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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,

The deviations of the respective observations from the mean are

–6, – 3, –2, –1, 0, 2, 3, 7

The absolute values of the deviations, i.e., are

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

The required mean deviation about the mean is

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,

The deviations of the respective observations from the mean are

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

The absolute values of the deviations, i.e. , are

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

The required mean deviation about the mean is

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

The deviations of the respective observations from the median, i.e.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 absolute values of the deviations,, 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

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

The deviations of the respective observations from the median, i.e.are

–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,, 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

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

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

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

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

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.

The absolute values of the deviations from median, i.e.are

|xiM|

2

0

2

3

5

8

fi

8

6

2

2

2

6

 

fi |xiM|

16

0

4

6

10

48

and

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.

observation = 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.are

  |xiM|

15

9

3

0

5

 fi

3

5

6

7

8

 fi |xiM|

45

45

18

0

40



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

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,

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

Answer:

The following table is formed.

Height in cms

Number of boys fi

Mid-point xi

fi xi

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

Here,

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

|xiMed.|
 

fi |xiMed.|

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 theor 25th item is 20 – 30.

Therefore, 20 – 30 is the median class.

It is known that,

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

∴ Median =

Thus, mean deviation about the median is given by,

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

|xiMed.|

fi |xiMed.|

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 theor 50th item is 35.5 – 40.5.

Therefore, 35.5 – 40.5 is the median class.

It is known that,

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

Thus, mean deviation about the median is given by,



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.

xi

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

Page No 371:

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.

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, n = 10

The following table is obtained.

xi

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

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

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,

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

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,

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

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

100

220

0

286

Mean,

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

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

0

0

0

0

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,



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

yi2

fiyi

fiyi2

0-10

5

5

–2

4

–10

20

10-20

8

15

–1

1

–8

8

20-30

15

25

0

0

0

0

30-40

16

35

1

1

16

16

40-50

6

45

2

4

12

24

50

10

68

Mean,

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

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

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

60

6

254

Mean,

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

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

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

 

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

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

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

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

150

–6

342

Here, h = 10, N = 150, A = 45

The standard deviation of group B is calculated as follows.

Marks

Group B

fi

Mid-point

xi

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

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

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.

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

The following table is obtained corresponding to shares X.

xi

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

0

0

49

–2

4

   

350

The prices of share Y are

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

The following table is obtained corresponding to shares Y.

yi

108

3

9

107

2

4

105

0

0

105

0

0

106

1

1

107

2

4

104

–1

1

103

–2

4

104

–1

1

101

–4

16

   

40


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 A = 100

∴ Standard deviation of the distribution of wages in firm

A ((σ1) =

Variance of the distribution of wages in firm = 121

∴ Standard deviation of the distribution of wages in firm

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

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?

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

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

25

50

130

Thus, the mean of both the teams is same.

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 x (in cm) and weight y

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

Which is more varying, the length or weight?

Answer:

Here, N = 50

∴ Mean,

Mean,

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, x, y.

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 + y2 – 2xy = 80 – 64 = 16

xy = ± 4 … (5)

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

x = 8 and y = 4, when xy = 4

x = 4 and y = 8, when xy = –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, x, y.

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 + y2 – 2xy = 100 – 96

⇒ (xy)2 = 4

xy = ± 2 … (5)

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

x = 8 and y = 6 when xy = 2

x = 6 and y = 8 when xy = – 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 x1, x2, x3, x4, x5, and x6.

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

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

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

Substituting the values of xi and in (2), we obtain

Therefore, variance of new observations =

Hence, the standard deviation of new observations is

Page No 380:

Question 4:

Given that is the mean and σ2 is the variance of n observations x1, x2xn. Prove that the mean and variance of the observations ax1, ax2, ax3axn are and a2 σ2, respectively (a ≠ 0).

Answer:

The given n observations are x1, x2xn.

Mean =

Variance = σ2

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

Therefore, mean of the observations, ax1, ax2axn, is .

Substituting the values of xiand in (1), we obtain

Thus, the variance of the observations, ax1, ax2axn, is a2 σ2.

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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

That is, incorrect sum of observations = 200

Correct sum of observations = 200 – 8 = 192

∴ Correct mean


Standard deviation, σ = 1ni=1nxi2 - 1n2i=1nxi22 = 1ni=1nxi2 - 1ni=1nxi22 = 1ni=1nxi2 - x¯2                      as, 1ni=1nx = x 2 = 120×Incorrecti=1nxi2 - 1024 = 120×Incorrecti=1nxi2 - 100120×Incorrecti=1nxi2 = 104Incorrecti=1nxi2 = 2080Now, correcti=1nxi2 = Incorrecti=1nxi2 - 82correcti=1nxi2 = 2080 - 64 = 2016 correct Standard Deviation = 1ncorrecti=1nxi2 - correct mean2correct Standard Deviation = 119×2016 - 192192correct Standard Deviation = 201619-192192correct Standard Deviation = 144019 = 121019correct 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

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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 .

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.

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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

Incorrect standard deviation

∴ Incorrect sum of observations = 2000

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



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