# Excercise 3.2 Data Handling - NCERT Solutions Class 7

Exercise 3.2

## Question 1

The scores in mathematics test (out of \(25\)) of \(15\) students are follows: -

\(=\left[ \begin{align} & \text{19, 25, 23, 20, 9, 20, 15, 10,} \\ & \,\,\,\,\,\text{ 5,16, 25, 20, 24, 12, 20}\text{.} \\ \end{align} \right]\)

Find the mode and median of the data. Are they same?

### Solution

**Video Solution**

**What is known?**

Scores in mathematics test of \(15\) students

**What is unknown?**

The mode and median of the data

**Reasoning:**

Mode – Mode of a given data is that value of observation which occurs for the most

number of times.

Median \(=\) middle of observation (in this case, \(8\)th observation)

**Steps:**

Scores of \(15\) students in mathematics test are \(=\left[ \begin{align} & \text{19, 25, 23, 20, 9, 20, 15, 10,} \\ & \,\,\,\,\,\text{ 5,16, 25, 20, 24, 12, 20}\text{.} \\ \end{align} \right]\)

Arranging scores in ascending order, we get

\[\left[ \begin{align} & \text{5, 9, 10, 12, 15, 16, 19, 20, } \\ & \,\,\text{20, 20, 20, 23, 24, 25, 25} \\ \end{align} \right]\]

Mode – Mode of a given data is that value of observation which occurs for the most

number of times.

Therefore\(, 20\) occurs most of the time.

\(∴\) Mode \(= 20.\)

Median \(=\) middle of observation (in this case, \(8\)^{th} observation)

\(∴\) Median \(= 20\)

Yes, mode and median of the given observations are same.

## Question 2

The runs scored in a cricket match by \(11\) players are as follows: -

\(6, 15, 120, 50, 100,80, 10,15,\\ 8, 10, 15\)

Find the mean, median and mode of the data. Are they same?

### Solution

**Video Solution**

**What is known?**

Runs scored in a cricket match by \(11\) Players

**What is unknown?**

The mean, median and mode of the data

**Reasoning: \(\begin{align}\text{Mean}=\frac{\text{Sum of all scores}}{\text{Total no}\text{. of players}}\end{align}\)**

Mode** \(=\) **Mode is the observation that occurs highest number of times

Median\(=\)Median is the middle observation

**Steps: **

Total number of players \(= 11\)

Scores of players \(= 6, 15, 120, 50, 100,80, 10,15, 8, \\ \quad10, 15\)

\[\begin{align}{\rm{Mean}}& = \frac{{{\text{Sum of all scores}}}}{{{\text{Total no}}.{\text{of players}}}} \\ &= \frac{{\left[ \begin{array}{l}{\rm{6}} + {\rm{8}} + {\rm{1}}0 + {\rm{15}} + \\{\rm{15}} + {\rm{15}} + {\rm{5}}0 + {\rm{8}}0 + \\{\rm{1}}00 + {\rm{12}}0 \\\end{array} \right]}}{{11}} \\ &= \frac{{{\rm{429}}}}{{11}} \\ &= 39 \\\end{align}\]

Thus, mean \(= 39.\)

Arranging the scores into ascending order, we get

\(6, 8, 10,10, 15, 15,15, 50, 80, \\ 100, 120 \)

Mode is the observation that occurs highest number of times

Here, \(15\) occurs \(3\) times

\(∴\) Mode \(=15.\)

Median is the middle observation

\(∴\) Median \(= 15\) (6^{th} observation)

Thus, Mean \(=\) \(39\) , Mode \(=\) \(15\) and median \(=\) \(15\)

No, the mean, mode and median are not same.

## Question 3

The weights (in kg) of \(15\) students of a class are: -

\(38, 42, 35, 37, 45, 50, 32, 43,43,\\ 40, 36, 38, 43, 38,47\)

(i) Find the median and mode of this data.

(ii) Is there more than one mode?

### Solution

**Video Solution**

**Reasoning:**

**What is known?**

The weights (in kg) of \(15\) students of a class

**What is unknown?**

The median and mode of this data.

**Reasoning:**

Mode \(=\) Mode is the observation that occurs highest number of times

Median \(=\) Median is the middle observation

**Steps:**

Total number of students \(=15\)

Weights of \(15\) students \(= 38, 42, 35, 37, 45, 50, 32, 43,\\ \quad43, 40, 36,38, 43, 38, 47.\)

Arranging in ascending order, we get \(32, 35, 36, 37,38, 38,38, 40, 42, \\43, 43,43, 45, 47, 50\)

(i) Mode is the observation that occurred highest number of times.

Thus\(, 38\) and \(43\) occur highest number of times.

\(∴\) Mode \(= 38\) and \(43.\)

Also, median \(= 40\) (8^{th} observation)

(ii) Yes, there are two modes.

## Question 4

Find the mode and median of the data:\( 13, 16, 12, 14, 19, 12, 14, 13, 14.\)

### Solution

**Video Solution**

**What is known?**

Given numbers

**What is unknown?**

The mode and median of the data

**Reasoning:**

Mode \(=\) Mode is the observation that occurs highest number of times

Median \(=\) Median is the middle observation

**Steps: **

Given data \(= 13, 16, 12, 14, 19, 12, 14, 13, 14\)

Arranging the data in ascending order, we get \(12, 12, 13, 13, 14, 14, 14, 16, 19.\)

\(∴\) Mode is the observation that occurs highest number of times.

\(∴\) Mode \(= 14\)

Also, median is the middle observation.

\(∴\) Median \(= 14\) (\(5\)^{th} observation)

## Question 5

Tell whether the statement is true or false:

(i) The mode is always one of the numbers in a data.

(ii) The mean is one of the numbers in a data.

(iii) The median is one of the numbers in a data.

(iv) The data \(6, 4, 3, 8, 9, 12, 13, 9,\) has mean \(9.\)

### Solution

**Video Solution**

**Steps: **

(i) **True**: Mode is the observations that occurs highest numbers of times. Therefore, it is one observation in a data.

(ii) **False**: Mean may or may not be one of the numbers in a data.

(iii) **True**: Median is the middle observations of the given data when it is arranged in ascending or descending order.

(iv) **False**: The given data \(6, 4, 3, 8, 9, 12, 13, 9.\)

\[\begin{align} \text{Mean} & =\frac{\left[ \begin{align} & 6+4+3+8+ \\& 9+12+13+9 \\\end{align} \right]}{8} \\ & =\frac{64}{8} \\ & =8 \\\end{align}\]

Therefore, mean is \(8.\)

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