RD Sharma Solutions for Class 11 Chapter 32 - Statistics

In earlier classes, we have learnt about methods of representing data graphically and in tubular form. Here, we shall study various methods of finding a representative value of the given data. RD Sharma Solutions for Class 11 Maths Chapter 32 Statistics are provided here for students to practice and prepare for their board exams. Students are advised to refer to the pdf of RD Sharma Class 11 Maths Solutions in order to help them grasp the shortcut techniques and important formulas. For students to understand the concepts easily, the solutions are solved in a stepwise manner, as each step carries marks in the annual exam. Students who aim to improve their academic performance can use RD Sharma as a key source of reference material to achieve their goals. Students can download the readily available pdf of RD Sharma solutions, from the below mentioned links.

Chapter 32 – Statistics contains seven exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

  • Measures of dispersion.
  • Range.
  • Mean deviation.
    • Mean deviation for ungrouped data or individual observations.
    • Mean deviation of a discrete frequency distribution.
    • Mean deviation of a grouped or continuous frequency distribution.
  • Limitations of mean deviation.
  • Variance and standard deviation.
    • Variance of individual observations.
    • Variance of a discrete frequency distribution.
    • Variance of a grouped or continuous frequency distribution.
  • Analysis of frequency distribution.

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Access answers to RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics

EXERCISE 32.1 PAGE NO: 32.6

1. Calculate the mean deviation about the median of the following observation :
(i) 3011, 2780, 3020, 2354, 3541, 4150, 5000

(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44

(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51

(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42

(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47

Solution:

(i) 3011, 2780, 3020, 2354, 3541, 4150, 5000

To calculate the Median (M), let us arrange the numbers in ascending order.

Median is the middle number of all the observation.

2354, 2780, 3011, 3020, 3541, 4150, 5000

So, Median = 3020 and n = 7

By using the formula to calculate Mean Deviation,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

xi

|di| = |xi – 3020|

3011

9

2780

240

3020

0

2354

666

3541

521

4150

1130

5000

1980

Total

4546

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/7 × 4546

= 649.42

∴ The Mean Deviation is 649.42.

(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44

To calculate the Median (M), let us arrange the numbers in ascending order.

Median is the middle number of all the observation.

34, 38, 42, 44, 46, 48, 54, 55, 63, 70

Here the Number of observations are Even then Median = (46+48)/2 = 47

Median = 47 and n = 10

By using the formula to calculate Mean Deviation,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

xi

|di| = |xi – 47|

38

9

70

23

48

1

34

13

42

5

55

8

63

16

46

1

54

7

44

3

Total

86

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/10 × 86

= 8.6

∴ The Mean Deviation is 8.6.

(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51

To calculate the Median (M), let us arrange the numbers in ascending order.

Median is the middle number of all the observation.

30, 34, 38, 40, 42, 44, 50, 51, 60, 66

Here the Number of observations are Even then Median = (42+44)/2 = 43

Median = 43 and n = 10

By using the formula to calculate Mean Deviation,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

xi

|di| = |xi – 43|

30

13

34

9

38

5

40

3

42

1

44

1

50

7

51

8

60

17

66

23

Total

87

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/10 × 87

= 8.7

∴ The Mean Deviation is 8.7.

(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42

To calculate the Median (M), let us arrange the numbers in ascending order.

Median is the middle number of all the observation.

22, 24, 25, 27, 28, 29, 30, 31, 41, 42

Here the Number of observations are Even then Median = (28+29)/2 = 28.5

Median = 28.5 and n = 10

By using the formula to calculate Mean Deviation,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

xi

|di| = |xi – 28.5|

22

6.5

24

4.5

30

1.5

27

1.5

29

0.5

31

2.5

25

3.5

28

0.5

41

12.5

42

13.5

Total

47

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/10 × 47

= 4.7

∴ The Mean Deviation is 4.7.

(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47

To calculate the Median (M), let us arrange the numbers in ascending order.

Median is the middle number of all the observation.

34, 38, 43, 44, 47, 48, 53, 55, 63, 70

Here the Number of observations are Even then Median = (47+48)/2 = 47.5

Median = 47.5 and n = 10

By using the formula to calculate Mean Deviation,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

xi

|di| = |xi – 47.5|

38

9.5

70

22.5

48

0.5

34

13.5

63

15.5

42

5.5

55

7.5

44

3.5

53

5.5

47

0.5

Total

84

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/10 × 84

= 8.4

∴ The Mean Deviation is 8.4.

2. Calculate the mean deviation from the mean for the following data :
(i) 4, 7, 8, 9, 10, 12, 13, 17

(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
(v) 57, 64, 43, 67, 49, 59, 44, 47, 61, 59

Solution:

(i) 4, 7, 8, 9, 10, 12, 13, 17

We know that,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [4 + 7 + 8 + 9 + 10 + 12 + 13 + 17]/8

= 80/8

= 10

Number of observations, ‘n’ = 8

xi

|di| = |xi – 10|

4

6

7

3

8

2

9

1

10

0

12

2

13

3

17

7

Total

24

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/8 × 24

= 3

∴ The Mean Deviation is 3.

(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

We know that,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [13 + 17 + 16 + 14 + 11 + 13 + 10 + 16 + 11 + 18 + 12 + 17]/12

= 168/12

= 14

Number of observations, ‘n’ = 12

xi

|di| = |xi – 14|

13

1

17

3

16

2

14

0

11

3

13

1

10

4

16

2

11

3

18

4

12

2

17

3

Total

28

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/12 × 28

= 2.33

∴ The Mean Deviation is 2.33.

(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

We know that,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44]/10

= 500/10

= 50

Number of observations, ‘n’ = 10

xi

|di| = |xi – 50|

38

12

70

20

48

2

40

10

42

8

55

5

63

13

46

4

54

4

44

6

Total

84

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/10 × 84

= 8.4

∴ The Mean Deviation is 8.4.

(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

We know that,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [36 + 72 + 46 + 42 + 60 + 45 + 53 + 46 + 51 + 49]/10

= 500/10

= 50

Number of observations, ‘n’ = 10

xi

|di| = |xi – 50|

36

14

72

22

46

4

42

8

60

10

45

5

53

3

46

4

51

1

49

1

Total

72

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/10 × 72

= 7.2

∴ The Mean Deviation is 7.2.

(v) 57, 64, 43, 67, 49, 59, 44, 47, 61, 59

We know that,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [57 + 64 + 43 + 67 + 49 + 59 + 44 + 47 + 61 + 59]/10

= 550/10

= 55

Number of observations, ‘n’ = 10

xi

|di| = |xi – 55|

57

2

64

9

43

12

67

12

49

6

59

4

44

11

47

8

61

6

59

4

Total

74

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/10 × 74

= 7.4

∴ The Mean Deviation is 7.4.

3. Calculate the mean deviation of the following income groups of five and seven members from their medians:

I

Income in ₹

II

Income in ₹

4000

3800

4200

4000

4400

4200

4600

4400

4800

4600

4800

5800

Solution:

Let us calculate the mean deviation for the first data set.

Since the data is arranged in ascending order,

4000, 4200, 4400, 4600, 4800

Median = 4400

Total observations = 5

We know that,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

Where, |di| = |xi – M|

xi

|di| = |xi – 4400|

4000

400

4200

200

4400

0

4600

200

4800

400

Total

1200

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/5 × 1200

= 240

Let us calculate the mean deviation for the second data set.

Since the data is arranged in ascending order,

3800, 4000, 4200, 4400, 4600, 4800, 5800

Median = 4400

Total observations = 7

We know that,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

Where, |di| = |xi – M|

xi

|di| = |xi – 4400|

3800

600

4000

400

4200

200

4400

0

4600

200

4800

400

5800

1400

Total

3200

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/7 × 3200

= 457.14

∴ The Mean Deviation of set 1 is 240 and set 2 is 457.14

4. The lengths (in cm) of 10 rods in a shop are given below:
40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2
(i) Find the mean deviation from the median.
(ii) Find the mean deviation from the mean also.

Solution:

(i) Find the mean deviation from the median

Let us arrange the data in ascending order,

15.2, 27.9, 30.2, 32.5, 40.0, 52.3, 52.8, 55.2, 72.9, 79.0

We know that,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

Where, |di| = |xi – M|

The number of observations are Even then Median = (40+52.3)/2 = 46.15

Median = 46.15

Number of observations, ‘n’ = 10

xi

|di| = |xi – 46.15|

40.0

6.15

52.3

6.15

55.2

9.05

72.9

26.75

52.8

6.65

79.0

32.85

32.5

13.65

15.2

30.95

27.9

19.25

30.2

15.95

Total

167.4

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/10 × 167.4

= 16.74

∴ The Mean Deviation is 16.74.

(ii) Find the mean deviation from the mean also.

We know that,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [40.0 + 52.3 + 55.2 + 72.9 + 52.8 + 79.0 + 32.5 + 15.2 + 27.9 + 30.2]/10

= 458/10

= 45.8

Number of observations, ‘n’ = 10

xi

|di| = |xi – 45.8|

40.0

5.8

52.3

6.5

55.2

9.4

72.9

27.1

52.8

7

79.0

33.2

32.5

13.3

15.2

30.6

27.9

17.9

30.2

15.6

Total

166.4

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/10 × 166.4

= 16.64

∴ The Mean Deviation is 16.64

5. In question 1(iii), (iv), (v) find the number of observations lying between \(\overline{X} – M.D\) and \(\overline{X} + M.D\), where M.D. is the mean deviation from the mean.

Solution:

(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51

We know that,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [34 + 66 + 30 + 38 + 44 + 50 + 40 + 60 + 42 + 51]/10

= 455/10

= 45.5

Number of observations, ‘n’ = 10

xi

|di| = |xi – 45.5|

34

11.5

66

20.5

30

15.5

38

7.5

44

1.5

50

4.5

40

5.5

60

14.5

42

3.5

51

5.5

Total

90

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/10 × 90

= 9

Now

\(\overline{X} – M.D\) = 45.5 – 9 = 36.5

\(\overline{X} + M.D\) = 45.5 + 9 = 54.5

So, There are total 6 observation between \(\overline{X} – M.D\) and \(\overline{X} + M.D\)

(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42

We know that,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [22 + 24 + 30 + 27 + 29 + 31 + 25 + 28 + 41 + 42]/10

= 299/10

= 29.9

Number of observations, ‘n’ = 10

xi

|di| = |xi – 29.9|

22

7.9

24

5.9

30

0.1

27

2.9

29

0.9

31

1.1

25

4.9

28

1.9

41

11.1

42

12.1

Total

48.8

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/10 × 48.8

= 4.88

Now

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 35

So, There are total 5 observation between RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 36
and RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 37

(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47

We know that,

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [38 + 70 + 48 + 34 + 63 + 42 + 55 + 44 + 53 + 47]/10

= 494/10

= 49.4

Number of observations, ‘n’ = 10

xi

|di| = |xi – 49.4|

38

11.4

70

20.6

48

1.4

34

15.4

63

13.6

42

7.4

55

5.6

44

5.4

53

3.6

47

2.4

Total

86.8

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/10 × 86.8

= 8.68

Now

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 40


EXERCISE 32.2 PAGE NO: 32.11

1. Calculate the mean deviation from the median of the following frequency distribution:

Heights in inches

58

59

60

61

62

63

64

65

66

No. of students

15

20

32

35

35

22

20

10

8

Solution:

To find the mean deviation from the median, firstly let us calculate the median.

We know, Median is the Middle term,

So, Median = 61

Let xi =Heights in inches

And, fi = Number of students

xi

fi

Cumulative Frequency

|di| = |xi – M|

= |xi – 61|

fi |di|

58

15

15

3

45

59

20

35

2

40

60

32

67

1

32

61

35

102

0

0

62

35

137

1

35

63

22

159

2

44

64

20

179

3

60

65

10

189

4

40

66

8

197

5

40

N = 197

Total = 336

N=197

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/197 × 336

= 1.70

∴ The mean deviation is 1.70.

2. The number of telephone calls received at an exchange in 245 successive on2-minute intervals is shown in the following frequency distribution:

Number of calls

0

1

2

3

4

5

6

7

Frequency

14

21

25

43

51

40

39

12

Compute the mean deviation about the median.

Solution:

To find the mean deviation from the median, firstly let us calculate the median.

We know, Median is the even term, (3+5)/2 = 4

So, Median = 8

Let xi =Number of calls

And, fi = Frequency

xi

fi

Cumulative Frequency

|di| = |xi – M|

= |xi – 61|

fi |di|

0

14

14

4

56

1

21

35

3

63

2

25

60

2

50

3

43

103

1

43

4

51

154

0

0

5

40

194

1

40

6

39

233

2

78

7

12

245

3

36

Total = 366

Total = 245

N = 245

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/245 × 336

= 1.49

∴ The mean deviation is 1.49.

3. Calculate the mean deviation about the median of the following frequency distribution:

xi

5

7

9

11

13

15

17

fi

2

4

6

8

10

12

8

Solution:

To find the mean deviation from the median, firstly let us calculate the median.

We know, N = 50

Median = (50)/2 = 25

So, the median Corresponding to 25 is 13

xi

fi

Cumulative Frequency

|di| = |xi – M|

= |xi – 61|

fi |di|

5

2

2

8

16

7

4

6

6

24

9

6

12

4

24

11

8

20

2

16

13

10

30

0

0

15

12

42

2

24

17

8

50

4

32

Total = 50

Total = 136

N = 50

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/50 × 136

= 2.72

∴ The mean deviation is 2.72.

4. Find the mean deviation from the mean for the following data:

(i)

xi

5

7

9

10

12

15

fi

8

6

2

2

2

6

Solution:

To find the mean deviation from the mean, firstly let us calculate the mean.

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 44

xi

fi

Cumulative Frequency (xifi)

|di| = |xi – Mean|

fi |di|

5

8

40

4

32

7

6

42

2

12

9

2

18

0

0

10

2

20

1

2

12

2

24

3

6

15

6

90

6

36

Total = 26

Total = 234

Total = 88

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 45

= 234/26

= 9

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 46

= 88/26

= 3.3

∴ The mean deviation is 3.3

(ii)

xi

5

10

15

20

25

fi

7

4

6

3

5

Solution:

To find the mean deviation from the mean, firstly let us calculate the mean.

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 47

xi

fi

Cumulative Frequency (xifi)

|di| = |xi – Mean|

fi |di|

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

Total = 25

Total = 350

Total = 158

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 48

= 350/25

= 14

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 49

= 158/25

= 6.32

∴ The mean deviation is 6.32

(iii)

xi

10

30

50

70

90

fi

4

24

28

16

8

Solution:

To find the mean deviation from the mean, firstly let us calculate the mean.

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 50

xi

fi

Cumulative Frequency (xifi)

|di| = |xi – Mean|

fi |di|

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

Total = 80

Total = 4000

Total = 1280

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 51

= 4000/80

= 50

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 52

= 1280/80

= 16

∴ The mean deviation is 16

5. Find the mean deviation from the median for the following data :

(i)

xi

15

21

27

30

fi

3

5

6

7

Solution:

To find the mean deviation from the median, firstly let us calculate the median.

We know, N = 21

Median = (21)/2 = 10.5

So, the median Corresponding to 10.5 is 27

xi

fi

Cumulative Frequency

|di| = |xi – M|

fi |di|

15

3

3

15

45

21

5

8

9

45

27

6

14

3

18

30

7

21

0

0

Total = 21

Total = 46

Total = 108

N = 21

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/21 × 108

= 5.14

∴ The mean deviation is 5.14

(ii)

xi

74

89

42

54

91

94

35

fi

20

12

2

4

5

3

4

Solution:

To find the mean deviation from the median, firstly let us calculate the median.

We know, N = 50

Median = (50)/2 = 25

So, the median Corresponding to 25 is 74

xi

fi

Cumulative Frequency

|di| = |xi – M|

fi |di|

74

20

4

39

156

89

12

6

32

64

42

2

10

20

80

54

4

30

0

0

91

5

42

15

180

94

3

47

17

85

35

4

50

20

60

Total = 50

Total = 189

Total = 625

N = 50

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/50 × 625

= 12.5

∴ The mean deviation is 12.5

(iii)

Marks obtained

10

11

12

14

15

No. of students

2

3

8

3

4

Solution:

To find the mean deviation from the median, firstly let us calculate the median.

We know, N = 20

Median = (20)/2 = 10

So, the median Corresponding to 10 is 12

xi

fi

Cumulative Frequency

|di| = |xi – M|

fi |di|

10

2

2

2

4

11

3

5

1

3

12

8

13

0

0

14

3

16

2

6

15

4

20

3

12

Total = 20

Total = 25

N = 20

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/20 × 25

= 1.25

∴ The mean deviation is 1.25


EXERCISE 32.3 PAGE NO: 32.16

1. Compute the mean deviation from the median of the following distribution:

Class

0-10

10-20

20-30

30-40

40-50

Frequency

5

10

20

5

10

Solution:

To find the mean deviation from the median, firstly let us calculate the median.

Median is the middle term of the Xi,

Here, the middle term is 25

So, Median = 25

Class Interval

xi

fi

Cumulative Frequency

|di| = |xi – M|

fi |di|

0-10

5

5

5

20

100

10-20

15

10

15

10

100

20-30

25

20

35

0

0

30-40

35

5

91

10

50

40-50

45

10

101

20

200

Total = 50

Total = 450

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/50 × 450

= 9

∴ The mean deviation is 9

2. Find the mean deviation from the mean for the following data:

(i)

Classes

0-100

100-200

200-300

300-400

400-500

500-600

600-700

700-800

Frequencies

4

8

9

10

7

5

4

3

Solution:

To find the mean deviation from the mean, firstly let us calculate the mean.

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 57

= 17900/50

= 358

Class Interval

xi

fi

Cumulative Frequency

|di| = |xi – M|

fi |di|

0-100

50

4

200

308

1232

100-200

150

8

1200

208

1664

200-300

250

9

2250

108

972

300-400

350

10

3500

8

80

400-500

450

7

3150

92

644

500-600

550

5

2750

192

960

600-700

650

4

2600

292

1168

700-800

750

3

2250

392

1176

Total = 50

Total = 17900

Total = 7896

N = 50

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/50 × 7896

= 157.92

∴ The mean deviation is 157.92

(ii)

Classes

95-105

105-115

115-125

125-135

135-145

145-155

Frequencies

9

13

16

26

30

12

Solution:

To find the mean deviation from the mean, firstly let us calculate the mean.

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 59

= 13630/106

= 128.58

Class Interval

xi

fi

Cumulative Frequency

|di| = |xi – M|

fi |di|

95-105

100

9

900

28.58

257.22

105-115

110

13

1430

18.58

241.54

115-125

120

16

1920

8.58

137.28

125-135

130

26

3380

1.42

36.92

135-145

140

30

4200

11.42

342.6

145-155

150

12

1800

21.42

257.04

N = 106

Total = 13630

Total = 1272.6

N = 106

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/106 × 1272.6

= 12.005

∴ The mean deviation is 12.005

3. Compute mean deviation from mean of the following distribution:

Marks

10-20

20-30

30-40

40-50

50-60

60-70

70-80

80-90

No. of students

8

10

15

25

20

18

9

5

Solution:

To find the mean deviation from the mean, firstly let us calculate the mean.

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 61

= 5390/110

= 49

Class Interval

xi

fi

Cumulative Frequency

|di| = |xi – M|

fi |di|

10-20

15

8

120

34

272

20-30

25

10

250

24

240

30-40

35

15

525

14

210

40-50

45

25

1125

4

100

50-60

55

20

1100

6

120

60-70

65

18

1170

16

288

70-80

75

9

675

26

234

80-90

85

5

425

36

180

N = 110

Total = 5390

Total = 1644

N = 110

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/110 × 1644

= 14.94

∴ The mean deviation is 14.94

4. The age distribution of 100 life-insurance policy holders is as follows:

Age (on nearest birthday)

17-19.5

20-25.5

26-35.5

36-40.5

41-50.5

51-55.5

56-60.5

61-70.5

No. of persons

5

16

12

26

14

12

6

5

Calculate the mean deviation from the median age.

Solution:

To find the mean deviation from the median, firstly let us calculate the median.

N = 96

So, N/2 = 96/2 = 48

The cumulative frequency just greater than 48 is 59, and the corresponding value of x is 38.25

So, Median = 38.25

Class Interval

xi

fi

Cumulative Frequency

|di| = |xi – M|

fi |di|

17-19.5

18.25

5

5

20

100

20-25.5

22.75

16

21

15.5

248

36-35.5

30.75

12

33

7.5

90

36-40.5

38.25

26

59

0

0

41-50.5

45.75

14

73

7.5

105

51-55.5

53.25

12

85

15

180

56-60.5

58.25

6

91

20

120

61-70.5

65.75

5

96

27.5

137.5

Total = 96

Total = 980.5

N = 96

\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)

= 1/96 × 980.5

= 10.21

∴ The mean deviation is 10.21

5. Find the mean deviation from the mean and from a median of the following distribution:

Marks

0-10

10-20

20-30

30-40

40-50

No. of students

5

8

15

16

6

Solution:

To find the mean deviation from the median, firstly let us calculate the median.

N = 50

So, N/2 = 50/2 = 25

The cumulative frequency just greater than 25 is 58, and the corresponding value of x is 28

So, Median = 28

By using the formula to calculate Mean,

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 64

= 1350/50

= 27

Class Interval

xi

fi

Cumulative Frequency

|di| = |xi – Median|

fi |di|

FiXi

|Xi – Mean|

Fi |Xi – Mean|

0-10

5

5

5

23

115

25

22

110

10-20

15

8

13

13

104

120

12

96

20-30

25

15

28

3

45

375

2

30

30-40

35

16

44

7

112

560

8

128

40-50

45

6

50

17

102

270

18

108

N = 50

Total = 478

Total = 1350

Total = 472

Mean deviation from Median = 478/50 = 9.56 


And, Mean deviation from Median = 472/50 = 9.44 


∴ The Mean Deviation from the median is 9.56 and from mean is 9.44.


EXERCISE 32.4 PAGE NO: 32.28

1. Find the mean, variance and standard deviation for the following data:
(i) 2, 4, 5, 6, 8, 17

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 65

(ii) 6, 7, 10, 12, 13, 4, 8, 12

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 66

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 67

2. The variance of 20 observations is 4. If each observation is multiplied by 2, find the variance of the resulting observations.

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 68

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 69

3. The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 70

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 71

4. The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 72

5. The mean and standard deviation of 6 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:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 73

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 74

∴ The mean of new observation is 24 and Standard deviation of new observation is 12.

6. The mean and variance of 8 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:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 75


EXERCISE 32.5 PAGE NO: 32.37

1. Find the standard deviation for the following distribution:

x:

4.5

14.5

24.5

34.5

44.5

54.5

64.5

f:

1

5

12

22

17

9

4

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 76

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 77

= 100 [1.857 – 0.0987]

= 100 [1.7583]

Var (X) = 175.83

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 78

2. Table below shows the frequency f with which ‘x’ alpha particles were radiated from a diskette

x:

0

1

2

3

4

5

6

7

8

9

10

11

12

f:

51

203

383

525

532

408

273

139

43

27

10

4

2

Calculate the mean and variance.

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 79

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 80

3. Find the mean, and standard deviation for the following data:
(i)

Year render:

10

20

30

40

50

60

No. of persons (cumulative)

15

32

51

78

97

109

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 81

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 82

(ii)

Marks:

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

Frequency:

1

6

6

8

8

2

2

3

0

2

1

0

0

0

1

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 83

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 84

4. Find the standard deviation for the following data:

(i)

x:

3

8

13

18

23

f:

7

10

15

10

6

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 85

(ii)

x:

2

3

4

5

6

7

f:

4

9

16

14

11

6

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 86

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 87


EXERCISE 32.6 PAGE NO: 32.41

1. Calculate the mean and S.D. for the following data:

Expenditure (in ₹):

0-10

10-20

20-30

30-40

40-50

Frequency:

14

13

27

21

15

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 88

2. Calculate the standard deviation for the following data:

Class:

0-30

30-60

60-90

90-120

120-150

150-180

180-210

Frequency:

9

17

43

82

81

44

24

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 89

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 90

3. Calculate the A.M. and S.D. for the following distribution:

Class:

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

Frequency:

18

16

15

12

10

5

2

1

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 91

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 92

4. A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50, the correct figure is 40. Find the correct mean and S.D.

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 93

5. Calculate the mean, median and standard deviation of the following distribution

Class-interval

31-35

36-40

41-45

46-50

51-55

56-60

61-65

66-70

Frequency:

2

3

8

12

16

5

2

3

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 94

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 95


EXERCISE 32.7 PAGE NO: 32.47

1. Two plants A and B of a factory show the following results about the number of workers and the wages paid to them

Plant A

Plant B

No. of workers

5000

6000

Average monthly wages

₹2500

₹2500

The variance of distribution of wages

81

100

In which plant A or B is there greater variability in individual wages?

Solution:

Variation of the distribution of wages in plant A (σ2 =18)

So, Standard deviation of the distribution A (σ – 9)


Similarly, the Variation of the distribution of wages in plant B (σ2 =100)

So, Standard deviation of the distribution B (σ – 10)

And, Average monthly wages in both the plants is 2500,

Since, the plant with a greater value of SD will have more variability in salary.

∴ Plant B has more variability in individual wages than plant A

2. The means and standard deviations of heights and weights of 50 students in a class are as follows:

Weights

Heights

Mean

63.2 kg

63.2 inch

Standard deviation

5.6 kg

11.5 inch

Which shows more variability, heights or weights?

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 96

3. The coefficient of variation of two distribution are 60% and 70%, and their standard deviations are 21 and 16 respectively. What is their arithmetic means?

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 97

= 22.86

∴ Means are 35 and 22.86

4. Calculate coefficient of variation from the following data:

Income (in ₹):

1000-1700

1700-2400

2400-3100

3100-3800

3800-4500

4500-5200

No. of families:

12

18

20

25

35

10

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 98

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 99

5. An analysis of the weekly 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

Average weekly wages

₹52.5

₹47.5

The variance of the distribution of wages

100

121


(i) Which firm A or B pays out the larger amount as weekly wages?
(ii) Which firm A or B has greater variability in individual wages?

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 100

6. The following are some particulars of the distribution of weights of boys and girls in a class:

Boys

Girls

Number

100

50

Mean weight

60 kg

45 kg

Variance

9

4


Which of the distributions is more variable?

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 101

Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics

Exercise 32.1 Solutions

Exercise 32.2 Solutions

Exercise 32.3 Solutions

Exercise 32.4 Solutions

Exercise 32.5 Solutions

Exercise 32.6 Solutions

Exercise 32.7 Solutions

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