NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics

# NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics

Edited By Ravindra Pindel | Updated on Sep 12, 2022 04:21 PM IST

NCERT Exemplar Class 11 Maths Solutions Chapter 15 includes solutions to questions of all types such as short answer questions, long answer questions, value-based questions, MCQ’s and HOTS. The solutions are given by experts in a simple manner and are solved step-by-step. If you who have doubts regarding the chapter Statistics, then refer to NCERT Exemplar Class 11 Maths chapter 15 solutions and follow the solutions for better understanding of the concept. Students can download the pdf for Class 11 Maths NCERT Exemplar Solutions Chapter 15 which consists of all the important questions. You can cover all the important topics and sub-topics using the NCERT Exemplar Class 11 Maths solutions chapter 15 PDF Download function. You can also get your queries solved by referring to the solutions solved step by step in an easily comprehensible format.
Also, check - NCERT Class 11 Maths Solutions for Other Chapters

Question:1

Find the mean deviation about the mean of the distribution:

We have to find the mean deviation about the mean of the distribution in this question.

Let us make a table of the given data and fill up the other columns after calculations

Hence, mean deviation becomes= $\frac{ \Sigma f_{i}(x_{i}-\bar{x_{i}})}{ \Sigma f_{i}}=\frac{25}{20}=1.25~ \\\\$

Question:2

Find the mean deviation about the median of the following distribution:

The data distribution is given in the question.

We have to find the mean deviation about the median of the distribution in this question.

Let us make a table of the given data and fill up the other columns after calculations

Hence, mean deviation becomes

Question:3

Calculate the mean deviation about the mean of the set of first n natural numbers when n is an odd number.

Set of first n natural numbers when n is an odd number is given

We have to find the mean deviation about the mean

We know that first natural numbers are 1, 2, 3 ……. , n .

It is given that n is odd number.

So, mean is

Question:4

Calculate the mean deviation about the mean of the set of first n natural numbers when n is an even number.

Set of first n natural numbers when n is an even number is given

We have to find the mean deviation about the mean

We know that first natural numbers are 1, 2, 3 ……. , n .

It is given that n is an even number.

So, mean is

Question:5

Find the standard deviation of the first n natural numbers.

$\\\\ \text{The sums are } \Sigma x_{i}=1+2+3+ \ldots \ldots +n=\frac{n \left( n+1 \right) }{2}~~~~ \\\\ \\ ~~ \Sigma x_{i}^{2}=1^{2}+2^{2}+3^{2}+ \ldots \ldots +n^{2}=\frac{n \left( n+1 \right) \left( 2n+1 \right) }{6}~~~ \\\\ \\ ~ Therefore, the standard deviation can be written as, \sigma =\sqrt[]{\frac{ \Sigma x_{i}^{2}}{n}- \left( \frac{ \Sigma x_{i}}{n} \right) ^{2}}~~~~ \\\\ \\ , \sigma =\sqrt[]{\frac{\frac{n \left( n+1 \right) \left( 2n+1 \right) }{6}}{n}- \left( \frac{\frac{n \left( n+1 \right) }{2}}{n} \right) ^{2}} = \sqrt[]{\frac{2n^{2}+n+2n+1}{6}-\frac{n^{2}+2n+1}{4}}~ = \sqrt[]{\frac{4n^{2}+6n+2-3n^{2}-6n-3}{12}} \\\\ \\ = \sqrt[]{\frac{n^{2}-1}{12}} \\\\$

Question:9

The frequency distribution:

where A is a positive integer, has a variance of 160. Determine the value of A.

A frequency distribution table is given where variance =160. We have to find the value of A, where A is a positive number. Let us make a table from the given data and fill out the other columns after calculation

And we know that variance is

$\\ \text{ Substituting the values from the above table and also given that variance}=160 \\ we~get~ 160=\frac{92A^{2}}{7}- \left( \frac{22A}{7} \right) ^{2}= \left( \frac{92A^{2}}{7}-\frac{484A^{2}}{49} \right) =\frac{7\ast92A^{2}-484A^{2}}{49}=\frac{160A^{2}}{49}~ \\\ ~~~ 160=\frac{160A^{2}}{49}~ \\\\ \\ ~~A^{2}=49 \\\\ \\ A=7 \\\\ \\ \text{ Hence, the value of A is 7} \\\\$

Question:10

For the frequency distribution:

Find the standard distribution.

A frequency distribution table is given and we have to find the standard deviation

Let us make a table from the given data and fill out the other columns after calculation

\begin{aligned} &\text { And we know that standard deviation is }\\ &\sigma=\sqrt{\frac{\Sigma f_{i} x_{i}^{2}}{n}-\left(\frac{\Sigma f_{i} x_{i}}{n}\right)^{2}} \end{aligned}

Substituting the value from the above table

$\sigma=\sqrt{\frac{1393}{60}-\left(\frac{277}{60}\right)^{2}}=\sqrt{23.23-21.34}=1.37$

Question:11

There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:

where x is a positive integer. Determine the mean and standard deviation of the marks.

It is given that there are 60 students in a class. The frequency distribution of the marks obtained by the students in a test is also given.

We have to find the mean and standard deviation of the marks.

It is given that there are 60 students in the class, so

And we know that standard deviation is

Question:12

The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.

The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. We have to find the overall standard deviation.

As per the given criteria. In the first set of samples number of sample bulbs ,

$\\ \\ \text{ Standard deviation }s_{1}=8 \text{hrs~ Mean life, }\overline{x_{1}}=650 \\\\ \\ \text{ And in second set of samples, number of sample bulbs }n_{2}=80 \text{ Standard deviation }s_{2}=7 hrs \\\\ \\ \text{~ Mean life }\overline{x_{2}}=660~ \\\\ \\ \text{We know the standard deviation for combined two series is } \\\\ \\ \sigma =\sqrt {\frac{n_{1}s_{1}^{2}+n_{2}s_{2}^{2}}{n_{1}+n_{2}}+ \left( n_{1}n_{2} \right) \frac{ \left( \overline{x_{1}}-\overline{x_{2}} \right) ^{2}}{ \left( n_{1}+n_{2} \right) ^{2}}} \\ \text{Substituting the corresponding values we get} \sigma =\sqrt {\frac{60 \left( 8 \right) ^{2}+80 \left( 7 \right) ^{2}}{60+80}+\frac{ \left( 60\ast80 \right) \left( 650-660 \right) ^{2}}{ \left( 20+80 \right) ^{2}}}= \sqrt {\frac{388}{7}+\frac{1200}{49}} = \\\\ =\sqrt {\frac{3916}{49}} = 8.9\\$

Question:13

Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all items and the sum of the squares of the items.

Mean and standard deviation of 100 items are 50 and 4 respectively
We have to find the sum of all items and the sum of the squares of the items.
As per the question, number of items n =100

Question:14

If for a distribution and the total number of item is 18, find the mean and standard deviation.

Substituting the corresponding values we get

$\\ \sigma=\sqrt{\frac{43}{18}-\left(\frac{3}{18}\right)^{2}}=\sqrt{2.39-(0.166)^{2}}=\sqrt{2.39-0.027}=\sqrt{2.363}=1.54$

Thus the mean and standard deviation of given items are 5.17 and 1.54 respectively.

Question:15

Find the mean and variance of the frequency distribution given below:

Frequency distribution is given. We have to find the mean and variance

Converting the ranges of x to groups the given table can be re written as

And, we know that variance is

Question:16

Calculate the mean deviation about the mean for the following frequency distribution:

A frequency distribution table is given, and we have to find the mean deviation about the mean
Let us make a table from the given data and fill out the other columns after calculation

Question:17

Calculate the mean deviation from the median of the following data:

A frequency distribution table is given and we have to find the mean deviation about the median

Let us make a table from the given data and fill out the other columns after calculation

Now here N=20, which is even

Here

Question:18

Determine the mean and standard deviation for the following distribution:
$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|} \hline \text { Marks } & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline \text { Frequency } & 1 & 6 & 6 & 8 & 8 & 2 & 2 & 3 & 0 & 2 & 1 & 0 & 0 & 0 & 1 \\ \hline \end{array}$

A frequency distribution table is given and we have to find the mean and standard deviation
Let us make a table from the given data and fill out the other columns after calculation

$\begin{array}{|l|l|l|} \hline \text { Marks }\left(x_{i}\right) & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & f_{i} x_{i} \\ \hline 2 & 1 & 2 \\ \hline 3 & 6 & 18 \\ \hline 4 & 6 & 24 \\ \hline 5 & 8 & 40 \\ \hline 6 & 8 & 48 \\ \hline 7 & 2 & 14 \\ \hline 8 & 2 & 16 \\ \hline 9 & 3 & 27 \\ \hline 10 & 0 & 0 \\ \hline 11 & 2 & 22 \\ \hline 12 & 1 & 12 \\ \hline 13 & 0 & 0 \\ \hline 14 & 0 & 0 \\ \hline 15 & 0 & 0 \\ \hline 16 & 1 & 16 \\ \hline \text { Total } & \mathrm{N}=40 & =239 \\ \hline \end{array}$
$\\ mean~\overline{X~}=\frac{ \Sigma f_{i}x_{i}}{ \Sigma f_{i}}=\frac{239}{40}=5.975\approx6 \\\\$
$\begin{array}{|l|l|l|l|l|l|} \hline \text { Marks }\left(x_{i}\right) & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & f_{i} x_{i} & d_{i}=x_{i}-\bar{x} & f_{i} d_{i} & f_{i} d_{i}^{2} \\ \hline 2 & 1 & 2 & -4 & -4 & 16 \\ \hline 3 & 6 & 18 & -3 & -18 & 54 \\ \hline 4 & 6 & 24 & -2 & -12 & 24 \\ \hline 5 & 8 & 40 & -1 & -8 & 8 \\ \hline 6 & 8 & 48 & 0 & 0 & 0 \\ \hline 7 & 2 & 14 & 1 & 2 & 2 \\ \hline 8 & 2 & 16 & 2 & 4 & 8 \\ \hline 9 & 3 & 27 & 3 & 9 & 27 \\ \hline 10 & 0 & 0 & 4 & 0 & 0 \\ \hline 11 & 2 & 22 & 5 & 10 & 50 \\ \hline 12 & 1 & 12 & 6 & 6 & 36 \\ \hline 13 & 0 & 0 & 7 & 0 & 0 \\ \hline 14 & 0 & 0 & 8 & 0 & 0 \\ \hline 15 & 0 & 0 & 9 & 0 & 0 \\ \hline 16 & 1 & 16 & 10 & 10 & 100 \\ \hline \text { Total } & \mathrm{N}=40 & =239 & & =-1 & =325 \\ \hline \end{array}$
And we know that standard deviation is
$\\ \sigma = \sqrt {\frac{ \Sigma f_{i}d_{i}^{2}}{n}- \left( \frac{ \Sigma f_{i}d_{i}}{n} \right) ^{2}}= \sqrt {\frac{325}{40}- \left( \frac{-1}{40} \right) ^{2}}=\sqrt {8.125- \left( 0.025 \right) ^{2}} =2.8721\\\\$

Question:19

The weights of coffee in 70 jars are shown in the following table:

Determine variance and standard deviation of the above distribution.

A frequency distribution table for the weights of coffee in 70 jars is given and we have to find the variance and standard deviation of the distribution.
Let us make a table from the given data and fill out the other columns after calculation

And we know that standard deviation is

Variance = 1.17

Question:20

Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d.

The first n terms of an A.P are given whose first term is a and common difference is d. We have to find mean and standard deviation.

The given AP in tabular form is as shown below,

Question:21

Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests.

Who is more intelligent and who is more consistent?

The marks obtained, out of 100, by 2 students Ravi and Hashina in 10 tests are given
We have to find who is more intelligent and who is more consistent
The marks of Ravi taken separately as follows

Here we have assumed 45 as mean
And we know that the standard deviation can be written as,

Here as

And we know that the standard deviation can be written as,

Question:22

Mean and standard deviation of 100 observations were found to be 40 and 10, respectively.
If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27
respectively.
We have to find the correct standard deviation
As per the given criteria, Number of observations, n=100
Mean of the given observations before correction,

But we know

Substituting the corresponding values, we get

Also given that standard deviation of the 100 observations is 10 before correction

Question:24

The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is
A. 2
B. 2.57
C. 3
D. 3.75

Given data is 3,10, 10, 4, 7 ,10 ,5 . They are total 7.

Question:26

Hence, mean deviation becomes

Question:27

Given that the marks obtained by 9 students in a mathematics test are 50, 69 , 20 ,33, 53, 39 ,40, 65, 59 As number of students =9 which is odd

Hence Mean deviation becomes

Question:28

Given data can be written in table form as,

Question:30

The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is
A. 50000
B. 250000
C. 252500
D. 255000

As per the question

Number of observation, n =100

$\\ \text{Mean of the given observations, }\overline{x}=50~ \\ \\ \text{But~we know, }\overline{X~}=\frac{ \Sigma x_{i}}{n}~ \\ \\ \text{Substituting the corresponding values, we get }50=\frac{ \Sigma x_{i}}{100}~~ \\ \\ \Sigma x_{i}=50 * 100=5000 \\ \\ \text{It is also given that the standard deviation of the 100 observations is 5 } \\\\ \sigma =5 \\ \\ \sigma = \sqrt {\frac{ \Sigma x_{i}^{2}}{n}- \left( \frac{ \Sigma x_{i}}{n} \right) ^{2}} \\ \\ \text{Substituting the corresponding values, we get }5=\sqrt {\frac{ \Sigma x_{i}^{2}}{100}- \left( \frac{5000}{100} \right) ^{2}}~ \\$

Question:31

Given observations are a, b, c, d, e

So, the mean of the 5 observations is

And the standard deviation of the 5 observations is

Now we will find the mean and standard deviation of the observations a+k , b+k, c+k, d+k, e+k we get

So, the mean of these 5 observations is

Question:32

$\\ \sigma _{1}=\sqrt {\frac{ \sum \left( kx_{i} \right) ^{2}~}{5}- \left( \frac{ \sum kx_{i}}{5} \right) ^{2}} \\ \\ =\sqrt {\frac{ \sum k^{2} \left( x_{i} \right) ^{2}~}{5}- \left( m_{1} \right) ^{2}} \\ \\ =\sqrt {\frac{k^{2} \sum \left( x_{i} \right) ^{2}~}{5}-k^{2}m^{2}} \\ \\ =\sqrt {k^{2} \left( \frac{ \sum \left( x_{i} \right) ^{2}~}{5}-m^{2} \right) } \\ \\ =k\sqrt {\frac{ \sum \left( x_{i} \right) ^{2}~}{5}-m^{2}} \\ \\ \sigma _{1}=k \sigma ~ \\ \\ \text{ Hence, the standard deviation of new set of observations k}x_{1}\text{, k}x_{2}\text{, k}x_{3}\text{, k}x_{4}\text{, k}x_{5} \: \: \:\ \ is\: \: \: \ \ k \sigma \\$

Question:34

Standard deviations for first 10 natural numbers is
A. 5.5
B. 3.87
C. 2.97
D. 2.87

Question:35

We know that the standard deviation of the first n natural numbers is
Now for first 10 natural numbers n=10
substituting this in the equation of standard deviation we get
Now when 1 is added to each numbers of 1 , 2, 3, 4, 5, 6,7 ,8 ,9, 10 ; we get new series as 1+1, 2+1, 3+1, 4+1, 5+1, 6+1, 7+1, 8+1, 9+1, 10+1
Now we know if standard deviation of x series is s, then standard deviation of k+x series is s,
So the S.D of 1+1, 2+1, 3+1, 4+1, 5+1, 6+1, 7+1, 8+1, 9+1, 10+1 series is also same as the S.D of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 series.

Now for variance we will square on both the sides

Question:36

First 10 positive integers are 1,2, 3, 4, 5,6 ,7, 8,9, 10 On multiplying each number by
-1 we get-1, -2, -3, -4 , -5, -6, -7, -8, -9 , -10
On adding 1 to each of the number, we get 0, -1, -2, -3, -4, -5, -6, -7, -8, -9

Question:37

The following information relates to a sample of size 60:

The variance is
A. 6.63
B. 16
C. 22
D. 44

Question:38

Question:39

Question:40

Fill in the blanks
Coefficient of variation =

Standard Deviation
We know that the Coefficient of variation can be written as

Question:41

Fill in the blanks
If is the mean of n values of x, then $\sum _{i=1}^{n} \left( x_{i}-\overline{x} \right)$ is always equal to _______.
If a has any value other than , then $\sum _{i=1}^{n} \left( x_{i}-\overline{x} \right) ^{2}$ is _________ than $\sum \left (X_i-a \right )^2$

Zero, less than Given x is mean of n values, then the sum of all n terms is denoted by
so difference of both these is laways equal to zero i.e.
And square of the above equation is also equal to zero, so
Now if "a" has the value other than x then

Question:42

Fill in the blanks:
If the variance of a data is 121, then the standard deviation of the data is _______.

11
We know the square root of variance is standard deviation
Taking square root on both sides, we get
So, if the variance of a data is 121, then the Standard deviation of the data is 11.

Question:43

Fill in the blanks
The standard deviation of a data is ___________ of any change in origin but is _____ on the change of scale.

Independent, dependent
Change of origin means some value has been added or subtracted in the observation. And we know the S.D does not change if any value is added or subtracted from the observations, So SD is independent of change in origin.
However, Standard deviation is only affected by a change in scale, that is, when some value is multiplied or divided to observations. Hence, the SD of any data is independent of any change in origin but dependent on any change of scale.

Question:44

The sum of the squares of the deviations of the values of the variable is _______ when taken about their arithmetic mean.

Minimum
The sum of the squares of the deviations of the values of the variable is minimum or least when taken about their arithmetic mean.

Question:45

Fill in the blanks
The mean deviation of the data is _______ when measured from the median.

Least
Mean deviation is sum of all deviations of a set of a data about the data’s mean. In addition, it is widely believed that the median is usually between the mean and the mode. So, the mean deviation of the data is least when measured from the median.

Question:46

Fill in the blanks
The standard deviation is _______ to the mean deviation taken from the arithmetic mean.

Greater than or equal to
SD is the difference between square of deviation of data about mean and square of mean. In addition, mean deviation is sum of all deviations of a set of data about the data’s mean. Hence, the SD is greater than or equal to the mean deviation taken from the arithmetic mean.

## Topics and Subtopics in NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics:

· Central Tendency

· Mean

· Median

· Mode

· Measure of Dispersion

· Range

· Quartile deviation

· Mean deviation

· Standard deviation

· Mean deviation for grouped data

· Mean deviation for ungrouped data

· Variance

· Analysis of Frequency Deviation

## More about NCERT Exemplar Class 11 Maths Solutions Chapter 15

By referring to NCERT Exemplar Class 11 Maths chapter 15 solutions students can easily gather the various concepts of Statistics. Statistics carries a significant number of marks according to the CBSE paper pattern and the students should concentrate on important concepts like standard deviation and mean deviation, as they are frequently asked in the examination.

The step-wise and straightforward format of the NCERT exemplar solutions for Class 11 Maths chapter 15 makes it easy for the students to understand the most complex concepts of Statistics in a quick and easygoing manner. The pdf is exclusively created by our experts for students who find the concepts of Statistics difficult.

Students can easily score marks in this chapter by thoroughly studying a few topics which are commonly asked in the final examinations. The students will understand the concept in the simplest way possible by going through the NCERT Exemplar Class 11 Maths solutions chapter 15 Statistics.

As most students tend to skip the entire chapter due to doubts in Statistics, our experts have formulated the NCERT Exemplar Class 11 Maths chapter 15 solutions in the most student-friendly manner.

## NCERT Solutions for Class 11 Mathematics Chapters

 Chapter 1 Sets Chapter 2 Relations and Functions Chapter 3 Trigonometric Functions Chapter 4 Principle of Mathematical Induction Chapter 5 Complex Numbers and Quadratic Equations Chapter 6 Linear Inequalities Chapter 7 Permutations and Combinations Chapter 8 Binomial Theorem Chapter 9 Sequences and Series Chapter 10 Straight lines Chapter 11 Conic Sections Chapter 12 Introduction to Three Dimensional Geometry Chapter 13 Limits and Derivatives Chapter 14 Mathematical Reasoning Chapter 16 Probability

## Important Topic to Cover from NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics:

Standard Deviation: Questions on Standard Deviation are frequently asked by the examiners. The students should go through the NCERT Exemplar Class 11 Maths chapter 15 solutions to solve all their doubts regarding standard deviation. The student should just follow the simple steps given in the solution and practice.

Mean Deviation: The students should not ignore solving the question of mean deviation as they are commonly asked in the exam. The students should memorize the formula for mean and standard deviation upon which the problem is based and solve the entire question by referring to the steps given in the Class 11 Maths NCERT Exemplar solutions chapter 15.

Mean, Median, Mode, and Range: These are the fundamental concepts of Statistics and can be asked in the examination for MCQ or short answer type questions.

Check Chapter-Wise NCERT Solutions of Book

 Chapter-1 Sets Chapter-2 Relations and Functions Chapter-3 Trigonometric Functions Chapter-4 Principle of Mathematical Induction Chapter-5 Complex Numbers and Quadratic equations Chapter-6 Linear Inequalities Chapter-7 Permutation and Combinations Chapter-8 Binomial Theorem Chapter-9 Sequences and Series Chapter-10 Straight Lines Chapter-11 Conic Section Chapter-12 Introduction to Three Dimensional Geometry Chapter-13 Limits and Derivatives Chapter-14 Mathematical Reasoning Chapter-15 Statistics Chapter-16 Probability

### NCERT Exemplar Class 11 Solutions

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1. Which are the important topics one should cover in Statistics?

Standard Deviation, Mean Deviation and Central Tendency are the important topics one should cover in Statistics.

2. Where can we download the Class 11 Maths NCERT Exemplar Solutions Chapter 15?

Students can get PDF by using the NCERT Exemplar Class 11 Maths solutions chapter 15 PDF Download function.

3. Are the solutions helpful for board examinations?

Yes, NCERT Exemplar Class 11 Maths solutions chapter 15 are helpful for board examinations.

4. Can these solutions be referred for cracking competitive examinations?

Yes, Class 11 Maths NCERT Exemplar solutions chapter 15 can be referred to when pursuing competitive examinations.

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