NCERT Solutions for Class 11 Maths Chapter 15 Statistics

 

NCERT Solutions for Class 11 Maths Chapter 15 Statistics: You have learnt the basic statistics like mean, mode and median known as "measures of the central tendency" in the previous class. In this chapter, you will learn important topics like measures of dispersion, mean deviation, standard deviation, quartile deviation, range and many more. In NCERT solutions for class 11 maths chapter 15 statistics, you will get questions related to all the above topics. You will also learn some methods for finding a representative value for the given data. This value is known as the measure of central tendency. This chapter is important for the CBSE class 11 final examination as well as in the various competitive exams like JEE Main, BITSAT, etc. As statistics deal with the collected data for specific purposes and in the digital world, everything that we do comes in the form of data. This is a very useful technique to interpret, analyze the data and make a conclusive decision based on the data. Solutions of NCERT for class 11 maths chapter 15 statistics will build your fundamentals of statistics which will be useful to learn advanced statistical techniques. There are 27 questions in 3 exercises in the NCERT textbook. All these questions are explained in CBSE NCERT solutions for class 11 maths chapter 15 statistics in a detailed manner. Check all NCERT solutions which will help you to understand the concepts in an easy way.
Topics of NCERT Grade 11 Maths Chapter-15 Statistics

15.1 Introduction

15.2 Measures of Dispersion

15.3 Range

15.4 Mean Deviation

15.5 Variance and Standard Deviation

15.6 Analysis of Frequency Distributions

 

The complete Solutions of NCERT Class 11 Mathematics Chapter 15 is provided below:

NCERT solutions for class 11 maths chapter 15 statistics-Exercise: 15.1

Question:1. Find the mean deviation about the mean for the data.

    \small 4,7,8,9,10,12,13,17

Answer:

Mean (\overline{x}) of the given data:

\overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{4+ 7+ 8+ 9+ 10+ 12+ 13+ 17}{8} = 10

The respective absolute values of the deviations from mean, |x_i - \overline{x}| are

6, 3, 2, 1, 0, 2, 3, 7

\therefore  \sum_{i=1}^{8}|x_i - 10| = 24

\therefore M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|

= \frac{24}{8} = 3

Hence, the mean deviation about the mean is 3.

Question:2. Find the mean deviation about the mean for the data.

   \small 38,70,48,40,42,55,63,46,54,44

Answer:

Mean (\overline{x}) of the given data:

\\ \overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{38+70+48+40+42+55+63+46+54+44}{10} \\ = \frac{500}{10} = 50

The respective absolute values of the deviations from mean, |x_i - \overline{x}| are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

\therefore  \sum_{i=1}^{8}|x_i - 50| = 84

\therefore M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|

= \frac{84}{10} = 8.4

Hence, the mean deviation about the mean is 8.4.

Question:3. Find the mean deviation about the median.

   \small 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Answer:

Number of observations, n = 12, which is even.

Arranging the values in ascending order:

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.

Now, Median (M)

\\ = \frac{(\frac{12}{2})^{th} observation + (\frac{12}{2}+ 1)^{th} observation}{2} \\ = \frac{13 + 14}{2} = \frac{27}{2}= 13.5

The respective absolute values of the deviations from median, |x_i - M| are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

\therefore  \sum_{i=1}^{8}|x_i - 13.5| = 28

\therefore M.D.(M) = \frac{1}{12}\sum_{i=1}^{n}|x_i - M|

= \frac{28}{12} = 2.33

Hence, the mean deviation about the median is 2.33.

Question:4. Find the mean deviation about the median.

    \small 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Answer:

Number of observations, n = 10, which is even.

Arranging the values in ascending order:

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

Now, Median (M)

\\ = \frac{(\frac{10}{2})^{th} observation + (\frac{10}{2}+ 1)^{th} observation}{2} \\ = \frac{46 + 49}{2} = \frac{95}{2}= 47.5

The respective absolute values of the deviations from median, |x_i - M| are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

\therefore  \sum_{i=1}^{8}|x_i - 47.5| = 70

\therefore M.D.(M) = \frac{1}{10}\sum_{i=1}^{n}|x_i - M|

= \frac{70}{10} = 7

Hence, the mean deviation about the median is 7.

Question:5 Find the mean deviation about the mean.

    \small \\x_i\hspace{1cm}5\hspace{1cm}10\hspace{1cm}15\hspace{1cm}20\hspace{1cm}25\\\small f_i\hspace{1cm}7\hspace{1cm}4\hspace{1.15cm}6\hspace{1.22cm}3\hspace{1.3cm}5

Answer:

x_i

f_i

f_ix_i

|x_i - \overline{x}|

f_i|x_i - \overline{x}|

5

7

35

9

63

10

4

40

4

16

15

6

90

1

6

20

3

60

6

18

25

5

125

11

55

 

\sum{f_i}

= 25

\sum f_ix_i

= 350

 

\sum f_i|x_i - \overline{x}|

=158

N = \sum_{i=1}^{5}{f_i} = 25 ; \sum_{i=1}^{5}{f_ix_i} = 350

\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{350}{12} = 14

Now, we calculate the absolute values of the deviations from mean, |x_i - \overline{x}|  and

\sum f_i|x_i - \overline{x}| = 158

\therefore M.D.(\overline{x}) = \frac{1}{25}\sum_{i=1}^{n}f_i|x_i - \overline{x}|

= \frac{158}{25} = 6.32

Hence, the mean deviation about the mean is 6.32

Question:6.   Find the mean deviation about the mean.

   \small \\x_i\hspace{1cm}10\hspace{1cm}30\hspace{1cm}50\hspace{1cm}70\hspace{1cm}90\\\small f_i\hspace{1cm}4\hspace{1.1cm}24\hspace{1.1cm}28\hspace{1.1cm}16\hspace{1.2cm}8

Answer:

x_i

f_i

f_ix_i

|x_i - \overline{x}|

f_i|x_i - \overline{x}|

10

4

40

40

160

30

24

720

20

480

50

28

1400

0

0

70

16

1120

20

320

90

8

720

40

320

 

\sum{f_i}

= 80

\sum f_ix_i

= 4000

 

\sum f_i|x_i - \overline{x}|

=1280

N = \sum_{i=1}^{5}{f_i} = 80 ; \sum_{i=1}^{5}{f_ix_i} = 4000

\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{4000}{80} = 50

Now, we calculate the absolute values of the deviations from mean, |x_i - \overline{x}|  and

\sum f_i|x_i - \overline{x}| = 1280

\therefore M.D.(\overline{x}) = \frac{1}{80}\sum_{i=1}^{5}f_i|x_i - \overline{x}|

= \frac{1280}{80} = 16

Hence, the mean deviation about the mean is 16

Question:7. Find the mean deviation about the median.

       \small x_i            \small 5            \small 7            \small 9            \small 10            \small 12            \small 15

      \small f_i             \small 8            \small 6            \small 2              \small 2             \small 2              \small 6

Answer:

x_i

f_i

c.f.

|x_i - M|

f_i|x_i - M|

5

8

8

2

16

7

6

14

0

0

9

2

16

2

4

10

2

18

3

6

12

2

20

5

10

15

6

26

8

48

Now, N = 26 which is even.

Median is the mean of 13^{th} and 14^{th} observations.

Both these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

Therefore, Median, M  = \frac{13^{th} observation + 14^{th} observation}{2} = \frac{7 + 7}{2} = \frac{14 }{2} = 7

Now, we calculate the absolute values of the deviations from median, |x_i - M|  and

\sum f_i|x_i - M| = 84

\therefore M.D.(M) = \frac{1}{26}\sum_{i=1}^{6}|x_i - M|

= \frac{84}{26} = 3.23

Hence, the mean deviation about the median is 3.23

Question:8 Find the mean deviation about the median.

   \small x_i            \small 15            \small 21            \small 27            \small 30            \small 35

    \small f_i            \small 3                \small 5             \small 6              \small 7             \small 8

Answer:

x_i

f_i

c.f.

|x_i - M|

f_i|x_i - M|

15

3

3

13.5

40.5

21

5

8

7.5

37.5

27

6

14

1.5

9

30

7

21

1.5

10.5

35

8

29

6.5

52

Now, N = 30, which is even.

Median is the mean of 15^{th} and 16^{th} observations.

Both these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.

Therefore, Median, M  = \frac{15^{th} observation + 16^{th} observation}{2} = \frac{30 + 30}{2} = 30

Now, we calculate the absolute values of the deviations from median, |x_i - M|  and

\sum f_i|x_i - M| = 149.5

\therefore M.D.(M) = \frac{1}{29}\sum_{i=1}^{5}|x_i - M|

= \frac{149.5}{29} = 5.1

Hence, the mean deviation about the median is 5.1

Question:9. Find the mean deviation about the mean.

Income per day in Rs

\small 0-100

\small 100-200

\small 200-300

\small 300-400

\small 400-500

\small 500-600

\small 600-700

\small 700-800

Number of persons

\small 4

\small 8

\small 9

\small 10

\small 7

\small 5

\small 4

\small 3

 

 

 

 

 

Answer:

Income

per day

Number of 

Persons f_i

Mid

Points x_i

f_ix_i

|x_i - \overline{x}|

f_i|x_i - \overline{x}|

0 -100

4

50

200

308

1232

100 -200

8

150

1200

208

1664

200-300

9

250

2250

108

972

300-400

10

350

3500

8

80

400-500

7

450

3150

92

644

500-600

5

550

2750

192

960

600-700

4

650

2600

292

1168

700-800

3

750

2250

392

1176

 

\sum{f_i}

=50

 

\sum f_ix_i

=17900

 

\sum f_i|x_i - \overline{x}|

=7896

 

N = \sum_{i=1}^{8}{f_i} = 50 ; \sum_{i=1}^{8}{f_ix_i} = 17900

\overline{x} = \frac{1}{N}\sum_{i=1}^{8}f_ix_i = \frac{17900}{50} = 358

Now, we calculate the absolute values of the deviations from mean, |x_i - \overline{x}|  and

\sum f_i|x_i - \overline{x}| = 7896

\therefore M.D.(\overline{x}) = \frac{1}{50}\sum_{i=1}^{8}f_i|x_i - \overline{x}|

= \frac{7896}{50} = 157.92

Hence, the mean deviation about the mean is 157.92

Question:10. Find the mean deviation about the mean.

Height in cms

\small 95-105

\small 105-115

\small 115-125

\small 125-135

\small 135-145

\small 145-155

Number of persond

\small 9

\small 13

\small 26

\small 30

\small 12

\small 10

Answer:

Height

in cms

Number of 

Persons f_i

Mid

Points x_i

f_ix_i

|x_i - \overline{x}|

f_i|x_i - \overline{x}|

95 -105

9

100

900

25.3

227.7

105 -115

13

110

1430

15.3

198.9

115-125

26

120

3120

5.3

137.8

125-135

30

130

3900

4.7

141

135-145

12

140

1680

14.7

176.4

145-155

10

150

1500

24.7

247

 

\sum{f_i}

=100

 

\sum f_ix_i

=12530

 

\sum f_i|x_i - \overline{x}|

=1128.8

 

N = \sum_{i=1}^{6}{f_i} = 100 ; \sum_{i=1}^{6}{f_ix_i} = 12530

\overline{x} = \frac{1}{N}\sum_{i=1}^{6}f_ix_i = \frac{12530}{100} = 125.3

Now, we calculate the absolute values of the deviations from mean, |x_i - \overline{x}|  and

\sum f_i|x_i - \overline{x}| = 1128.8

\therefore M.D.(\overline{x}) = \frac{1}{100}\sum_{i=1}^{6}f_i|x_i - \overline{x}|

= \frac{1128.8}{100} = 11.29

Hence, the mean deviation about the mean is 11.29

Question:11. Find the mean deviation about median for the following data : 

Marks

\small 0-10

\small 10-20

\small 20-30

\small 30-40

\small 40-50

\small 50-60

Number of girls

\small 6

\small 8

\small 14

\small 16

\small 4

\small 2

Answer:

Marks

Number of 

Girls f_i

Cumulative

Frequency c.f.

Mid

Points x_i

|x_i - M|

f_i|x_i - M|

0-10

6

6

5

22.85

137.1

10-20

8

14

15

12.85

102.8

20-30

14

28

25

2.85

39.9

30-40

16

44

35

7.15

114.4

40-50

4

48

45

17.15

68.6

50-60

2

50

55

27.15

54.3

 

 

 

 

 

\sum f_i|x_i - M|

=517.1

Now, N = 50, which is even.

The class interval containing \left (\frac{N}{2} \right)^{th} or 25^{th} item is 20-30. Therefore, 20-30 is the median class.

We know,

Median = l + \frac{\frac{N}{2}- C}{f}\times h

Here, l = 20, C = 14, f = 14, h = 10 and N = 50

Therefore, Median = 20 + \frac{25 - 14}{14}\times 10 = 20 + 7.85 = 27.85

Now, we calculate the absolute values of the deviations from median, |x_i - M|  and

\sum f_i|x_i - M| = 517.1

\therefore M.D.(M) = \frac{1}{50}\sum_{i=1}^{6}f_i|x_i - M|

= \frac{517.1}{50} = 10.34

Hence, the mean deviation about the median is 10.34

Question:12 Calculate the mean deviation about median age for the age distribution of \small 100 persons given below:

Age (in years)

\small 16-20

\small 21-25

\small 26-30

\small 31-35

\small 36-40

\small 41-45

\small 46-50

\small 51-55

Number

\small 5

\small 6

\small 12

\small 14

\small 26

\small 12

\small 16

\small 9

[Hint  Convert the given data into continuous frequency distribution by subtracting 0.5from the lower limit and adding 0.5 to the upper limit of each class interval]

Answer:

Age

(in years)

Number

 f_i

Cumulative

Frequency c.f.

Mid

Points x_i

|x_i - M|

f_i|x_i - M|

15.5-20.5

5

5

18

20

100

20.5-25.5

6

11

23

15

90

25.5-30.5

12

23

28

10

120

30.5-35.5

14

37

33

5

70

35.5-40.5

26

63

38

0

0

40.5-45.5

12

75

43

5

60

45.5-50.5

16

91

48

10

160

50.5-55.5

9

100

53

15

135

 

 

 

 

 

\sum f_i|x_i - M|

=735

Now, N = 100, which is even.

The class interval containing \left (\frac{N}{2} \right)^{th} or 50^{th} item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.

We know,

Median = l + \frac{\frac{N}{2}- C}{f}\times h

Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100

Therefore, Median = 35.5 + \frac{50 - 37}{26}\times 5 = 35.5 + 2.5 = 38

Now, we calculate the absolute values of the deviations from median, |x_i - M|  and

\sum f_i|x_i - M| = 735

\therefore M.D.(M) = \frac{1}{100}\sum_{i=1}^{8}f_i|x_i - M|

= \frac{735}{100} = 7.35

Hence, the mean deviation about the median is 7.35

NCERT solutions for class 11 maths chapter 15 statistics-Exercise: 15.2

Question:1. Find the mean and variance for each of the data.

       \small 6, 7, 10, 12, 13, 4, 8, 12

Answer:

Mean (\overline{x}) of the given data:

\overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{6+ 7+ 10+ 12+ 13+ 4+ 8+ 12}{8} = \frac{72}{8} = 9

The respective values of the deviations from mean, (x_i - \overline{x}) are

-3, -2, 1 3 4 -5 -1 3

\therefore  \sum_{i=1}^{8}(x_i - 10)^2 = 74

\therefore \sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2

\frac{1}{8}\sum_{i=1}^{8}(x_i - \overline{x})^2= \frac{74}{8} = 9.25

Hence,  Mean = 9 and Variance = 9.25

Question:2. Find the mean and variance for each of the data.

  First n natural numbers.

Answer:

Mean (\overline{x}) of first n natural numbers:

\overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}

We know, Variance\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2 

\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}\left (x_i - \frac{n+1}{2} \right )^2

We know that (a-b)^2 = a^2 - 2ab + b^2

\\ \therefore n\sigma^2 = \sum_{i=1}^{n}x_i^2 + \sum_{i=1}^{n}(\frac{n+1}{2})^2 - 2\sum_{i=1}^{n}x_i\frac{n+1}{2} \\ = \frac{n(n+1)(2n+1)}{6} + \frac{(n+1)^2}{4}\times n - 2.\frac{(n+1)}{2}.\frac{n(n+1)}{2}

\\ \implies \sigma^2 = \frac{(n+1)(2n+1)}{6} + \frac{(n+1)^2}{4} - \frac{(n+1)^2}{2}

\\ = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} \\ = (n+1)\left [\frac{4n+2 - 3n -3}{12} \right ] \\ = (n+1).\frac{(n-1)}{12} \\ = \frac{n^2-1}{12}

Hence, Mean = \frac{n+1}{2} and Variance = \frac{n^2-1}{12}

Question:3. Find the mean and variance for each of the data 

 First 10 multiples of 3

Answer:

First 10 multiples of 3 are:

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Mean (\overline{x}) of the above values:

\\ \overline{x} = \frac{1}{10}\sum_{i=1}^{10}x_i = \frac{3+ 6+ 9+ 12+ 15+ 18+ 21+ 24+ 27+ 30}{10} \\ = 3.\frac{\frac{10(10+1)}{2}}{10} = 16.5

The respective values of the deviations from mean, (x_i - \overline{x}) are

-13.5, -10.5, -7.5, -4.5, -1.5, 1.5, 4.5, 7.5, 10.5, 13.5

\therefore  \sum_{i=1}^{10}(x_i - 16.5)^2 = 742.5

\therefore \sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2

\frac{1}{10}\sum_{i=1}^{10}(x_i - \overline{x})^2= \frac{742.5}{10} = 74.25

Hence,  Mean = 16.5 and Variance = 74.25

Question:4. Find the mean and variance for each of the data.

\small x_i

6

10

14

18

24

28

30

\small f_i

2

4

7

12

8

4

3

 

Answer:

x_i

f_i

f_ix_i

(x_i - \overline{x})

(x_i - \overline{x})^2

f_i(x_i - \overline{x})^2

2

12

-13

169

338

10

4

40

-9

81

324

14

7

98

-5

25

175

18

12

216

-1

1

12

24

8

192

5

25

200

28

4

112

9

81

324

30

3

90

13

169

363

 

\sum{f_i}

= 40

\sum f_ix_i

= 760

 

 

\sum f_i(x_i - \overline{x})^2

=1736

N = \sum_{i=1}^{7}{f_i} = 40 ; \sum_{i=1}^{7}{f_ix_i} = 760

\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{760}{40} = 19

We know, Variance, \sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2

\implies \sigma^2 = \frac{1736}{40} = 43.4

Hence, Mean = 19 and Variance = 43.4

Question:​​​​​​​5. Find the mean and variance for each of the data.

\small x_i

92

93

97

98

102

104

109

\small f_i

3

2

3

2

6

3

3

 

Answer:

x_i

f_i

f_ix_i

(x_i - \overline{x})

(x_i - \overline{x})^2

f_i(x_i - \overline{x})^2

92 

3

276

-8

64

192

93

2

186

-7

49

98

97

3

291

-3

9

27

98

2

196

-2

4

8

102

6

612

2

4

24

104

3

312

4

16

48

109

3

327

9

81

243

 

\sum{f_i}

= 22

\sum f_ix_i

= 2200

 

 

\sum f_i(x_i - \overline{x})^2

=640

N = \sum_{i=1}^{7}{f_i} = 22 ; \sum_{i=1}^{7}{f_ix_i} = 2200

\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{2200}{22} = 100

We know, Variance, \sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2

\implies \sigma^2 = \frac{640}{22} = 29.09

Hence, Mean = 100 and Variance = 29.09

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

\small x_i

60

61

62

63

64

65

66

67

68

\small f_i

2

1

12

29

25

12

10

4

5

 

Answer:

Let the assumed mean, A = 64 and h = 1

x_i

f_i

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

y_i^2

f_iy_i

f_iy_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

 

\sum{f_i}

=100

 

 

\sum f_iy_i

= 0

\sum f_iy_i ^2

=286

N = \sum_{i=1}^{9}{f_i} = 100 ; \sum_{i=1}^{9}{f_iy_i} = 0

Mean,

\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =64 + \frac{0}{100} = 64

We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

\\ \implies \sigma^2 = \frac{1}{(100)^2}\left [100(286) - (0)^2 \right ] \\ = \frac{28600}{10000} = 2.86

We know,  Standard Deviation = \sigma = \sqrt{Variance}

\therefore \sigma = \sqrt{2.86} = 1.691

Hence, Mean = 64 and Standard Deviation = 1.691

Question:​​​​​​​7.Find the mean and variance for the following frequency distributions.

Classes

0-30

30-60

60-90

90-120

120-150

150-180

180-210

Frequencies

2

3

5

10

3

5

2

 

Answer:

Classes

Frequency

f_i

Mid point 

x_i

f_ix_i

(x_i - \overline{x})

(x_i - \overline{x})^2

f_i(x_i - \overline{x})^2

0-30

2

15

30

-92

8464

16928

30-60

3

45

135

-62

3844

11532

60-90

5

75

375

-32

1024

5120

90-120

10

105

1050

2

4

40

120-150

3

135

405

28

784

2352

150-180

5

165

825

58

3364

16820

180-210

2

195

390

88

7744

15488

 

\sum{f_i} = N

= 30

 

\sum f_ix_i

= 3210

 

 

\sum f_i(x_i - \overline{x})^2

=68280

 

\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{3210}{30} = 107

We know, Variance, \sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2

\implies \sigma^2 = \frac{68280}{30} = 2276

Hence, Mean = 107 and Variance = 2276

Question:​​​​​​​8.  Find the mean and variance for the following frequency distributions.

Classes

0-10

10-20

20-30

30-40

40-50

Frequencies

5

8

15

16

6

 

Answer:

Classes

Frequency

f_i

Mid-point 

x_i

f_ix_i

(x_i - \overline{x})

(x_i - \overline{x})^2

f_i(x_i - \overline{x})^2

0-10

5

5

25

-22

484

2420

10-20

8

15

120

-12

144

1152

20-30

15

25

375

-2

4

60

30-40

16

35

560

8

64

1024

40-50

6

45

270

18

324

1944

 

\sum{f_i} = N

= 50

 

\sum f_ix_i

= 1350

 

 

\sum f_i(x_i - \overline{x})^2

=6600

 

\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{1350}{50} = 27

We know, Variance, \sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2

\implies \sigma^2 = \frac{6600}{50} = 132

Hence, Mean = 27 and Variance = 132

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

Height in cms

70-75

75-80

80-85

85-90

90-95

95-100

100-105

105-110

110-115

No. of students 

3

4

7

7

15

9

6

6

3

 

 

 

 

 

Answer:

Let the assumed mean, A = 92.5 and h = 5

Height

in cms

Frequency

f_i

Midpoint

x_i

\dpi{100} y_i = \frac{x_i-A}{h}

y_i^2

f_iy_i

f_iy_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

 

\sum{f_i} =N = 60

 

 

 

\sum f_iy_i

= 6

\sum f_iy_i ^2

=254

Mean,

\overline{y} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =92.5 + \frac{6}{60}\times5 = 93

We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

\\ \implies \sigma^2 = \frac{1}{(60)^2}\left [60(254) - (6)^2 \right ] \\ = \frac{1}{(60)^2}\left [15240 - 36 \right ] \\ = \frac{15204}{144} = 105.583

We know,  Standard Deviation = \sigma = \sqrt{Variance}

\therefore \sigma = \sqrt{105.583} = 10.275

Hence, Mean = 93, Variance = 105.583 and Standard Deviation = 10.275

Question: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 32.5-36.5,36.5-40.4,40.5-44.5,44.5-48.5,48.5-52.5  and then proceed.]

Answer:

Let the assumed mean, A = 92.5 and h = 5

Diameters

No. of

circles f_i

Midpoint

x_i

\dpi{100} y_i = \frac{x_i-A}{h}

y_i^2

f_iy_i

f_iy_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

 

\sum{f_i} =N = 100

 

 

 

\sum f_iy_i

= 25

\sum f_iy_i ^2

=199

Mean,

\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =42.5 + \frac{25}{100}\times4 = 43.5

We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

\\ \implies \sigma^2 = \frac{1}{(100)^2}\left [100(199) - (25)^2 \right ]\times4^2 \\ = \frac{1}{625}\left [19900 - 625 \right ] \\ = \frac{19275}{625} = 30.84

We know,  Standard Deviation = \sigma = \sqrt{Variance}

\therefore \sigma = \sqrt{30.84} = 5.553

Hence, Mean = 43.5, Variance = 30.84 and Standard Deviation = 5.553

CBSE NCERT solutions for class 11 maths chapter 15 statistics-Exercise: 15.3

Question:​​​​​​​1.From the data given below state which group is more variable, A or B?

Marks

10-20

20-30

30-40

40-50

50-60

60-70

70-80

Group A

9

17

32

33

40

10

9

Group B

10

20

30

25

43

15

7

 

Answer:

The group having a higher coefficient of variation will be more variable.

Let the assumed mean, A = 45 and h = 10

For Group A

Marks

Group A

f_i

Midpoint

x_i

\dpi{100} y_i = \frac{x_i-A}{h}

= \frac{x_i-45}{10}

y_i^2

f_iy_i

f_iy_i^2

10-20

9

15

-3

9

-27

81

20-30

17

25

-2

4

-34

68

30-40

32

35

-1

1

-32

32

40-50

33

45

0

0

0

0

50-60

40

55

1

1

40

40

60-70

10

65

2

4

20

40

70-80

9

75

3

9

27

81

 

\sum{f_i} =N = 150

 

 

 

\sum f_iy_i

= -6

\sum f_iy_i ^2

=342

Mean,

\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =45 + \frac{-6}{150}\times10 = 44.6

We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

\\ \implies \sigma^2 = \frac{1}{(150)^2}\left [150(342) - (-6)^2 \right ]\times10^2 \\ = \frac{1}{15^2}\left [51264 \right ] \\ =227.84

We know,  Standard Deviation = \sigma = \sqrt{Variance}

\therefore \sigma = \sqrt{227.84} = 15.09

Coefficient of variation = \frac{\sigma}{\overline x}\times100

C.V.(A) = \frac{15.09}{44.6}\times100 = 33.83

 Similarly, 

For Group B

Marks

Group A

f_i

Midpoint

x_i

\dpi{100} y_i = \frac{x_i-A}{h}

= \frac{x_i-45}{10}

y_i^2

f_iy_i

f_iy_i^2

10-20

10

15

-3

9

-30

90

20-30

20

25

-2

4

-40

80

30-40

30

35

-1

1

-30

30

40-50

25

45

0

0

0

0

50-60

43

55

1

1

43

43

60-70

15

65

2

4

30

60

70-80

7

75

3

9

21

72

 

\sum{f_i} =N = 150

 

 

 

\sum f_iy_i

= -6

\sum f_iy_i ^2

=375

Mean,

\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =45 + \frac{-6}{150}\times10 = 44.6

We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

\\ \implies \sigma^2 = \frac{1}{(150)^2}\left [150(375) - (-6)^2 \right ]\times10^2 \\ = \frac{1}{15^2}\left [56214 \right ] \\ =249.84

We know,  Standard Deviation = \sigma = \sqrt{Variance}

\therefore \sigma = \sqrt{249.84} = 15.80

Coefficient of variation = \frac{\sigma}{\overline x}\times100

C.V.(B) = \frac{15.80}{44.6}\times100 = 35.42

Since C.V.(B) > C.V.(A)

Therefore, Group B is more variable.

Question:2 From the prices of shares X and Y below, find out which is more stable in value:      

X

35

54

52

53

56

58

52

50

51

49

Y

108

107

105

105

106

107

104

103

104

101

 

Answer:

X(x_i)

Y(y_i)

x_i^2

y_i^2

35

108

1225

11664

54

107

2916

11449

52

105

2704

11025

53

105

2809

11025

56

106

8136

11236

58

107

3364

11449

52

104

2704

10816

50

103

2500

10609

51

104

2601

10816

49

101

2401

10201

=510

= 1050

=26360

=110290

For X,

Mean , \overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{510}{10} = 51

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

\\ \implies \sigma^2 = \frac{1}{(10)^2}\left [10(26360) - (510)^2 \right ] \\ = \frac{1}{100}.\left [263600 - 260100 \right ] \\ \\ = 35

We know,  Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{35} = 5.91

C.V.(X) = \frac{\sigma}{\overline x}\times100 = \frac{5.91}{51}\times100 = 11.58 

Similarly, For Y,

Mean , \overline{y} = \frac{1}{n}\sum_{i=1}^{n}y_i = \frac{1050}{10} = 105

Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum y_i^2 - (\sum y_i)^2 \right ]

\\ \implies \sigma^2 = \frac{1}{(10)^2}\left [10(110290) - (1050)^2 \right ] \\ = \frac{1}{100}.\left [1102900 - 1102500 \right ] \\ \\ = 4

We know,  Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{4} = 2

C.V.(Y) = \frac{\sigma}{\overline y}\times100 = \frac{2}{105}\times100 = 1.904

Since C.V.(Y) < C.V.(X)

Hence Y is more stable.  

Question:3(i) An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:
 

 

Firm A

Firm B

No. of wage earners

586

648

Mean of monthly wages

Rs\hspace {1mm} 5253

Rs\hspace {1mm} 5253

Variance of the distribution of wages

100

121

Which firm A or B pays larger amount as monthly wages?

Answer:

Given, Mean of monthly wages of firm A = 5253

Number of wage earners = 586

Total amount paid = 586 x 5253 = 30,78,258

Again, Mean of monthly wages of firm B = 5253

Number of wage earners = 648

Total amount paid = 648 x 5253 = 34,03,944

Hence firm B pays larger amount as monthly wages.

Question:3(ii) An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

 

Firm A

Firm B

No. of wage earners

586

648

Mean of monthly wages

Rs\hspace {1mm} 5253

Rs\hspace {1mm} 5253

Variance of the distribution of wages

100

121

 Which firm, A or B, shows greater variability in individual wages?

Answer:

Given, Variance of firm A = 100

Standard Deviation = \sigma_A = \sqrt{Variance}= \sqrt{100} = 10

Again, Variance of firm B = 121

Standard Deviation = \sigma_B = \sqrt{Variance}= \sqrt{121} = 11

Since \sigma_B>\sigma_A, firm B has greater variability in individual wages.