drawing a green marble) = Number of green marbles = 3.

Total number of possible outcomes = 15.

$P(\text{Green}) = \frac{3}{15}$

$P(\text{Green}) = \frac{\cancel{3}^{1}}{\cancel{15}_{5}} = \frac{1}{5}$

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(d) Probability of drawing a marble that is not blue

Let Not B be the event of drawing a marble that is not blue.

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

Number of favourable outcomes (drawing a marble that is not blue) = Number of red marbles + Number of green marbles

Number of favourable outcomes = $5 + 3 = 8$

Total number of possible outcomes = 15.

$P(\text{Not Blue}) = \frac{8}{15}$

Alternatively, the probability of not drawing a blue marble is $1$ minus the probability of drawing a blue marble.

$P(\text{Not Blue}) = 1 - P(\text{Blue})$

$P(\text{Not Blue}) = 1 - \frac{7}{15} = \frac{15}{15} - \frac{7}{15} = \frac{15-7}{15} = \frac{8}{15}$

Given:

Average daily temperatures (in $^\circ\text{C}$) for the first two weeks of a month.

Week 1 temperatures: $30, 32, 31, 33, 30, 32, 34$

Week 2 temperatures: $31, 30, 33, 34, 32, 33, 35$

Number of days in each week = $7$.

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To Plot:

A double bar graph to compare the temperatures of the two weeks.

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To Find:

The mean temperature for Week 1.

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Solution:

We will first calculate the mean temperature for Week 1, and then describe how to construct the double bar graph.

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Mean Temperature for Week 1

The mean ($\overline{x}$) of a dataset is the sum of all observations divided by the number of observations.

Formula: $\overline{x} = \frac{\sum x_i}{n}$

For Week 1, the observations are: $30, 32, 31, 33, 30, 32, 34$

Sum of Week 1 temperatures ($ \sum x $):

$30 + 32 + 31 + 33 + 30 + 32 + 34 = 222$

Number of days ($n$) = $7$

Mean temperature for Week 1 ($\overline{x}_{\text{Week 1}}$):

$\overline{x}_{\text{Week 1}} = \frac{222}{7}$

$\overline{x}_{\text{Week 1}} \approx 31.71^\circ\text{C}$ (rounded to two decimal places)

The mean temperature for Week 1 was approximately $31.71^\circ\text{C}$.

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Construction of the Double Bar Graph

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

2. Label the x-axis as "Day" and mark points for each day from 1 to 7.

3. Label the y-axis as "Temperature ($^\circ\text{C}$)".

4. Choose an appropriate scale for the y-axis. The temperatures range from 30 to 35. A scale starting slightly below 30 (e.g., 25 or 28) and going up to 40, with intervals of 1 or 2 units, would be suitable to show the variations clearly.

5. For each day on the x-axis, draw two adjacent bars. One bar will represent the temperature on that day in Week 1, and the other will represent the temperature on that day in Week 2.

6. Use different colours or patterns to distinguish the bars for Week 1 and Week 2 (e.g., blue for Week 1, red for Week 2).

7. The height of the bar for Week 1 temperature on a given day should correspond to its value according to the chosen scale on the y-axis.

8. Similarly, the height of the adjacent bar for Week 2 temperature on the same day should correspond to its value.

9. Leave a uniform gap between different pairs of bars (representing different days).

10. Provide a key or legend to indicate which colour/pattern represents Week 1 and which represents Week 2.

For example, for Day 1, draw a bar up to 30 for Week 1 and an adjacent bar up to 31 for Week 2. Repeat this for Days 2 through 7.

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Interpretation from the Graph (Optional but helpful):

Once constructed, the double bar graph allows for a visual comparison of daily temperatures between the two weeks. You can easily see which week was warmer on a specific day, track the general temperature trend across the week for each week, and identify days with the biggest temperature difference between the weeks.

Explanation of calculating the Median for an even number of observations:

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The median is the middle value in a dataset that has been arranged in either ascending or descending order. It is a measure of central tendency that is less affected by extreme values than the mean.

When the number of observations ($n$) in a dataset is even, there isn't a single middle value. Instead, there are two values that fall in the middle of the ordered list.

The process for finding the median in this case is as follows:

1. Arrange the data: First, arrange all the observations in either ascending (from smallest to largest) or descending (from largest to smallest) order.

2. Identify the middle terms: Since $n$ is even, the two middle terms will be located at the positions $\frac{n}{2}$ and $(\frac{n}{2} + 1)$ in the ordered list.

3. Calculate the average: The median is the average (arithmetic mean) of these two middle terms.

Median = $\frac{(\frac{n}{2}\text{-th term}) + ((\frac{n}{2}+1)\text{-th term})}{2}$

This average of the two central values serves as the median, representing the point that divides the dataset into two equal halves.

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Given:

The dataset is: $25, 28, 22, 35, 32, 30, 28, 26$.

The number of observations is $n=8$, which is an even number.

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To Find:

The median of the given data.

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Solution:

1. Arrange the data in ascending order:

$22, 25, 26, 28, 28, 30, 32, 35$

2. Identify the middle terms:

The number of observations $n=8$. The positions of the middle terms are $\frac{n}{2}$ and $(\frac{n}{2} + 1)$.

$\frac{n}{2} = \frac{8}{2} = 4$-th term

$(\frac{n}{2} + 1) = (4 + 1) = 5$-th term

From the ordered list ($22, 25, 26, 28, 28, 30, 32, 35$), the 4th term is $28$ and the 5th term is $28$.

3. Calculate the average of the middle terms:

Median = $\frac{\text{4th term} + \text{5th term}}{2}$

Median = $\frac{28 + 28}{2}$

Median = $\frac{56}{2}$

Median = $28$

The median of the given data is $28$.

Given:

A box contains slips numbered from $1$ to $10$.

The slips are: $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$.

Total number of slips = $10$.

A slip is drawn at random from the box.

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To Find:

The probability of drawing:

(a) An odd number.

(b) A number greater than $6$.

(c) A one-digit number.

(d) The number $11$.

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Solution:

The total number of possible outcomes when drawing a slip is the total number of slips, which is $10$.

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

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

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(a) Probability of drawing an odd number

Let E$_1$ be the event of drawing an odd number.

The odd numbers between 1 and 10 are: $1, 3, 5, 7, 9$.

Number of favourable outcomes (drawing an odd number) = $5$.

Total number of possible outcomes = $10$.

$P(\text{Odd number}) = \frac{\text{Number of odd numbers}}{\text{Total number of slips}}$

$P(\text{Odd number}) = \frac{5}{10}$

$P(\text{Odd number}) = \frac{\cancel{5}^{1}}{\cancel{10}_{2}} = \frac{1}{2}$

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(b) Probability of drawing a number greater than 6

Let E$_2$ be the event of drawing a number greater than 6.

The numbers greater than 6 from the slips are: $7, 8, 9, 10$.

Number of favourable outcomes (drawing a number greater than 6) = $4$.

Total number of possible outcomes = $10$.

$P(\text{Number > 6}) = \frac{\text{Number of numbers greater than 6}}{\text{Total number of slips}}$

$P(\text{Number > 6}) = \frac{4}{10}$

$P(\text{Number > 6}) = \frac{\cancel{4}^{2}}{\cancel{10}_{5}} = \frac{2}{5}$

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(c) Probability of drawing a one-digit number

Let E$_3$ be the event of drawing a one-digit number.

The one-digit numbers from the slips are: $1, 2, 3, 4, 5, 6, 7, 8, 9$.

Number of favourable outcomes (drawing a one-digit number) = $9$.

Total number of possible outcomes = $10$.

$P(\text{One-digit number}) = \frac{\text{Number of one-digit numbers}}{\text{Total number of slips}}$

$P(\text{One-digit number}) = \frac{9}{10}$

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(d) Probability of drawing the number 11

Let E$_4$ be the event of drawing the number 11.

The slips are numbered from 1 to 10. The number 11 is not present on any slip in the box.

Number of favourable outcomes (drawing the number 11) = $0$.

Total number of possible outcomes = $10$.

$P(\text{Number 11}) = \frac{\text{Number of times 11 appears}}{\text{Total number of slips}}$

$P(\text{Number 11}) = \frac{0}{10} = 0$

This is an impossible event.

Given:

The runs scored by a batsman in $10$ innings are: $45, 30, 55, 60, 20, 70, 30, 40, 50, 30$.

Number of innings ($n$) = $10$.

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To Find:

The mean, median, and mode of the runs scored.

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Solution:

We will calculate the Mean, Median, and Mode for the given set of runs.

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1. Mean

The mean ($\overline{x}$) is the sum of all observations divided by the number of observations.

Formula: $\overline{x} = \frac{\text{Sum of all observations}}{\text{Number of observations}}$

Sum of runs ($ \sum x $):

$45 + 30 + 55 + 60 + 20 + 70 + 30 + 40 + 50 + 30 = 430$

Number of observations ($n$) = $10$

Mean ($\overline{x}$):

$\overline{x} = \frac{430}{10}$

$\overline{x} = 43$

The mean run scored is $43$.

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2. Median

To find the median, we first arrange the runs in ascending order.

Sorted runs:

$20, 30, 30, 30, 40, 45, 50, 55, 60, 70$

The number of observations is $n = 10$, which is an even number.

For an even number of observations, the median is the average of the $(\frac{n}{2})$-th term and the $(\frac{n}{2} + 1)$-th term.

Median position = $(\frac{10}{2})$-th term and $(\frac{10}{2} + 1)$-th term

Median position = 5th term and 6th term.

From the sorted list ($20, 30, 30, 30, 40, 45, 50, 55, 60, 70$), the 5th term is $40$ and the 6th term is $45$.

Median = $\frac{\text{5th term} + \text{6th term}}{2}$

Median = $\frac{40 + 45}{2} = \frac{85}{2} = 42.5$

The median run scored is $42.5$.

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3. Mode

The mode is the observation that occurs most frequently in the data.

Let's count the frequency of each run:

$20$: 1 time

$30$: 3 times

$40$: 1 time

$45$: 1 time

$50$: 1 time

$55$: 1 time

$60$: 1 time

$70$: 1 time

The run $30$ appears most frequently (3 times).

The mode run scored is $30$.

Given:

The attendance of a class for 5 days is as follows:

Monday: 35

Tuesday: 38

Wednesday: 32

Thursday: 39

Friday: 36

Number of days = 5.

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To Plot:

A bar graph to represent the given attendance data.

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To Find:

The mean attendance for the 5 days.

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Solution:

We will first calculate the mean attendance, and then describe how to construct the bar graph.

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Mean Attendance

The mean ($\overline{x}$) is the sum of all observations divided by the number of observations.

Formula: $\overline{x} = \frac{\text{Sum of all observations}}{\text{Number of observations}}$

The observations are the attendance for each day: $35, 38, 32, 39, 36$.

Sum of attendance for 5 days ($ \sum x $):

$35 + 38 + 32 + 39 + 36 = 180$

Number of days ($n$) = $5$

Mean attendance ($\overline{x}$):

$\overline{x} = \frac{180}{5}$

$\overline{x} = 36$

The mean attendance for the 5 days was $36$.

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Construction of the Bar Graph

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

2. Label the x-axis as "Day" and mark points for each day (Monday, Tuesday, Wednesday, Thursday, Friday).

3. Label the y-axis as "Number of Students Present" (Attendance).

4. Choose an appropriate scale for the y-axis. The attendance ranges from 32 to 39. A scale starting slightly below 30 (e.g., 25 or 30) and going up to 40 or 45, with intervals of 1 or 2 units, would be suitable.

5. Draw a bar for each day. The width of the bars should be uniform, and there should be a uniform gap between the bars.

6. The height of the bar for each day should correspond to the attendance number for that day, according to the chosen scale on the y-axis.

Based on the data:

• For Monday, draw a bar up to 35 on the y-axis.
• For Tuesday, draw a bar up to 38 on the y-axis.
• For Wednesday, draw a bar up to 32 on the y-axis.
• For Thursday, draw a bar up to 39 on the y-axis.
• For Friday, draw a bar up to 36 on the y-axis.

7. Label each bar with the corresponding day.

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Interpretation from the Graph (Optional but helpful):

The bar graph visually represents the attendance trend throughout the week. You can easily see which day had the highest attendance (Thursday) and which day had the lowest attendance (Wednesday).

Let's explain the concepts of 'random experiment', 'sample space', and 'event' in the context of drawing a card from a standard deck of 52 playing cards.

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1. Random Experiment:

A random experiment is an experiment or a process where the outcome cannot be predicted with certainty beforehand, but the set of all possible outcomes is known. The experiment can be repeated under identical conditions.

In the context of drawing a card from a standard deck:

The act of drawing a single card at random from a well-shuffled standard deck of 52 playing cards is a random experiment.

• We know the set of all possible cards that could be drawn (any of the 52 cards).
• Before drawing, we cannot say for sure which specific card will be drawn (it depends on chance).
• We can repeat this experiment (drawing a card, noting it, and perhaps replacing it and reshuffling) multiple times.

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2. Sample Space:

The sample space (usually denoted by $S$ or $\Omega$) of a random experiment is the set of all possible outcomes of the experiment. Each individual outcome in the sample space is called a sample point.

In the context of drawing a card from a standard deck:

The sample space is the set of all possible cards that can be drawn. A standard deck consists of 52 unique cards (13 ranks in each of 4 suits: Hearts, Diamonds, Clubs, Spades).

The sample space $S$ contains all 52 cards:

$S = \{ \text{Ace of Hearts, 2 of Hearts, ..., King of Hearts,} $

$\phantom{S = \{}\text{Ace of Diamonds, ..., King of Diamonds,} $

$\phantom{S = \{}\text{Ace of Clubs, ..., King of Clubs,} $

$\phantom{S = \{}\text{Ace of Spades, ..., King of Spades} \}$

The total number of outcomes in the sample space is $|S| = 52$.

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3. Event:

An event is a subset of the sample space. It is a collection of one or more outcomes from the random experiment. An event occurs if the outcome of the random experiment is one of the outcomes in the event set.

In the context of drawing a card from a standard deck:

An event is any specific collection of cards from the deck.

Here are some examples of events:

• Event A: Drawing a red card.

This event consists of all the Hearts and Diamonds. The number of outcomes in Event A is $26$ (13 Hearts + 13 Diamonds).

$A \subset S$, where $A = \{ \text{All Hearts, All Diamonds} \}$.

$|A| = 26$.

• Event B: Drawing a King.

This event consists of the King of Hearts, King of Diamonds, King of Clubs, and King of Spades. The number of outcomes in Event B is $4$.

$B \subset S$, where $B = \{ \text{King of Hearts, King of Diamonds, King of Clubs, King of Spades} \}$.

$|B| = 4$.

• Event C: Drawing the Ace of Spades.

This is an event with a single outcome (a simple event). The number of outcomes in Event C is $1$.

$C \subset S$, where $C = \{ \text{Ace of Spades} \}$.

$|C| = 1$.

The probability of an event E occurring is often calculated as $P(E) = \frac{\text{Number of outcomes in E}}{\text{Total number of outcomes in S}}$, provided all outcomes are equally likely.