NCERT Solutions for Class 11 Maths Chapter 15 – Statistics

NCERT Solutions for Class 11 Maths Chapter 15 – Statistics

Page No 360:

Question 1:

Find the mean deviation about the mean for the data

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

ANSWER:

The given data is

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

Mean of the data, ncert solutions for class 11 maths chapter 15 statistics

The deviations of the respective observations from the mean 1 ncert solutions for class 11 maths chapter 15 statisticsare

–6, – 3, –2, –1, 0, 2, 3, 7

The absolute values of the deviations, i.e.2 ncert solutions for class 11 maths chapter 15 statistics, are

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

The required mean deviation about the mean is

3 ncert solutions for class 11 maths chapter 15 statistics

Page No 360:

Question 2:

Find the mean deviation about the mean for the data

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

ANSWER:

The given data is

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

Mean of the given data,

4 ncert solutions for class 11 maths chapter 15 statistics

The deviations of the respective observations from the mean 1 ncert solutions for class 11 maths chapter 15 statisticsare

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

The absolute values of the deviations, i.e. 2 ncert solutions for class 11 maths chapter 15 statistics, are

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

The required mean deviation about the mean is

5 ncert solutions for class 11 maths chapter 15 statistics

Page No 360:

Question 3:

Find the mean deviation about the median for the data.

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

ANSWER:

The given data is

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

Here, the numbers of observations are 12, which is even.

Arranging the data in ascending order, we obtain

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

6 ncert solutions for class 11 maths chapter 15 statistics

The deviations of the respective observations from the median, i.e.7 ncert solutions for class 11 maths chapter 15 statisticsare

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

The absolute values of the deviations,8 ncert solutions for class 11 maths chapter 15 statistics, 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

The required mean deviation about the median is

9 ncert solutions for class 11 maths chapter 15 statistics

Page No 360:

Question 4:

Find the mean deviation about the median for the data

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

ANSWER:

The given data is

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

Here, the number of observations is 10, which is even.

Arranging the data in ascending order, we obtain

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

10 ncert solutions for class 11 maths chapter 15 statistics

The deviations of the respective observations from the median, i.e.7 ncert solutions for class 11 maths chapter 15 statisticsare

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

The absolute values of the deviations,8 ncert solutions for class 11 maths chapter 15 statistics, are

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

Thus, the required mean deviation about the median is

11 ncert solutions for class 11 maths chapter 15 statistics

Page No 360:

Question 5:

Find the mean deviation about the mean for the data.

xi

5

10

15

20

25

fi

7

4

6

3

5

ANSWER:

xi

fi

fi xi

12 ncert solutions for class 11 maths chapter 15 statistics

13 ncert solutions for class 11 maths chapter 15 statistics

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

14 ncert solutions for class 11 maths chapter 15 statistics

15 ncert solutions for class 11 maths chapter 15 statistics

Page No 360:

Question 6:

Find the mean deviation about the mean for the data

xi

10

30

50

70

90

fi

4

24

28

16

8

ANSWER:

xi

fi

fi xi

12 ncert solutions for class 11 maths chapter 15 statistics

13 ncert solutions for class 11 maths chapter 15 statistics

10

4

40

40

160

30

24

720

20

480

50

28

1400

70

16

1120

20

320

90

8

720

40

320

80

4000

1280

16 ncert solutions for class 11 maths chapter 15 statistics

Page No 360:

Question 7:

Find the mean deviation about the median for the data.

xi

5

7

9

10

12

15

fi

8

6

2

2

2

6

ANSWER:

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

xi

fi

c.f.

5

8

8

7

6

14

9

2

16

10

2

18

12

2

20

15

6

26

Here, N = 26, which is even.

Median is the mean of 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

17 ncert solutions for class 11 maths chapter 15 statistics

The absolute values of the deviations from median, i.e.18 ncert solutions for class 11 maths chapter 15 statisticsare

|xi – M|

2

2

3

5

8

fi

8

6

2

2

2

6

fi |xi – M|

16

4

6

10

48

19 ncert solutions for class 11 maths chapter 15 statisticsand20 ncert solutions for class 11 maths chapter 15 statistics

21 ncert solutions for class 11 maths chapter 15 statistics

Page No 360:

Question 8:

Find the mean deviation about the median for the data

xi

15

21

27

30

35

fi

3

5

6

7

8

ANSWER:

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

xi fi

c.f.

15

3

3

21

5

8

27

6

14

30

7

21

35

8

29

Here, N = 29, which is odd.

22 ncert solutions for class 11 maths chapter 15 statisticsobservation = 15th observation

This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.

∴ Median = 30

The absolute values of the deviations from median, i.e.18 ncert solutions for class 11 maths chapter 15 statisticsare

  |xi – M|

15

9

3

5

 fi

3

5

6

7

8

 fi |xi – M|

45

45

18

40

23 ncert solutions for class 11 maths chapter 15 statistics

∴ 24 ncert solutions for class 11 maths chapter 15 statistics25 ncert solutions for class 11 maths chapter 15 statistics

Page No 361:

Question 9:

Find the mean deviation about the mean for the data.

Income per day

Number of persons

0-100

4

100-200

8

200-300

9

300-400

10

400-500

7

500-600

5

600-700

4

700-800

3

ANSWER:

The following table is formed.

Income per day

Number of persons fi

Mid-point xi

fi xi

12 ncert solutions for class 11 maths chapter 15 statistics

13 ncert solutions for class 11 maths chapter 15 statistics

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

50

17900

7896

Here,26 ncert solutions for class 11 maths chapter 15 statistics

27 ncert solutions for class 11 maths chapter 15 statistics

28 ncert solutions for class 11 maths chapter 15 statistics

Page No 361:

Question 10:

Find the mean deviation about the mean for the data

Height in cms

Number of boys

95-105

9

105-115

13

115-125

26

125-135

30

135-145

12

145-155

10

Page No 361:

Question 11:

Find the mean deviation about median for the following data:

Marks

Number of girls

0-10

6

10-20

8

20-30

14

30-40

16

40-50

4

50-60

2

ANSWER:

The following table is formed.

Marks

Number of girls fi

Cumulative frequency (c.f.)

Mid-point 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 the32 ncert solutions for class 11 maths chapter 15 statisticsor 25th item is 20 – 30.

Therefore, 20 – 30 is the median class.

It is known that,

33 ncert solutions for class 11 maths chapter 15 statistics

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

∴ Median =34 ncert solutions for class 11 maths chapter 15 statistics35 ncert solutions for class 11 maths chapter 15 statistics

Thus, mean deviation about the median is given by,

36 ncert solutions for class 11 maths chapter 15 statistics

Page No 361:

Question 12:

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

Age

Number

16-20

5

21-25

6

26-30

12

31-35

14

36-40

26

41-45

12

46-50

16

51-55

9

ANSWER:

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows.

Age

Number fi

Cumulative frequency (c.f.)

Mid-point 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

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 the37 ncert solutions for class 11 maths chapter 15 statisticsor 50th item is 35.5 – 40.5.

Therefore, 35.5 – 40.5 is the median class.

It is known that,

33 ncert solutions for class 11 maths chapter 15 statistics

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

38 ncert solutions for class 11 maths chapter 15 statistics 39 ncert solutions for class 11 maths chapter 15 statistics

Thus, mean deviation about the median is given by,

40 ncert solutions for class 11 maths chapter 15 statistics

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, 41 ncert solutions for class 11 maths chapter 15 statistics

The following table is obtained.

xi

42 ncert solutions for class 11 maths chapter 15 statistics

43 ncert solutions for class 11 maths chapter 15 statistics

6

–3

9

7

–2

4

10

–1

1

12

3

9

13

4

16

4

–5

25

8

–1

1

12

3

9

74

44 ncert solutions for class 11 maths chapter 15 statistics

Page No 371:

Question 2:

Find the mean and variance for the first natural numbers

ANSWER:

The mean of first n natural numbers is calculated as follows.

45 ncert solutions for class 11 maths chapter 15 statistics

46 ncert solutions for class 11 maths chapter 15 statistics

Page No 371:

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, = 10

47 ncert solutions for class 11 maths chapter 15 statistics

The following table is obtained.

xi

42 ncert solutions for class 11 maths chapter 15 statistics

43 ncert solutions for class 11 maths chapter 15 statistics

3

–13.5

182.25

6

–10.5

110.25

9

–7.5

56.25

12

–4.5

20.25

15

–1.5

2.25

18

1.5

2.25

21

4.5

20.25

24

7.5

56.25

27

10.5

110.25

30

13.5

182.25

742.5

48 ncert solutions for class 11 maths chapter 15 statistics

Page No 371:

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.

xi

f i

fixi

49 ncert solutions for class 11 maths chapter 15 statistics

43 ncert solutions for class 11 maths chapter 15 statistics

50 ncert solutions for class 11 maths chapter 15 statistics

6

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

11

121

363

40

760

1736

Here, N = 40, 51 ncert solutions for class 11 maths chapter 15 statistics

52 ncert solutions for class 11 maths chapter 15 statistics

53 ncert solutions for class 11 maths chapter 15 statistics

Page No 371:

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.

xi

f i

fixi

49 ncert solutions for class 11 maths chapter 15 statistics

43 ncert solutions for class 11 maths chapter 15 statistics

50 ncert solutions for class 11 maths chapter 15 statistics

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

22

2200

640

Here, N = 22, 54 ncert solutions for class 11 maths chapter 15 statistics

55 ncert solutions for class 11 maths chapter 15 statistics

56 ncert solutions for class 11 maths chapter 15 statistics

Page No 371:

Question 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

ANSWER:

The data is obtained in tabular form as follows.

xi

fi

57 ncert solutions for class 11 maths chapter 15 statistics

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

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

286

Mean, 58 ncert solutions for class 11 maths chapter 15 statistics59 ncert solutions for class 11 maths chapter 15 statistics

60 ncert solutions for class 11 maths chapter 15 statistics

Page No 371:

Question 7:

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

ANSWER:

Class

Frequency fi

Mid-point xi

61 ncert solutions for class 11 maths chapter 15 statistics

yi2

fiyi

fiyi2

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

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, 62 ncert solutions for class 11 maths chapter 15 statistics63 ncert solutions for class 11 maths chapter 15 statistics

64 ncert solutions for class 11 maths chapter 15 statistics

Page No 372:

Question 8:

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

ANSWER:

Class

Frequency

fi

Mid-point xi

65 ncert solutions for class 11 maths chapter 15 statistics

yi2

fiyi

fiyi2

0-10

5

5

–2

4

–10

20

10-20

8

15

–1

1

–8

8

20-30

15

25

30-40

16

35

1

1

16

16

40-50

6

45

2

4

12

24

50

10

68

Mean, 66 ncert solutions for class 11 maths chapter 15 statistics67 ncert solutions for class 11 maths chapter 15 statistics

68 ncert solutions for class 11 maths chapter 15 statistics

Page No 372:

Question 9:

Find the mean, variance and standard deviation using short-cut method

Height

in cms

No. of children

70-75

3

75-80

4

80-85

7

85-90

7

90-95

15

95-100

9

100-105

6

105-110

6

110-115

3

ANSWER:

Class Interval

Frequency fi

Mid-point xi

69 ncert solutions for class 11 maths chapter 15 statistics

yi2

fiyi

fiyi2

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

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, 70 ncert solutions for class 11 maths chapter 15 statistics71 ncert solutions for class 11 maths chapter 15 statistics

72 ncert solutions for class 11 maths chapter 15 statistics

Page No 372:

Question 10:

The diameters of circles (in mm) drawn in a design are given below:

Diameters

No. of children

33-36

15

37-40

17

41-44

21

45-48

22

49-52

25

ANSWER:

Class Interval

Frequency fi

Mid-point xi

73 ncert solutions for class 11 maths chapter 15 statistics

fi2

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

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

Here, N = 100, h = 4

Let the assumed mean, A, be 42.5.

Mean, 66 ncert solutions for class 11 maths chapter 15 statistics74 ncert solutions for class 11 maths chapter 15 statistics

75 ncert solutions for class 11 maths chapter 15 statistics

Page No 375:

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:

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

Marks

Group A fi

Mid-point xi

76 ncert solutions for class 11 maths chapter 15 statistics

yi2

fiyi

fiyi2

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

50-60

40

55

1

1

40

40

60-70

10

65

2

4

20

40

70-80

9

75

3

9

27

81

150

–6

342

Here, = 10, N = 150, A = 45

77 ncert solutions for class 11 maths chapter 15 statistics

The standard deviation of group B is calculated as follows.

Marks

Group B

fi

Mid-point

xi

78 ncert solutions for class 11 maths chapter 15 statistics

yi2

fiyi

fiyi2

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

50-60

43

55

1

1

43

43

60-70

15

65

2

4

30

60

70-80

7

75

3

9

21

63

150

–6

366

79 ncert solutions for class 11 maths chapter 15 statistics

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.

Page No 375:

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:

The prices of the shares X are

35, 54, 52, 53, 56, 58, 52, 50, 51, 49

Here, the number of observations, N = 10

80 ncert solutions for class 11 maths chapter 15 statistics

The following table is obtained corresponding to shares X.

xi

42 ncert solutions for class 11 maths chapter 15 statistics

43 ncert solutions for class 11 maths chapter 15 statistics

35

–16

256

54

3

9

52

1

1

53

2

4

56

5

25

58

7

49

52

1

1

50

–1

1

51

49

–2

4

350

81 ncert solutions for class 11 maths chapter 15 statistics

The prices of share Y are

108, 107, 105, 105, 106, 107, 104, 103, 104, 101

82 ncert solutions for class 11 maths chapter 15 statistics

The following table is obtained corresponding to shares Y.

yi

83 ncert solutions for class 11 maths chapter 15 statistics

84 ncert solutions for class 11 maths chapter 15 statistics

108

3

9

107

2

4

105

105

106

1

1

107

2

4

104

–1

1

103

–2

4

104

–1

1

101

–4

16

40

ncert solutions for class 11 maths chapter 15 statistics
C.V. of prices of shares X is greater than the C.V. of prices of shares Y.

Thus, the prices of shares Y are more stable than the prices of shares X.

Page No 375:

Question 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 wage 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?

ANSWER:

(i) Monthly wages of firm A = Rs 5253

Number of wage earners in firm A = 586

∴Total amount paid = Rs 5253 × 586

Monthly wages of firm B = Rs 5253

Number of wage earners in firm B = 648

∴Total amount paid = Rs 5253 × 648

Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.

(ii) Variance of the distribution of wages in firm A85 ncert solutions for class 11 maths chapter 15 statistics = 100

∴ Standard deviation of the distribution of wages in firm

A ((σ1) = 86 ncert solutions for class 11 maths chapter 15 statistics

Variance of the distribution of wages in firm 87 ncert solutions for class 11 maths chapter 15 statistics= 121

∴ Standard deviation of the distribution of wages in firm 88 ncert solutions for class 11 maths chapter 15 statistics

The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater standard deviation will have more variability.

Thus, firm B has greater variability in the individual wages.

Page No 376:

Question 4:

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

No. of goals scored

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?

ANSWER:

The mean and the standard deviation of goals scored by team A are calculated as follows.

No. of goals scored

No. of matches

fixi

xi2

fixi2

1

1

9

9

1

9

2

7

14

4

28

3

5

15

9

45

4

3

12

16

48

25

50

130

89 ncert solutions for class 11 maths chapter 15 statistics

Thus, the mean of both the teams is same.

90 ncert solutions for class 11 maths chapter 15 statistics

The standard deviation of team B is 1.25 goals.

The average number of goals scored by both the teams is same i.e., 2. Therefore, the team with lower standard deviation will be more consistent.

Thus, team A is more consistent than team B.

Page No 376:

Question 5:

The sum and sum of squares corresponding to length (in cm) and weight y

(in gm) of 50 plant products are given below:

91 ncert solutions for class 11 maths chapter 15 statistics

Which is more varying, the length or weight?

ANSWER:

92 ncert solutions for class 11 maths chapter 15 statistics

Here, N = 50

∴ Mean, 93 ncert solutions for class 11 maths chapter 15 statistics

94 ncert solutions for class 11 maths chapter 15 statistics

95 ncert solutions for class 11 maths chapter 15 statistics

96 ncert solutions for class 11 maths chapter 15 statistics

97 ncert solutions for class 11 maths chapter 15 statistics

Mean, 98 ncert solutions for class 11 maths chapter 15 statistics

99 ncert solutions for class 11 maths chapter 15 statistics

Thus, C.V. of weights is greater than the C.V. of lengths. Therefore, weights vary more than the lengths.

Page No 380:

Question 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.

ANSWER:

Let the remaining two observations be x and y.

Therefore, the observations are 6, 7, 10, 12, 12, 13, xy.

100 ncert solutions for class 11 maths chapter 15 statistics

From (1), we obtain

x2 + y2 + 2xy = 144 …(3)

From (2) and (3), we obtain

2xy = 64 … (4)

Subtracting (4) from (2), we obtain

x2 + y– 2xy = 80 – 64 = 16

⇒ x – y = ± 4 … (5)

Therefore, from (1) and (5), we obtain

x = 8 and y = 4, when x – y = 4

x = 4 and y = 8, when x – y = –4

Thus, the remaining observations are 4 and 8.

Page No 380:

Question 2:

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

ANSWER:

Let the remaining two observations be x and y.

The observations are 2, 4, 10, 12, 14, xy.

101 ncert solutions for class 11 maths chapter 15 statistics

From (1), we obtain

x2 + y2 + 2xy = 196 … (3)

From (2) and (3), we obtain

2xy = 196 – 100

⇒ 2xy = 96 … (4)

Subtracting (4) from (2), we obtain

x2 + y– 2xy = 100 – 96

⇒ (x – y)2 = 4

⇒ x – y = ± 2 … (5)

Therefore, from (1) and (5), we obtain

x = 8 and y = 6 when x – y = 2

x = 6 and y = 8 when x – y = – 2

Thus, the remaining observations are 6 and 8.

Page No 380:

Question 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.

ANSWER:

Let the observations be x1x2x3x4x5, and x6.

It is given that mean is 8 and standard deviation is 4.

102 ncert solutions for class 11 maths chapter 15 statistics

If each observation is multiplied by 3 and the resulting observations are yi, then

103 ncert solutions for class 11 maths chapter 15 statistics

From (1) and (2), it can be observed that,

104 ncert solutions for class 11 maths chapter 15 statistics

Substituting the values of xi and 105 ncert solutions for class 11 maths chapter 15 statistics in (2), we obtain

106 ncert solutions for class 11 maths chapter 15 statistics

Therefore, variance of new observations = 107 ncert solutions for class 11 maths chapter 15 statistics

Hence, the standard deviation of new observations is 108 ncert solutions for class 11 maths chapter 15 statistics

Page No 380:

Question 4:

Given that 105 ncert solutions for class 11 maths chapter 15 statisticsis the mean and σ2 is the variance of observations x1x2 … xn. Prove that the mean and variance of the observations ax1ax2ax3 …axare 109 ncert solutions for class 11 maths chapter 15 statistics and a2 σ2, respectively (a ≠ 0).

ANSWER:

The given n observations are x1x2 … xn.

Mean = 105 ncert solutions for class 11 maths chapter 15 statistics

Variance = σ2

110 ncert solutions for class 11 maths chapter 15 statistics

If each observation is multiplied by a and the new observations are yi, then

111 ncert solutions for class 11 maths chapter 15 statistics

Therefore, mean of the observations, ax1ax2 … axn, is 112 ncert solutions for class 11 maths chapter 15 statistics.

Substituting the values of xiand 105 ncert solutions for class 11 maths chapter 15 statisticsin (1), we obtain

113 ncert solutions for class 11 maths chapter 15 statistics

Thus, the variance of the observations, ax1ax2 … axn, is a2 σ2.

Page No 380:

Question 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.

ANSWER:

(i) Number of observations (n) = 20

Incorrect mean = 10

Incorrect standard deviation = 2

114 ncert solutions for class 11 maths chapter 15 statistics

That is, incorrect sum of observations = 200

Correct sum of observations = 200 – 8 = 192

∴ Correct mean 115 ncert solutions for class 11 maths chapter 15 statistics 116 ncert solutions for class 11 maths chapter 15 statistics

Standard deviation, σ = 1nni=1xi2  1n2(ni=1xi)2−−−−−−−−−−−−−−−−−−−−−−√

2 = 1nni=1xi2  (1nni=1xi)2−−−−−−−−−−−−−−−−−−−−−−√

2 = 1nni=1xi2  (x¯)2−−−−−−−−−−−−−−−√[as, 1nni=1x = x ]

2 = 120×Incorrectni=1xi2  (10)2−−−−−−−−−−−−−−−−−−−−−−−−−−√

4 = 120×Incorrectni=1xi2  100120×Incorrectni=1xi2 = 104

Incorrectni=1xi2 = 2080Now, correctni=1xi2 = Incorrectni=1xi2  (8)2

correctni=1xi2 = 2080  64 = 2016

 correct Standard Deviation = 1ncorrectni=1xi2  (correct mean)2−−−−−−√

correct Standard Deviation = 119×2016  (19219)2−−−−−−−−−−−−−−−−√

correct Standard Deviation = 201619(19219)2−−−−−−−−−−−√

correct Standard Deviation = 144019 = 121019

correct Standard Deviation = 12 × 3.16219 = 1.997Standard deviation, σ = 1n∑i=1nxi2 – 1n2∑i=1nxi2

⇒2 = 1n∑i=1nxi2 – 1n∑i=1nxi2⇒2 = 1n∑i=1nxi2 – x¯2    as, 1n∑i=1nx = x ⇒2 = 120×Incorrect∑i=1nxi2 – 102

⇒4 = 120×Incorrect∑i=1nxi2 – 100⇒120×Incorrect∑i=1nxi2 = 104

⇒Incorrect∑i=1nxi2 = 2080Now, correct∑i=1nxi2 = Incorrect∑i=1nxi2 – 82

⇒correct∑i=1nxi2 = 2080 – 64 = 2016∴ correct Standard Deviation = 1ncorrect∑i=1nxi2 – correct mean2

⇒correct Standard Deviation = 119×2016 – 192192

⇒correct Standard Deviation = 201619-192192

⇒correct Standard Deviation = 144019 = 121019

⇒correct Standard Deviation = 12 × 3.16219 = 1.997

(ii) When 8 is replaced by 12,

Incorrect sum of observations = 200

∴ Correct sum of observations = 200 – 8 + 12 = 204

117 ncert solutions for class 11 maths chapter 15 statistics

Page No 380:

Question 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?

ANSWER:

Standard deviation of Mathematics = 12

Standard deviation of Physics = 15

Standard deviation of Chemistry = 20

The coefficient of variation (C.V.) is given by 118 ncert solutions for class 11 maths chapter 15 statistics.

119 ncert solutions for class 11 maths chapter 15 statistics

The subject with greater C.V. is more variable than others.

Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in Mathematics.

Page No 380:

Question 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.

ANSWER:

Number of observations (n) = 100

Incorrect mean 120 ncert solutions for class 11 maths chapter 15 statistics

Incorrect standard deviation 121 ncert solutions for class 11 maths chapter 15 statistics

122 ncert solutions for class 11 maths chapter 15 statistics

∴ Incorrect sum of observations = 2000

⇒ Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940

123 ncert solutions for class 11 maths chapter 15 statistics

124 ncert solutions for class 11 maths chapter 15 statistics

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