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RD Chapter 32- Statistics Ex-32.2 Interview Questions Answers

Question 1 :

Calculate the meandeviation 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

Answer 1 :

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

= 1/197 × 336

= 1.70

Themean deviation is 1.70.

Question 2 :

The number oftelephone calls received at an exchange in 245 successive on2-minute intervalsis 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 meandeviation about the median.

Answer 2 :

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

= 1/245 × 336

= 1.49

Themean deviation is 1.49.

Question 3 :

Calculate the meandeviation about the median of the following frequency distribution:

xi

5

7

9

11

13

15

17

fi

2

4

6

8

10

12

8

Answer 3 :

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

= 1/50 × 136

= 2.72

Themean deviation is 2.72.

Question 4 :

Find the meandeviation from the mean for the following data:

(i)

xi

5

7

9

10

12

15

fi

8

6

2

2

2

6

(ii)

xi

5

10

15

20

25

fi

7

4

6

3

5

(iii)

xi

10

30

50

70

90

fi

4

24

28

16

8

Answer 4 :

(i)

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

By using the formula,

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

= 234/26

= 9

= 88/26

= 3.3

Themean deviation is 3.3

(ii)

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

By using the formula,

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

= 350/25

= 14

= 158/25

= 6.32

Themean deviation is 6.32

(iii)

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

By using the formula,

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

= 4000/80

= 50

= 1280/80

= 16

Themean deviation is 16

Question 5 :

Find the meandeviation from the median for the following data :

(i)

xi

15

21

27

30

fi

3

5

6

7

(ii)

xi

74

89

42

54

91

94

35

fi

20

12

2

4

5

3

4

(iii)

Marks obtained

10

11

12

14

15

No. of students

2

3

8

3

4

Answer 5 :

(i)

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

= 1/21 × 108

= 5.14

Themean deviation is 5.14

(ii)

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

= 1/50 × 625

= 12.5

Themean deviation is 12.5

(iii)

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

= 1/20 × 25

= 1.25

Themean deviation is 1.25


Selected

 

RD Chapter 32- Statistics Ex-32.2 Contributors

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