NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.2

NCERT Solutions for Class 11 Maths Chapter 15 Statistics, contains solutions for all exercise 15.2 questions. By practising these questions, students can build their confidence level. NCERT Solutions also help the students to excel their skills in statistics. Download the NCERT Solutions of Class 11 maths now and start practising.

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Access other exercise solutions of Class 11 Maths Chapter 15 Statistics

Exercise 15.1 Solutions : 12 Questions

Exercise 15.3 Solutions : 5 Questions

Miscellaneous Exercise Solutions: 7 Questions

Access Solutions for Class 11 Maths Chapter 15.2 Exercise

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.

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