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CBSE Class 11 Maths Chapter 13 Statistics – NCERT Solutions 2025–26

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Get a free PDF of detailed solutions for Class 11 Maths Chapter 13 Statistics Exercise 13.2

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.2 (Ex 13.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 13 Statistics Exercise 13.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 11

Subject:

Class 11 Maths

Chapter Name:

Chapter 13 - Statistics

Exercise:

Exercise - 13.2

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

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Access NCERT Solutions for Class 11 Maths Chapter 13 – Statistics

Exercise – 13.2

1. Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12.

Ans: Given data: 6, 7, 10, 12, 13, 4, 8, 12.

Mean can be given as: 

\[\bar x = \frac{{\sum\limits_{i = 1}^8 {{x_i}} }}{{\text{n}}} = \frac{{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}}{8} = \frac{{72}}{8} = 9\].

The following table is obtained:

${x_i}$

$\left( {{x_i} - \bar x} \right)$

${\left( {{x_i} - \bar x} \right)^2}$

6

$6 - 9 =  - 3$

9

7

$7 - 9 =  - 2$

4

10

$10 - 9 = 1$

1

12

$12 - 9 = 3$

9

13

$13 - 9 = 4$

16

4

$4 - 9 =  - 5$

25

8

$8 - 9 =  - 1$

1

12

$12 - 9 = 3$

9



74


Variance can be given as: 

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

\[{\sigma ^2} = \frac{1}{8} \times 74\]

${\sigma ^2} = 9.25$.

Therefore, mean is 9 and variance is 9.25.


2. Find the mean and variance for the first $n$ natural numbers.

Ans: The mean of first $n$ natural numbers can be calculated as:

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

\[\therefore Mean = \frac{{\frac{{n(n + 1)}}{2}}}{n} = \frac{{n + 1}}{2}\]

Variance, \[{\sigma ^2} = \frac{1}{n}\mathop \sum \limits_{i = 1}^n {\left( {{x_i} - \bar x} \right)^2}\] 

\[ \Rightarrow \frac{1}{n}\mathop \sum \limits_{i = 1}^n {\left( {{x_i} - \left( {\frac{{n + 1}}{2}} \right)} \right)^2}\]

As we know,${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$. Hence:

\[ \Rightarrow \frac{1}{n}\mathop \sum \limits_{i = 1}^{\text{n}} {\text{x}}_{\text{i}}^2 - \frac{1}{{\text{n}}}\mathop \sum \limits_{{\text{i}} = 1}^{\text{n}} 2\left( {\frac{{{\text{n}} + 1}}{{\text{n}}}} \right){{\text{x}}_i} + \frac{1}{n}\mathop \sum \limits_{{\text{i}} = 1}^{\text{n}} {\left( {\frac{{{\text{n}} + 1}}{2}} \right)^2}\]

\[ \Rightarrow \frac{1}{n}\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \left( {\frac{{n + 1}}{2}} \right)\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right] + \frac{{{{\left( {n + 1} \right)}^2}}}{{4n}} \times n\]

\[ \Rightarrow \frac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \frac{{{{\left( {n + 1} \right)}^2}}}{2} + \frac{{{{\left( {n + 1} \right)}^2}}}{4}\]

Taking LCM, we get:

\[ \Rightarrow \frac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \frac{{{{\left( {n + 1} \right)}^2}}}{4}\]

Taking $(n + 1)$ as common, we get:

$ \Rightarrow (n + 1)\left[ {\frac{{4n + 2 - 3n - 3}}{{12}}} \right]$

$ \Rightarrow \frac{{(n + 1)(n - 1)}}{{12}}$

Now, we know, $(a + b)(a - b) = {a^2} - {b^2}$. Hence,

$ \Rightarrow \frac{{{n^2} - 1}}{{12}}$

Therefore, mean is $\frac{{n + 1}}{2}$ and variance is $\frac{{{n^2} - 1}}{{12}}$.


3. Find the mean and variance for the first 10 multiples of 3.

Ans: First 10 multiples of 3 can be written as: 3, 9, 12, 15, 18, 21, 24, 27, 30. Hence, $n = 10$.

Mean, \[\bar x = \frac{{\sum\limits_{i = 1}^8 {{x_i}} }}{{\text{n}}} = \frac{{3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30}}{{10}} = \frac{{165}}{{10}} = 16.5\]

The following table is obtained:

${x_i}$

$\left( {{x_i} - \bar x} \right)$

${\left( {{x_i} - \bar x} \right)^2}$

3

$3 - 16.5 =  - 13.5$

182.25

6

$6 - 16.5 =  - 10.5$

110.25

9

$9 - 16.5 =  - 7.5$

56.25

12

$12 - 16.5 =  - 4.5$

20.25

15

$15 - 16.5 =  - 1.5$

2.25

18

$18 - 16.5 = 1.5$

2.25

21

$21 - 16.5 = 4.5$

20.25

24

$24 - 16.5 = 7.5$

56.25

27

$27 - 16.5 = 10.5$

110.25

30

$30 - 16.5 = 13.5$

182.25



742.5


Variance can be given as: 

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

\[{\sigma ^2} = \frac{1}{{10}} \times 742.5\]

${\sigma ^2} = 74.25$.

Therefore, mean is $16.5$ and variance is $74.25$.


4. Find the mean and variance for the data:

${x_i}$

6

10

14

18

24

28

30

${f_i}$

2

4

7

12

8

4

3


Ans: Calculating the given data we get the following table:

${x_i}$

${f_i}$

${f_i}{x_i}$

${x_i} - \bar x$

${\left( {{x_i} - \bar x} \right)^2}$

${f_i}{\left( {{x_i} - \bar x} \right)^2}$

6

2

12

$6 - 19 =  - 13$

169

338

10

4

40

$10 - 19 =  - 9$

81

324

14

7

98

$14 - 19 =  - 5$

25

175

18

12

216

$18 - 19 =  - 1$

1

12

24

8

192

$24 - 19 = 5$

25

200

28

4

112

$28 - 19 = 9$

81

324

30

3

90

$30 - 19 = 11$

121

363


40

760



1736


Here, \[{\text{N}} = 40\], \[\mathop \sum \limits_{i = 1}^7 {f_i}{x_i} = 760\], \[\mathop \sum \limits_{i = 1}^7 {f_i}{(x - \bar x)^2} = 1736\]

Mean, \[\bar x = \frac{{\sum\limits_{i = 1}^7 {{f_i}{x_i}} }}{{{\text{\;N}}}} = \frac{1}{{40}} \times 760 = 19\]

Variance, ${\sigma ^2} = \frac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{{({x_i} - \bar x)}^2}}  = \frac{1}{{40}} \times 1736 = 43.4$

Therefore, mean is $19$ and variance is $43.4$.


5. Find the mean and variance for the data:

${x_i}$

92

93

97

98

102

104

109

${f_i}$

3

2

3

2

6

3

3


Ans: Calculating the given data we get the following table:

${x_i}$

${f_i}$

${f_i}{x_i}$

${x_i} - \bar x$

${\left( {{x_i} - \bar x} \right)^2}$

${f_i}{\left( {{x_i} - \bar x} \right)^2}$

92

3

276

$92 - 100 =  - 8$

64

192

93

2

186

$93 - 100 =  - 7$

49

98

97

3

291

$97 - 100 =  - 3$

9

27

98

2

196

$98 - 100 =  - 2$

4

8

102

6

612

$102 - 100 = 2$

4

24

104

3

312

$104 - 100 = 4$

16

48

109

3

327

$109 - 100 = 9$

81

243


22

2200



640


Here, \[{\text{N}} = 22\], \[\mathop \sum \limits_{i = 1}^7 {f_i}{x_i} = 2200\], \[\mathop \sum \limits_{i = 1}^7 {f_i}{(x - \bar x)^2} = 640\]

Mean, \[\bar x = \frac{{\sum\limits_{i = 1}^7 {{f_i}{x_i}} }}{{{\text{\;N}}}} = \frac{1}{{22}} \times 2200 = 100\]

Variance, ${\sigma ^2} = \frac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{{({x_i} - \bar x)}^2}}  = \frac{1}{{22}} \times 640 = 29.09$

Therefore, mean is $100$ and variance is $29.09$.


6. Find the mean and standard deviation using short-cut method.

${x_i}$

60

61

62

63

64

65

66

67

68

${f_i}$

2

1

12

29

25

12

10

4

5


Ans: Suppose mean $A = 54$ and here, $h = 1$.

Calculating the given data, we get the following table:

${x_i}$

${f_i}$

${y_i} = \frac{{{x_i} - A}}{h}$

$y_i^2$

${f_i}{y_i}$

${f_i}y_i^2$

60

2

-4

16

-8

32

61

1

- 3

9

-3

9

62

12

- 2

4

-24

48

63

29

- 1

1

-29

29

64

25

0

0

0

0

65

12

1

1

12

12

66

10

2

4

20

40

67

4

3

9

12

36

68

5

4

16

20

80


100

220


0

286


Mean, \[\bar x = A + \frac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{{{\text{\;N}}}} \times h\]

               $ = 64 + \frac{0}{{100}} \times 1$

               $ = 64 + 0$

               $ = 64$

Variance, ${\sigma ^2} = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^9 {{f_i}y_i^2}  - {{\left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)}^2}} \right]$

                       $ = \frac{{{{(1)}^2}}}{{{{(100)}^2}}}\left[ {100(286) - {{(0)}^2}} \right]$

                       $ = 2.86$

Hence, standard deviation, $\sigma  = \sqrt {2.86}  = 1.69$

Therefore, mean is $64$ and standard deviation is $1.69$.


7. Find the mean and variance for the following frequency distribution.

Classes

0-30

30-60

60-90

90-120

120-150

150-180

180-210

Frequencies

2

3

5

10

3

5

2


Ans: Suppose mean $A = 105$ and here, $h = 30$.

Calculating the given data, we get the following table:

Class

Frequency ${f_i}$

Mid-point ${x_i}$

${y_i} = \frac{{{x_i} - A}}{h}$

$y_i^2$

${f_i}{y_i}$

${f_i}y_i^2$

0-30

2

15

-3

9

-6

18

30-60

3

45

- 2

4

- 6

12

60-90

5

75

- 1

1

- 5

5

90-120

10

105

0

0

0

0

120-150

3

135

1

1

3

3

150-180

5

165

2

4

10

20

180-210

2

195

3

9

6

18


30




2

76


Mean, \[\bar x = A + \frac{{\sum\limits_{i = 1}^7 {{f_i}{y_i}} }}{{{\text{\;N}}}} \times h\]

               $ = 105 + \frac{2}{{30}} \times 30$

               $ = 105 + 2$

               $ = 107$

Variance, ${\sigma ^2} = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^7 {{f_i}y_i^2}  - {{\left( {\sum\limits_{i = 1}^7 {{f_i}{y_i}} } \right)}^2}} \right]$

                       $ = \frac{{{{(30)}^2}}}{{{{(30)}^2}}}\left[ {30(76) - {{(2)}^2}} \right]$

                       $ = 2280 - 4$

                       $ = 2276$

Therefore, mean is $107$ and variance is $2276$.


8. Find the mean and variance for the following frequency distribution.

Classes

0-10

10-20

20-30

30-40

40-50

Frequencies

5

8

15

16

6


Ans: Suppose mean $A = 25$ and here, $h = 10$.

Calculating the given data, we get the following table:

Class

Frequency ${f_i}$

Mid-point ${x_i}$

${y_i} = \frac{{{x_i} - A}}{h}$

$y_i^2$

${f_i}{y_i}$

${f_i}y_i^2$

0-10

5

5

-2

4

-10

20

10-20

8

15

-1

1

-8

8

20-30

15

25

0

0

0

0

30-40

16

35

1

1

16

16

40-50

6

45

2

4

12

24


50




10

68


Mean, \[\bar x = A + \frac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{{{\text{\;N}}}} \times h\]

               $ = 25 + \frac{{10}}{{50}} \times 10$

               $ = 25 + 2$

               $ = 27$

Variance, ${\sigma ^2} = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^5 {{f_i}y_i^2}  - {{\left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)}^2}} \right]$

                       $ = \frac{{{{(10)}^2}}}{{{{(50)}^2}}}\left[ {50(68) - {{(10)}^2}} \right]$

                       $ = \frac{1}{{25}}\left[ {50 \times 68 - {{(10)}^2}} \right]$

                       $ = \frac{{3300}}{{25}}$

                       $ = 132$

Therefore, mean is $27$ and variance is $132$.


9. Find the mean, variance and standard deviation using short-cut method.

Height in cm

70-75

75-80

80-85

85-90

90-95

95-100

100-105

105-110

110-115

No. of children

2

1

12

29

25

12

10

4

5


Ans: Suppose mean $A = 92.5$ and here, $h = 5$.

Calculating the given data, we get the following table:


Class

Frequency ${f_i}$

Mid-point ${x_i}$

${y_i} = \frac{{{x_i} - A}}{h}$

$y_i^2$

${f_i}{y_i}$

${f_i}y_i^2$

70-75

3

72.5

- 4

16

-12

48

75-80

4

77.5

- 3

9

- 12

36

80-85

7

82.5

- 2

4

- 14

28

85-90

7

87.5

- 1

1

- 7

7

90-95

15

92.5

0

0

0

0

95-100

9

97.5

1

1

9

9

100-105

6

102.5

2

4

12

24

105-110

6

107.5

3

9

18

54

110-115

3

112.5

4

16

12

48


60




6

254


Mean, \[\bar x = A + \frac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{{{\text{\;N}}}} \times h\]

               $ = 92.5 + \frac{6}{{60}} \times 5$

               $ = 92.5 + 0.5$

               $ = 93$

Variance, ${\sigma ^2} = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^9 {{f_i}y_i^2}  - {{\left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)}^2}} \right]$

                       $ = \frac{{{{(5)}^2}}}{{{{(60)}^2}}}\left[ {60(254) - {{(6)}^2}} \right]$

                       $ = \frac{{25}}{{3600}}\left( {15204} \right)$

                       $ = 105.58$

Hence, standard deviation, $\sigma  = \sqrt {105.58}  = 10.27$

Therefore, mean is $93$, variance is $105.58$ and standard deviation is $10.27$.


10. The diameters of circles (in mm) drawn in a design are given below:

Diameters

33-36

37-40

41-44

45-48

49-52

No. of circles

15

17

21

22

25


Calculate the standard deviation and mean diameter of the circles.

(Hint: First make the data continuous by making the classes as 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5 – 48.5, 48.5 – 52.5 and then proceed.)

Ans: Suppose mean $A = 42.5$ and here, $h = 4$.

Calculating the given data, we get the following table:

Class Interval

Frequency ${f_i}$

Mid-point ${x_i}$

${y_i} = \frac{{{x_i} - A}}{h}$

$y_i^2$

${f_i}{y_i}$

${f_i}y_i^2$

32.5-36.5

15

34.5

-2

4

-30

60

36.5-40.5

17

38.5

-1

1

-17

17

40.5-44.5

21

42.5

0

0

0

0

44.5-48.5

22

46.5

1

1

22

22

48.5-52.5

25

50.5

2

4

50

100


100




25

199


Mean, \[\bar x = A + \frac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{{{\text{\;N}}}} \times h\]

               $ = 42.5 + \frac{{25}}{{100}} \times 4$

               $ = 42.5 + 1$

               $ = 43.5$

Variance, ${\sigma ^2} = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^5 {{f_i}y_i^2}  - {{\left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)}^2}} \right]$

                       $ = \frac{{{{(4)}^2}}}{{{{(100)}^2}}}\left[ {100(19) - {{(25)}^2}} \right]$

                       $ = \frac{{16}}{{10000}}\left( {19275} \right)$

                       $ = 30.84$

Hence, standard deviation, $\sigma  = \sqrt {30.84}  = 5.55$

Therefore, mean is $43.5$, variance is $30.84$ and standard deviation is $5.55$.


All Topics Under NCERT Class 11 Maths for Exercise 13.2

The topics and subtopics covered under exercise 13.2 Maths Class 11 NCERT are given below.

Section 13.4.3: Limitations of mean deviation

Section 13.5 Variance and standard deviation

Section 13.5.1: Standard deviation

Section 13.5.2: Standard deviation of a discrete frequency distribution 

Section  13.5.3: Standard deviation of a continuous frequency distribution 

Section  13.5.4: Evaluating variance and standard deviation using the shortcut method


Which types of questions are asked in exercise 13.2 class 11 NCERT Maths?

In exercise 13.2 of Class 11 Mathematics, 10 questions are asked related to Statistics. 

The first six questions (1 - 6) are given to find the mean and variance based on the data provided. In questions 7 and 8, we have to find the mean and variance of the frequency distribution. In question 9, we have to evaluate the mean, standard deviation and variance using the short-cut method. 

In the last question, we have to calculate the mean and standard deviation of the diameters of the circles. For that, we will first make the data i.e., diameter in continuous classes.


NCERT Solutions for Class 11 Maths Chapter 13 Statistics Exercise 13.2

Opting for the NCERT solutions for Ex 13.2 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.2 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 13 Exercise 13.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 13 Exercise 13.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 13 Exercise 13.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.


Access Exercise wise NCERT Solutions for Chapter 13 Maths Class 11


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