NCERT Solutions for Class 11 Maths Chapter 15 – Statistics Ex 15.2
Page No 371:
Question 1:
Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12
Answer:
6, 7, 10, 12, 13, 4, 8, 12
Mean,
The following table is obtained.
 x_{i}6–397–2410–111239134164–5258–11123974
Question 2:
Find the mean and variance for the first n natural numbers
Answer:
The mean of first n natural numbers is calculated as follows.
Question 3:
Find the mean and variance for the first 10 multiples of 3
Answer:
The first 10 multiples of 3 are
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Here, number of observations, n = 10
The following table is obtained.
 x_{i}3–13.5182.256–10.5110.259–7.556.2512–4.520.2515–1.52.25181.52.25214.520.25247.556.252710.5110.253013.5182.25742.5
Question 4:
Find the mean and variance for the data
xi  6  10  14  18  24  28  30 
f i  2  4  7  12  8  4  3 
Answer:
The data is obtained in tabular form as follows.
 x_{i}f if_{i}x_{i}6212–1316933810440–98132414798–5251751812216–11122481925252002841129813243039011121363407601736
Here, N = 40,
Question 5:
Find the mean and variance for the data
xi  92  93  97  98  102  104  109 
f i  3  2  3  2  6  3  3 
Answer:
The data is obtained in tabular form as follows.
 x_{i}f if_{i}x_{i}923276–864192932186–74998973291–3927982196–248102661224241043312416481093327981243222200640
Here, N = 22,
Question 6:
Find the mean and standard deviation using shortcut method.
x_{i}  60  61  62  63  64  65  66  67  68 
f_{i}  2  1  12  29  25  12  10  4  5 
Answer:
The data is obtained in tabular form as follows.
 x_{i}f_{i}y_{i}^{2}f_{i}y_{i}f_{i}y_{i}^{2}602–416–832611–39–396212–24–24486329–11–2929642500006512111212661024204067439123668541620801002200286
Mean,
Question 7:
Find the mean and variance for the following frequency distribution.
Classes

030

3060

6090

90120

120150

150180

180210

Frequencies

2

3

5

10

3

5

2

Answer:
 ClassFrequency f_{i}Midpoint x_{i}y_{i}^{2}f_{i}y_{i}f_{i}y_{i}^{2}030215–39–6183060345–24–6126090575–11–559012010105000012015031351133150180516524102018021021953961830276
Mean,
Page No 372:
Question 8:
Find the mean and variance for the following frequency distribution.
Classes  010  1020  2030  3040  4050 
Frequencies  5  8  15  16  6 
Answer:
Class

Frequency
f_{i}

Midpoint x_{i}

y_{i}^{2}

f_{i}y_{i}

f_{i}y_{i}^{2}
 
010

5

5

–2

4

–10

20

1020

8

15

–1

1

–8

8

2030

15

25

0

0

0

0

3040

16

35

1

1

16

16

4050

6

45

2

4

12

24

50

10

68

Mean,
Question 9:
Find the mean, variance and standard deviation using shortcut method
Height
in cms

No. of children

7075

3

7580

4

8085

7

8590

7

9095

15

95100

9

100105

6

105110

6

110115

3

Answer:
Class Interval

Frequency f_{i}

Midpoint x_{i}

y_{i}^{2}

f_{i}y_{i}

f_{i}y_{i}^{2}
 
7075

3

72.5

–4

16

–12

48

7580

4

77.5

–3

9

–12

36

8085

7

82.5

–2

4

–14

28

8590

7

87.5

–1

1

–7

7

9095

15

92.5

0

0

0

0

95100

9

97.5

1

1

9

9

100105

6

102.5

2

4

12

24

105110

6

107.5

3

9

18

54

110115

3

112.5

4

16

12

48

60

6

254

Mean,
Question 10:
The diameters of circles (in mm) drawn in a design are given below:
Diameters 
No. of children

3336

15

3740

17

4144

21

4548

22

4952

25

Answer:
Class Interval

Frequency f_{i}

Midpoint x_{i}

f_{i}^{2}

f_{i}y_{i}

f_{i}y_{i}^{2}
 
32.536.5

15

34.5

–2

4

–30

60

36.540.5

17

38.5

–1

1

–17

17

40.544.5

21

42.5

0

0

0

0

44.548.5

22

46.5

1

1

22

22

48.552.5

25

50.5

2

4

50

100

100

25

199

Here, N = 100, h = 4
Let the assumed mean, A, be 42.5.
Mean,
Page No 375:
Question 1:
From the data given below state which group is more variable, A or B?
Marks

1020

2030

3040

4050

5060

6070

7080

Group A

9

17

32

33

40

10

9

Group B

10

20

30

25

43

15

7

Answer:
Firstly, the standard deviation of group A is calculated as follows.
Marks

Group A f_{i}

Midpoint x_{i}

y_{i}^{2}

f_{i}y_{i}

f_{i}y_{i}^{2}
 
1020

9

15

–3

9

–27

81

2030

17

25

–2

4

–34

68

3040

32

35

–1

1

–32

32

4050

33

45

0

0

0

0

5060

40

55

1

1

40

40

6070

10

65

2

4

20

40

7080

9

75

3

9

27

81

150

–6

342

Here, h = 10, N = 150, A = 45
The standard deviation of group B is calculated as follows.
Marks

Group B
f_{i}

Midpoint
x_{i}

y_{i}^{2}

f_{i}y_{i}

f_{i}y_{i}^{2}
 
1020

10

15

–3

9

–30

90

2030

20

25

–2

4

–40

80

3040

30

35

–1

1

–30

30

4050

25

45

0

0

0

0

5060

43

55

1

1

43

43

6070

15

65

2

4

30

60

7080

7

75

3

9

21

63

150

–6

366

Since the mean of both the groups is same, the group with greater standard deviation will be more variable.
Thus, group B has more variability in the marks.