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RD Chapter 32- Statistics Ex-32.3 Interview Questions Answers

Question 1 :

Compute the meandeviation from the median of the following distribution:

Class

0-10

10-20

20-30

30-40

40-50

Frequency

5

10

20

5

10

Answer 1 :

To find the mean deviation from the median, firstly let uscalculate 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=

= 1/50 × 450

= 9

Themean deviation is 9

Question 2 :

Find the meandeviation 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

(ii)

Classes

95-105

105-115

115-125

125-135

135-145

145-155

Frequencies

9

13

16

26

30

12

Answer 2 :

(i)

To find the mean deviation from the mean, firstly let uscalculate the mean.

By using the formula,

= 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=

n= 1/50 × 7896

= 157.92

Themean deviation is 157.92

(ii)

To find the mean deviation from the mean, firstly let uscalculate the mean.

By using the formula,

= 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=

= 1/106 × 1272.6

= 12.005

Themean deviation is 12.005

Question 3 :

Compute meandeviation 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

Answer 3 :

To find the mean deviation from the mean, firstly let uscalculate 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=

= 1/110 × 1644

= 14.94

Themean deviation is 14.94

Question 4 :

The agedistribution 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 meandeviation from the median age.

Answer 4 :

To find the mean deviation from the median, firstly let uscalculate the median.

N = 96

So, N/2 = 96/2 = 48

The cumulative frequency just greater than 48 is 59, and thecorresponding 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=

= 1/96 × 980.5

= 10.21

Themean deviation is 10.21

Question 5 :

Find the meandeviation 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

Answer 5 :

To find the mean deviation from the median, firstly let uscalculate the median.

N = 50

So, N/2 = 50/2 = 25

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

So, Median = 28

By using the formula to calculate Mean,

= 1350/50

= 27

Class Interval

xi

fi

Cumulative Frequency

|di| = |xi – Median|

fi |di|

FiXi

|Xi – Mean|

F|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 


TheMean Deviation from the median is 9.56 and from mean is 9.44.


Selected

 

RD Chapter 32- Statistics Ex-32.3 Contributors

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