NCERT Solutions for Class 11 Maths Chapter 15 Statistics

The NCERT Solutions For Class 11 Maths Chapter 15 Statistics are available as a pdf on this page. The NCERT Solutions are authored by the most experienced educators in the teaching industry making the solution of every problem straightforward and justifiable. Every solution written in the pdf given underneath empowers the student to get ready for the test and accomplish it. These solutions assist a class 11 student with mastering the idea of Limits and Derivatives.

Download PDF of NCERT Solutions for Class 11 Maths Chapter 15 Statistics

Access NCERT Solutions for Class 11 Maths Chapter 15 Statistics

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Get detailed solution for all the questions listed under below exercises:

Exercise 15.1 Solutions : 12 Questions

Exercise 15.2 Solutions : 10 Questions

Exercise 15.3 Solutions : 5 Questions

Miscellaneous Exercise Solutions: 7 Questions

Access Solutions for Class 11 Maths Chapter 15 Exercise

Exercise 15.1 Page: 360

Find the mean deviation about the mean for the data in Exercises 1 and 2.

1. 4, 7, 8, 9, 10, 12, 13, 17

Solution:-

First we have to find (x̅) of the given data

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 1

So, the respective values of the deviations from mean,

i.e., xi – x̅ are, 10 – 4 = 6, 10 – 7 = 3, 10 – 8 = 2, 10 – 10 = 0,

10 – 12 = – 2, 10 – 13 = – 3, 10 – 17 = – 7

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

Now absolute values of the deviations,

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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 2

MD = sum of deviations/ number of observations

= 24/8

= 3

So, the mean deviation for the given data is 3.

2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Solution:-

First we have to find (x̅) of the given data

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 3

So, the respective values of the deviations from mean,

i.e., xi – x̅ are, 50 – 38 = -12, 50 -70 = -20, 50 – 48 = 2, 50 – 40 = 10, 50 – 42 = 8,

50 – 55 = – 5, 50 – 63 = – 13, 50 – 46 = 4, 50 – 54 = -4, 50 – 44 = 6

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

Now absolute values of the deviations,

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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 4

MD = sum of deviations/ number of observations

= 84/10

= 8.4

So, the mean deviation for the given data is 8.4.

Find the mean deviation about the median for the data in Exercises 3 and 4.

3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Solution:-

First we have to arrange the given observations into ascending order,

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

The number of observations is 12

Then,

Median = ((12/2)th observation + ((12/2)+ 1)th observation)/2

(12/2)th observation = 6th = 13

(12/2)+ 1)th observation = 6 + 1

= 7th = 14

Median = (13 + 14)/2

= 27/2

= 13.5

So, the absolute values of the respective deviations from the median, i.e., |xi – 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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 5

Mean Deviation,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 6

= (1/12) × 28

= 2.33

So, the mean deviation about the median for the given data is 2.33.

4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Solution:-

First we have to arrange the given observations into ascending order,

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

The number of observations is 10

Then,

Median = ((10/2)th observation + ((10/2)+ 1)th observation)/2

(10/2)th observation = 5th = 46

(10/2)+ 1)th observation = 5 + 1

= 6th = 49

Median = (46 + 49)/2

= 95

= 47.5

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are

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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 7

Mean Deviation,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 8

= (1/10) × 70

= 7

So, the mean deviation about the median for the given data is 7.

Find the mean deviation about the mean for the data in Exercises 5 and 6.

5.

xi

5

10

15

20

25

fi

7

4

6

3

5

Solution:-

Let us make the table of the given data and append other columns after calculations.

Xi

fi

fixi

|xi – x̅|

fi |xi – x̅|

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

25

350

158

The sum of calculated data,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 9

The absolute values of the deviations from the mean, i.e., |xi – x̅|, as shown in the table.

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 10

6.

xi

10

30

50

70

90

fi

4

24

28

16

8

Solution:-

Let us make the table of the given data and append other columns after calculations.

Xi

fi

fixi

|xi – x̅|

fi |xi – x̅|

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

80

4000

1280

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 11

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 12

Find the mean deviation about the median for the data in Exercises 7 and 8.

7.

xi

5

7

9

10

12

15

fi

8

6

2

2

2

6

Solution:-

Let us make the table of the given data and append other columns after calculations.

Xi

fi

c.f.

|xi – M|

fi |xi – 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 the 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

Then,

Median = (13th observation + 14th observation)/2

= (7 + 7)/2

= 14/2

= 7

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are shown in the table.

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 13

8.

xi

15

21

27

30

35

fi

3

5

6

7

8

Solution:-

Let us make the table of the given data and append other columns after calculations.

Xi

fi

c.f.

|xi – M|

fi |xi – 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 the 15th and 16th observations. Both of these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.

Then,

Median = (15th observation + 16th observation)/2

= (30 + 30)/2

= 60/2

= 30

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are shown in the table.

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 14

Find the mean deviation about the mean for the data in Exercises 9 and 10.

9.

Income per day in ₹

0 – 100

100 – 200

200 – 300

300 – 400

400 – 500

500 – 600

600 – 700

700 – 800

Number of persons

4

8

9

10

7

5

4

3

Solution:-

Let us make the table of the given data and append other columns after calculations.

Income per day in ₹

Number of persons fi

Mid – points

xi

fixi

|xi – x̅|

fi|xi – x̅|

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

1160

700 – 800

3

750

2250

392

1176

50

17900

7896

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 15

10.

Height in cms

95 – 105

105 – 115

115 – 125

125 – 135

135 – 145

145 – 155

Number of boys

9

13

26

30

12

10

Solution:-

Let us make the table of the given data and append other columns after calculations.

Height in cms

Number of boys fi

Mid – points

xi

fixi

|xi – x̅|

fi|xi – x̅|

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

100

12530

1128.8

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 16

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

Marks

0 -10

10 -20

20 – 30

30 – 40

40 – 50

50 – 60

Number of girls

6

8

14

16

4

2

Solution:-

Let us make the table of the given data and append other columns after calculations.

Marks

Number of Girls fi

Cumulative frequency (c.f.)

Mid – points

xi

|xi – Med|

fi|xi – Med|

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

50

517.1

The class interval containing Nth/2 or 25th item is 20-30

So, 20-30 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 20, c = 14, f = 14, h = 10 and n = 50

Median = 20 + (((25 – 14))/14) × 10

= 20 + 7.85

= 27.85

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 17

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

Age

(in years)

16 – 20

21 – 25

26 – 30

31 – 35

36 – 40

41 – 45

46 – 50

51 – 55

Number

5

6

12

14

26

12

16

9

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

Solution:-

The given data is converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding the 0.5 to the upper limit of each class intervals and append other columns after calculations.

Age

Number fi

Cumulative frequency (c.f.)

Mid – points

xi

|xi – Med|

fi|xi – Med|

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

100

735

The class interval containing Nth/2 or 50th item is 35.5 – 40.5

So, 35.5 – 40.5 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 35.5, c = 37, f = 26, h = 5 and N = 100

Median = 35.5 + (((50 – 37))/26) × 5

= 35.5 + 2.5

= 38

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 18


Exercise 15.2 Page: 371

Find the mean and variance for each of the data in Exercise 1 to 5.

1. 6, 7, 10, 12, 13, 4, 8, 12

Solution:-

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 19

So, x̅ = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8

= 72/8

= 9

Let us make the table of the given data and append other columns after calculations.

Xi

Deviations from mean

(xi – x̅)

(xi – x̅)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

We know that Variance,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 20

σ2 = (1/8) × 74

= 9.2

∴Mean = 9 and Variance = 9.25

2. First n natural numbers

Solution:-

We know that Mean = Sum of all observations/Number of observations

∴Mean, x̅ = ((n(n + 1))2)/n

= (n + 1)/2

and also WKT Variance,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 21

By substitute that value of x̅ we get,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 22

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 23

WKT, (a + b)(a – b) = a2 – b2

σ2 = (n2 – 1)/12

∴Mean = (n + 1)/2 and Variance = (n2 – 1)/12

3. First 10 multiples of 3

Solution:-

First we have to write the first 10 multiples of 3,

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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 24

So, x̅ = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10

= 165/10

= 16.5

Let us make the table of the data and append other columns after calculations.

Xi

Deviations from mean

(xi – x̅)

(xi – x̅)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

Then, Variance

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 25

= (1/10) × 742.5

= 74.25

∴Mean = 16.5 and Variance = 74.25

4.

xi

6

10

14

18

24

28

30

fi

2

4

7

12

8

4

3

Solution:-

Let us make the table of the given data and append other columns after calculations.

Xi

fi

fixi

Deviations from mean

(xi – x̅)

(xi – x̅)2

fi(xi – x̅)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

N = 40

760

1736

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 26

5.

xi

92

93

97

98

102

104

109

fi

3

2

3

2

6

3

3

Solution:-

Let us make the table of the given data and append other columns after calculations.

Xi

fi

fixi

Deviations from mean

(xi – x̅)

(xi – x̅)2

fi(xi – x̅)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

N = 22

2200

640

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 27

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

xi

60

61

62

63

64

65

66

67

68

fi

2

1

12

29

25

12

10

4

5

Solution:-

Let the assumed mean A = 64. Here h = 1

We obtain the following table from the given data.

Xi

Frequency fi

Yi = (xi – A)/h

Yi2

fiyi

fiyi2

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

0

286

Mean,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 28

Where A = 64, h = 1

So, x̅ = 64 + ((0/100) × 1)

= 64 + 0

= 64

Then, variance,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 29

σ2 = (12/1002) [100(286) – 02]

= (1/10000) [28600 – 0]

= 28600/10000

= 2.86

Hence, standard deviation = σ = √2.886

= 1.691

∴ Mean = 64 and Standard Deviation = 1.691

Find the mean and variance for the following frequency distributions in Exercises 7 and 8.

7.

Classes

0 – 30

30 – 60

60 – 90

90 – 120

120 – 150

150 – 180

180 – 210

Frequencies

2

3

5

10

3

5

2

Solution:-

Let us make the table of the given data and append other columns after calculations.

Classes

Frequency fi

Mid – points

xi

fixi

(xi – x̅)

(xi – x̅)2

fi(xi – x̅)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

N = 30

3210

68280

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 30

8.

Classes

0 – 10

10 – 20

20 – 30

30 – 40

40 –50

Frequencies

5

8

15

16

6

Solution:-

Let us make the table of the given data and append other columns after calculations.

Classes

Frequency fi

Mid – points

xi

fixi

(xi – x̅)

(xi – x̅)2

fi(xi – x̅)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

N = 50

1350

6600

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 31

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 32

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

Frequencies

3

4

7

7

15

9

6

6

3

Solution:-

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

Let us make the table of the given data and append other columns after calculations.

Height (class)

Number of children

Frequency fi

Midpoint

Xi

Yi = (xi – A)/h

Yi2

fiyi

fiyi2

70 – 75

2

72.5

-4

16

-12

48

75 – 80

1

77.5

-3

9

-12

36

80 – 85

12

82.5

-2

4

-14

28

85 – 90

29

87.5

-1

1

-7

7

90 – 95

25

92.5

0

0

0

0

95 – 100

12

97.5

1

1

9

9

100 – 105

10

102.5

2

4

12

24

105 – 110

4

107.5

3

9

18

54

110 – 115

5

112.5

4

16

12

48

N = 60

6

254

Mean,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 33

Where, A = 92.5, h = 5

So, x̅ = 92.5 + ((6/60) × 5)

= 92.5 + ½

= 92.5 + 0.5

= 93

Then, Variance,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 34

σ2 = (52/602) [60(254) – 62]

= (1/144) [15240 – 36]

= 15204/144

= 1267/12

= 105.583

Hence, standard deviation = σ = √105.583

= 10.275

∴ Mean = 93, variance = 105.583 and Standard Deviation = 10.275

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.]

Solution:-

Let the assumed mean, A = 42.5 and h = 4

Let us make the table of the given data and append other columns after calculations.

Height (class)

Number of children

(Frequency fi)

Midpoint

Xi

Yi = (xi – A)/h

Yi2

fiyi

fiyi2

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

N = 100

25

199

Mean,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 35

Where, A = 42.5, h = 4

So, x̅ = 42.5 + (25/100) × 4

= 42.5 + 1

= 43.5

Then, Variance,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 34

σ2 = (42/1002)[100(199) – 252]

= (1/625) [19900 – 625]

= 19275/625

= 771/25

= 30.84

Hence, standard deviation = σ = √30.84

= 5.553

∴ Mean = 43.5, variance = 30.84 and Standard Deviation = 5.553.


Exercise 15.3 Page: 375

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

Solution:-

For comparing the variability or dispersion of two series, we calculate the coefficient of variance for each series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other.

Co-efficient of variation (C.V.) = (σ/ x̅) × 100

Where, σ = standard deviation, x̅ = mean

For Group A

Marks

Group A

fi

Mid-point

Xi

Yi = (xi – A)/h

(Yi)2

fiyi

fi(yi)2

10 – 20

9

15

((15 – 45)/10) = -3

(-3)2

= 9

– 27

81

20 – 30

17

25

((25 – 45)/10) = -2

(-2)2

= 4

– 34

68

30 – 40

32

35

((35 – 45)/10) = – 1

(-1)2

= 1

– 32

32

40 – 50

33

45

((45 – 45)/10) = 0

02

0

0

50 – 60

40

55

((55 – 45)/10) = 1

12

= 1

40

40

60 – 70

10

65

((65 – 45)/10) = 2

22

= 4

20

40

70 – 80

9

75

((75 – 45)/10) = 3

32

= 9

27

81

Total

150

-6

342

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 37

Where A = 45,

and yi = (xi – A)/h

Here h = class size = 20 – 10

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 38

σ2 = (102/1502) [150(342) – (-6)2]

= (100/22500) [51,300 – 36]

= (100/22500) × 51264

= 227.84

Hence, standard deviation = σ = √227.84

= 15.09

∴ C.V for group A = (σ/ x̅) × 100

= (15.09/44.6) × 100

= 33.83

Now, for group B.

Marks

Group B

fi

Mid-point

Xi

Yi = (xi – A)/h

(Yi)2

fiyi

fi(yi)2

10 – 20

10

15

((15 – 45)/10) = -3

(-3)2

= 9

– 30

90

20 – 30

20

25

((25 – 45)/10) = -2

(-2)2

= 4

– 40

80

30 – 40

30

35

((35 – 45)/10) = – 1

(-1)2

= 1

– 30

30

40 – 50

25

45

((45 – 45)/10) = 0

02

0

0

50 – 60

43

55

((55 – 45)/10) = 1

12

= 1

43

43

60 – 70

15

65

((65 – 45)/10) = 2

22

= 4

30

160

70 – 80

7

75

((75 – 45)/10) = 3

32

= 9

21

189

Total

150

-6

592

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 39

Where A = 45,

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 40

σ2 = (102/1502) [150(592) – (-6)2]

= (100/22500) [88,800 – 36]

= (100/22500) × 88,764

= 394.50

Hence, standard deviation = σ = √394.50

= 19.86

∴ C.V for group B = (σ/ x̅) × 100

= (19.86/44.6) × 100

= 44.53

By comparing C.V. of group A and group B.

C.V of Group B > C.V. of Group A

So, Group B is more variable.

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

Solution:-

From the given data,

Let us make the table of the given data and append other columns after calculations.

X (xi)

Y (yi)

Xi2

Yi2

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

Total = 510

1050

26360

110290

We have to calculate Mean for x,

Mean x̅ = ∑xi/n

Where, n = number of terms

= 510/10

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 41= 51

Then, Variance for x =

= (1/102)[(10 × 26360) – 5102]

= (1/100) (263600 – 260100)

= 3500/100

= 35

WKT Standard deviation = √variance

= √35

= 5.91

So, co-efficient of variation = (σ/ x̅) × 100

= (5.91/51) × 100

= 11.58

Now, we have to calculate Mean for y,

Mean ȳ = ∑yi/n

Where, n = number of terms

= 1050/10

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 42= 105

Then, Variance for y =

= (1/102)[(10 × 110290) – 10502]

= (1/100) (1102900 – 1102500)

= 400/100

= 4

WKT Standard deviation = √variance

= √4

= 2

So, co-efficient of variation = (σ/ x̅) × 100

= (2/105) × 100

= 1.904

By comparing C.V. of X and Y.

C.V of X > C.V. of Y

So, Y is more stable than X.

3. 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 wages earners

586

648

Mean of monthly wages

Rs 5253

Rs 5253

Variance of the distribution of wages

100

121

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

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

Solution:-

From the given table,

Mean monthly wages of firm A = Rs 5253

and Number of wage earners = 586

Then,

Total amount paid = 586 × 5253

= Rs 3078258

Mean monthly wages of firm B = Rs 5253

Number of wage earners = 648

Then,

Total amount paid = 648 × 5253

= Rs 34,03,944

(i) So, firm B pays larger amount as monthly wages.

(ii) Variance of firm A = 100

We know that, standard deviation (σ)= √100

=10

Variance of firm B = 121

Then,

Standard deviation (σ)=√(121 )

=11

Hence the standard deviation is more in case of Firm B that means in firm B there is greater variability in individual wages.

4. The following is the record of goals scored by team A in a football session:

No. of goals scored

0

1

2

3

4

No. of matches

1

9

7

5

3

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?

Solution:-

From the given data,

Let us make the table of the given data and append other columns after calculations.

Number of goals scored xi

Number of matches fi

fixi

Xi2

fixi2

0

1

0

0

0

1

9

9

1

9

2

7

14

4

28

3

5

15

9

45

4

3

12

16

48

Total

25

50

130

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 43

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 44

Since C.V. of firm B is greater

∴ Team A is more consistent.

5. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 45

Which is more varying, the length or weight?

Solution:-

First we have to calculate Mean for Length x,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 46

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 47

Miscellaneous Exercise Page: 380

1. The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Solution:-

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 48

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 49

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 50

2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.

Solution:-

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 51

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 52

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 53

Hence, from equation (i) and (v) we have:

When x – y = 2 then x = 8 and y = 6

And, when x – y = – 2 then x = 6 and y = 8

∴ the remaining observations are 6 and 8

3. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Solution:-

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 54

We know that,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 55

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 56

4. Given that x̅ is the mean and σ2 is the variance of n observations x1, x2, …,xn . Prove that the mean and variance of the observations ax1, ax2, ax3, …., axn are ax̅ and a2σ2, respectively, (a ≠ 0).

Solution:-

From the question it is given that, n observations are x1, x2,…..xn

Mean of the n observation = x̅

Variance of the n observation = σ2

As we know that,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 57

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 58

5. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. (ii) If it is replaced by 12

Solution:-

(i) If wrong item is omitted,

From the question it is given that,

The number of observations i.e. n = 20

The incorrect mean = 20

The incorrect standard deviation = 2

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 59

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 61
NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 60

(ii) If it is replaced by 12,

From the question it is given that,

The number of incorrect sum observations i.e. n = 200

The correct sum of observations n = 200 – 8 + 12

n = 204

Then, correct mean = correct sum/20

= 204/20

= 10.2

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 62

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 63

6. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

Subject

Mathematics

Physics

Chemistry

Mean

42

32

40.9

Standard deviation

12

15

20

Which of the three subjects shows the highest variability in marks and which shows the lowest?

Solution:-

From the question it is given that,

Mean of Mathematics = 42

Standard deviation of Mathematics = 12

Mean of Physics = 32

Standard deviation of physics = 15

Mean of Chemistry = 40.9

Standard deviation of chemistry = 20

As we know that,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 64

7. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Solution:-

From the question it is given that,

The total number of observations (n) = 100

Incorrect mean, (x̅) = 20

And, Incorrect standard deviation (σ) = 3

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 65

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 66

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 67


NCERT Solutions for Class 11 Maths Chapter 15 – Statistics

The topics of Class 11 Chapter 15 – Statistics of NCERT Solution are as follows:

15.1 Introduction

This section talks about the concept of central tendency, mean and median [during even and the odd number of observations] with examples. It introduces the concept of measure of dispersion.

The values which cluster around the middle or centre of the distribution are measures of central tendency. They are mean, median and mode.

In a class, mean can be used to find the average marks scored by the students.

When calculating the height of students, the median can be used to find the middlemost value.

15.2 Measures of Dispersion

This section defines measures of dispersion, different measures of dispersion [range, quartile deviation, mean deviation, standard deviation] 

Measures of dispersion explain the relationship with measures of central tendency. For example, the spread of data tells how well the mean represents the data. If the spread is large, then mean is not representative of data.

15.3 Range

This section defines the range, its formula and an example.

The range gives the variability of scores using the maximum and minimum values in the set.

In a cricket match, Batsman A range = 121 – 0 = 121 and Batsman B range = 52 – 22 = 30

Range A > Range B. So, the scores are spread in case of A whereas for B they are close to each other.

15.4 Mean Deviation

This section defines mean deviation, the formula for mean deviation.

The concept of mean deviation can be used by biologists in the comparison of different animal weights and decide what would be a healthy weight.

15.4.1Mean deviation for ungrouped data

The process of obtaining the mean deviation for ungrouped data is elaborated in this section.

Find the mean, deviations from the mean, absolute deviations and substitute the values in the mean deviation formula and arrive at the answer.

15.4.2 Mean deviation for grouped data

The process of obtaining mean deviation for discrete and continuous frequency distributions are elaborated in this section with solved examples.

15.4.3 Limitations of mean deviation

  • If in a series the degree of variability is very high, then median will not be a representative of data. Hence the mean deviation calculated about such median cannot be relied on.
  • If the sum of deviations from the mean is greater than the sum of deviations from the median, then the mean deviation about mean is not very specific.
  • The absolute mean deviation calculated can’t be subjected to further algebraic treatment. It can’t be used as an appropriate measure of dispersion.

15.5 Variance and standard deviation

15.5.1 Standard Deviation

This section involves the variance and standard deviation definition, formula and solved examples on the discrete and continuous frequency distribution. 

A science test was taken by a class of students. The mean score of the test was 85% on calculation. The teacher found the standard deviation of other scores and noticed that a very small standard deviation exists which suggests that most of the students scored very close to 85%.

15.5.2 Shortcut method to find Variance and standard deviation

This section deals with the simpler way of calculating the standard deviation with a few illustrations. 

15.6 Analysis of Frequency Distributions

This section deals with the process of comparing the variability of two series having the same mean, coefficient of variation with few solved problems.

A few points on Chapter 15 Statistics

  • Range is defined as the difference between the maximum and minimum value of the given data.
  • If there exists series with equal means, then the series with lesser standard deviation is more consistent or less scattered. 
  • The addition or subtraction of a positive number to each data point of the data set will not affect the variance.

The solutions give substitute strategies and clarifications to take care of problems which makes the student feel sure while taking the test. Additionally, taking care of many muddled issues upgrades the information on mathematical ability in the students. The solutions cater to all the vital inquiries which a student should and ought to have aced to show up for the test. The BYJU’S subject specialists who have written these solutions have complete knowledge about the question paper setting and the marks distributed across the chapters.

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