NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.2) Exercise 15.2

NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.2) Exercise 15.2

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Access NCERT Solutions for Class 11 Maths Chapter 15 – Statistics part-1

Access NCERT Solutions for Class 11 Maths Chapter 15 – Statistics

Exercise – 15.2

  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.


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

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

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

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

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


  1. 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 $74.25$.


  1. 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 $43.4$.


  1. 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 $29.09$.


  1. 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 $1.69$.


  1. 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 $2276$.


  1. 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 $132$.


  1. 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 $10.27$.


  1. 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 $5.55$.


NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.2

Opting for the NCERT solutions for Ex 15.2 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 15.2 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 15 Exercise 15.2 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 15 Exercise 15.2, 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 15 Exercise 15.2 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. 

FAQs (Frequently Asked Questions)

1. What are the topics included in Chapter 15 of NCERT Class11 Maths Textbook?

Ans. NCERT Chapter 15 of Class 11 Maths includes all the important Statistics topics that are important for the final exams. It is categorized into the following exercises for better understanding of the students:

  • Chapter 15 Exercise 15.1 - Introduction 

  • Chapter 15 Exercise 15.2 - Measures of Dispersion

  • Chapter 15 Exercise 15.3 - Range

  • Chapter 15 Exercise 15.4 - Mean Deviation

  • Chapter 15 Exercise 15.5 - Variance and Standard Deviation 

  • Chapter 15 Exercise 15.6 - Analysis of Frequency Distributions

2. Can I download the NCERT Solutions for Class 11 Maths from Vedantu site?

Ans: Yes, you can download the NCERT Solutions for Class 11 Maths from Vedantu's website. We have also provided the Class 11 Maths Chapter 15 solutions on this page in PDF format. Vedantu is an online learning platform which provides free PDF downloads of solved NCERT Solutions along with previous year papers for various state boards and CBSE exams. These solutions are solved by expert teachers on Vedantu.com.

Also, register yourself to the Vedantu app to get the latest information, study materials and PDFs CBSE and various state board examinations.

3. Why should I opt for the NCERT Solutions for Class 11 Maths offered by Vedantu?

Ans: The NCERT Solutions offered by Vedantu for Class 11 Maths are solved by our best expert teachers who have years of valuable experience in the similar field. They provide detailed chapter-wise solutions and step-by-step instructions for easy learning. Solving these NCERT solutions regularly will also help the candidates to have a better knowledge about the topics and thus help them in boosting their confidence. For more information, follow Vedantu. 

4. Which books shall I refer along with NCERT Class 11 Maths Textbook?

Ans. Although NCERT textbooks provide detailed information and solutions for every chapters, it is advisable to the students to refer to R.S. Agarwal and R.D. Sharma textbooks for better understanding. There are proper chapter-wise solutions provided in the textbook which are categorised as very short, short, long, multiple-type and high order thinking type questions. Therefore, CBSE students can refer to both R.S. Agarwal and R. D. Sharma books for gaining better concepts regarding any topic. The students can also download free PDFs of solved R.S.Agarwal's questions from the Vedantu’s website. These solutions are solved by our expert teachers who are having years of experience in this subject.

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