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Chapter 15 Statistics Ex-15.1 Interview Questions Answers

Question 1 : Find the mean deviation about the mean for the data in Exercises 1 and 2.
4, 7, 8, 9, 10, 12, 13, 17

Answer 1 :

First we have to find (x̅) of the given data
 
So, the respective values of the deviations from mean,
i.e., xi – x̅ are, 10 – 4 = 6, 10 – 7 = 3, 10 – 8 = 2, 10 – 9 = 1, 10 – 10 = 0,
10 – 12 = – 2, 10 – 13 = – 3, 10 – 17 = – 7
6, 3, 2, 1, 0, -2, -3, -7
Now absolute values of the deviations,
6, 3, 2, 1, 0, 2, 3, 7
 
MD = sum of deviations/ number of observations
= 24/8
= 3
So, the mean deviation for the given data is 3.

Question 2 : Find the mean deviation about the mean for the data in Exercises 1 and 2.
38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Answer 2 :

First we have to find (x̅) of the given data
 
So, the respective values of the deviations from mean,
i.e., xi – x̅ are, 50 – 38 = -12, 50 -70 = -20, 50 – 48 = 2, 50 – 40 = 10, 50 – 42 = 8,
50 – 55 = – 5, 50 – 63 = – 13, 50 – 46 = 4, 50 – 54 = -4, 50 – 44 = 6
-12, 20, -2, -10, -8, 5, 13, -4, 4, -6
Now absolute values of the deviations,
12, 20, 2, 10, 8, 5, 13, 4, 4, 6
 
MD = sum of deviations/ number of observations
= 84/10
= 8.4
So, the mean deviation for the given data is 8.4.

Question 3 : Find the mean deviation about the median for the data in Exercises 3 and 4.
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Answer 3 :

First we have toarrange the given observations into ascending order,

10, 11, 11, 12, 13,13, 14, 16, 16, 17, 17, 18.

The number ofobservations is 12

Then,

Median = ((12/2)th observation+ ((12/2)+ 1)th observation)/2

(12/2)th observation= 6th = 13

(12/2)+ 1)th observation= 6 + 1

= 7th =14

Median = (13 + 14)/2

= 27/2

= 13.5

So, the absolutevalues of the respective deviations from the median, i.e., |xi –M| are

3.5, 2.5, 2.5, 1.5,0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

Mean Deviation,

= (1/12) × 28

= 2.33

So, the mean deviationabout the median for the given data is 2.33.

Question 4 : Find the mean deviation about the median for the data in Exercises 3 and 4.
36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Answer 4 :

First we have toarrange the given observations into ascending order,

36, 42, 45, 46, 46,49, 51, 53, 60, 72.

The number ofobservations is 10

Then,

Median = ((10/2)th observation+ ((10/2)+ 1)th observation)/2

(10/2)th observation= 5th = 46

(10/2)+ 1)th observation= 5 + 1

= 6th =49

Median = (46 + 49)/2

= 95

= 47.5

So, the absolutevalues of the respective deviations from the median, i.e., |xi –M| are

11.5, 5.5, 2.5, 1.5,1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Mean Deviation,

= (1/10) × 70

= 7

So, the mean deviationabout the median for the given data is 7.

Question 5 : Find the mean deviation about the mean for the data in Exercises 5 and 6.

xi

5

10

15

20

25

fi

7

4

6

3

5

Answer 5 :

Let us make the tableof the given data and append other columns after calculations.

Xi

fi

fixi

|xi – x̅|

fi |xi – x̅|

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

25

350

158

The sum of calculateddata,

The absolute values ofthe deviations from the mean, i.e., |xi – x̅|, as shown in thetable.

Question 6 : Find the mean deviation about the mean for the data in Exercises 5 and 6.

xi

10

30

50

70

90

fi

4

24

28

16

8

Answer 6 : Let us make the table of the given data and append other columns after calculations.

Xi

fi

fixi

|xi – x̅|

fi |xi – x̅|

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

80

4000

1280

Question 7 : Find the mean deviation about the median for the data in Exercises 7 and 8.

xi

5

7

9

10

12

15

fi

8

6

2

2

2

6

Answer 7 : Let us make the table of the given data and append other columns after calculations.

Xi

fi

c.f.

|xi – M|

fi |xi – M|

5

8

8

2

16

7

6

14

0

0

9

2

16

2

4

10

2

18

3

6

12

2

20

5

10

15

6

26

8

48

Now, N = 26, which iseven.

Median is the mean ofthe 13th and 14th observations. Both of theseobservations lie in the cumulative frequency 14, for which the correspondingobservation is 7.

Then,

Median = (13th observation+ 14th observation)/2

= (7 + 7)/2

= 14/2

= 7

So, the absolutevalues of the respective deviations from the median, i.e., |xi –M| are shown in the table.

Question 8 : Find the mean deviation about the median for the data in Exercises 7 and 8.

xi

15

21

27

30

35

fi

3

5

6

7

8

Answer 8 :

Let us make the tableof the given data and append other columns after calculations.

Xi

fi

c.f.

|xi – M|

fi |xi – M|

15

3

3

15

45

21

5

8

9

45

27

6

14

3

18

30

7

21

0

0

35

8

29

5

40

Now, N = 29, which isodd.

So 29/2 = 14.5

The cumulativefrequency greater than 14.5 is 21, for which the corresponding observation is30.

Then,

Median = (15th observation+ 16th observation)/2

= (30 + 30)/2

= 60/2

= 30

So, the absolutevalues of the respective deviations from the median, i.e., |xi –M| are shown in the table.

Question 9 : Find the mean deviation about the mean for the data in Exercises 9 and 10.

Income per day in ₹

0 – 100

100 – 200

200 – 300

300 – 400

400 – 500

500 – 600

600 – 700

700 – 800

Number of persons

4

8

9

10

7

5

4

3

Answer 9 :

Let us make the tableof the given data and append other columns after calculations.

Income per day in ₹

Number of persons fi

Mid – points

xi

fixi

|xi – x̅|

fi|xi – x̅|

0 – 100

4

50

200

308

1232

100 – 200

8

150

1200

208

1664

200 – 300

9

250

2250

108

972

300 – 400

10

350

3500

8

80

400 – 500

7

450

3150

92

644

500 – 600

5

550

2750

192

960

600 – 700

4

650

2600

292

1160

700 – 800

3

750

2250

392

1176

50

17900

7896

Question 10 : Find the mean deviation about the mean for the data in Exercises 9 and 10.

Height in cms

95 – 105

105 – 115

115 – 125

125 – 135

135 – 145

145 – 155

Number of boys

9

13

26

30

12

10

Answer 10 :

Let us make the tableof the given data and append other columns after calculations.

Height in cms

Number of boys fi

Mid – points

xi

fixi

|xi – x̅|

fi|xi – x̅|

95 – 105

9

100

900

25.3

227.7

105 – 115

13

110

1430

15.3

198.9

115 – 125

26

120

3120

5.3

137.8

125 – 135

30

130

3900

4.7

141

135 – 145

12

140

1680

14.7

176.4

145 – 155

10

150

1500

24.7

247

100

12530

1128.8


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Chapter 15 Statistics Ex-15.1 Contributors

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