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RD Chapter 32- Statistics Ex-32.1 Interview Questions Answers

Question 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

Answer 1 :

(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=

xi

|di| = |xi – 3020|

3011

9

2780

240

3020

0

2354

666

3541

521

4150

1130

5000

1980

Total

4546

MD=
= 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=

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=
= 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=

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=
= 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=

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=
= 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=

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=
= 1/10 × 84
= 8.4
∴ The Mean Deviation is 8.4.

Question 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

Answer 2 :

(i) 4, 7,8, 9, 10, 12, 13, 17

We know that,

MD=

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=

= 1/8 × 24

= 3

TheMean Deviation is 3.

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

We know that,

MD=

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=

= 1/12 × 28

= 2.33

TheMean Deviation is 2.33.

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

We know that,

MD=

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=

= 1/10 × 84

= 8.4

TheMean Deviation is 8.4.

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

We know that,

MD=

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=

= 1/10 × 72

= 7.2

TheMean Deviation is 7.2.

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

We know that,

MD=

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=

= 1/10 × 74

= 7.4

TheMean Deviation is 7.4.

Question 3 :

Calculate the meandeviation of the following income groups of five and seven members from theirmedians:

I

Income in ₹

II

Income in ₹

4000

3800

4200

4000

4400

4200

4600

4400

4800

4600

4800

5800

Answer 3 :

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=

Where, |di| = |xi – M|

xi

|di| = |xi – 4400|

4000

400

4200

200

4400

0

4600

200

4800

400

Total

1200

MD=

= 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=

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=

= 1/7 × 3200

= 457.14

TheMean Deviation of set 1 is 240 and set 2 is 457.14

Question 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.

Answer 4 :

(i) Findthe 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=

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=

= 1/10 × 167.4

= 16.74

TheMean Deviation is 16.74.

(ii) Findthe mean deviation from the mean also.

We know that,

MD=

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=

= 1/10 × 166.4

= 16.64

The Mean Deviation is 16.64

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

Answer 5 :

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

We know that,

MD=

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=

= 1/10 × 90

= 9

Now

M.D = 45.5 – 9 = 36.5

+M.D = 45.5 + 9 = 54.5

So,There are total 6 observation between M.D and +M.D

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

We know that,

MD=

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=

= 1/10 × 48.8

= 4.88

Now

So, There are total 5 observation between and

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

We know that,

MD=

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=

= 1/10 × 86.8

= 8.68

Now


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RD Chapter 32- Statistics Ex-32.1 Contributors

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