RD Sharma Solutions for Class 11 Chapter 32 - Statistics Exercise 32.3

This exercise mainly discusses mean deviation of a grouped or continuous frequency distribution. To help students score well in the board exams, experts at BYJU’S have solved the exercise wise problems using shortcut methods. They can solve these problems without any time constraint. These solutions help students clear their confusion about the concepts discussed here. Students can download the pdf of solutions based on their requirements. RD Sharma Class 11 Maths Solutions pdf is readily available, where students can download easily and start practising offline.

Download the pdf of RD Sharma Solutions for Class 11 Maths Exercise 32.3 Chapter 32 – Statistics

 

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

Exercise 32.1 Solutions

Exercise 32.2 Solutions

Exercise 32.4 Solutions

Exercise 32.5 Solutions

Exercise 32.6 Solutions

Exercise 32.7 Solutions

Access answers to RD Sharma Solutions for Class 11 Maths Exercise 32.3 Chapter 32 – Statistics

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.

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