RD Sharma Solutions for Class 11 Chapter 32 - Statistics Exercise 32.2

Here in this exercise, we shall discuss mean deviation of a discrete frequency distribution, along with an algorithm to calculate the mean deviation. Students who find difficulty in solving maths problems can follow the solutions prepared by the experts, who have formulated the solutions in the simplest way, for any student to understand easily. The main aim of preparing solutions is to help students solve textbook problems without any difficulty. Students who are not able to clear doubts during class hours can clear them by referring to the RD Sharma Class 11 Maths Solutions. Students can effortlessly download the solutions pdf, from the links given below.

Download the pdf of RD Sharma Solutions for Class 11 Maths Exercise 32.2 Chapter 32 – Statistics

 

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Access answers to RD Sharma Solutions for Class 11 Maths Exercise 32.2 Chapter 32 – Statistics

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

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