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Chapter 15 Statistics Ex-15.2 Interview Questions Answers

Question 1 : Find the mean and variance for the data 
6, 7, 10, 12, 13, 4, 8, 12

Answer 1 :


So, x̅ = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8
= 72/8
= 9
Let us make the table of the given data and append other columns after calculations.

Xi

Deviations from mean

(xi – x̅)

(xi – x̅)2

6

6 – 9 = -3

9

7

7 – 9 = -2

4

10

10 – 9 = 1

1

12

12 – 9 = 3

9

13

13 – 9 = 4

16

4

4 – 9 = – 5

25

8

8 – 9 = – 1

1

12

12 – 9 = 3

9

74

We know that Variance,

σ2 =(1/8) × 74

= 9.2

Mean = 9 and Variance= 9.25

Question 2 : Find the mean and variance for the data 

First n natural numbers

Answer 2 :

We know that Mean =Sum of all observations/Number of observations

Mean, x̅ = ((n(n +1))2)/n

= (n + 1)/2

and also WKT Variance,

By substitute thatvalue of x̅ we get,

WKT, (a + b)(a – b) =a2 – b2

σ2 =(n2 – 1)/12

Mean = (n + 1)/2 andVariance = (n2 – 1)/12

Question 3 : Find the mean and variance for the data 
First 10 multiples of 3

Answer 3 :

First we have to writethe first 10 multiples of 3,

3, 6, 9, 12, 15, 18,21, 24, 27, 30

So, x̅ = (3 + 6 + 9 +12 + 15 + 18 + 21 + 24 + 27 + 30)/10

= 165/10

= 16.5

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

Xi

Deviations from mean

(xi – x̅)

(xi – x̅)2

3

3 – 16.5 = -13.5

182.25

6

6 – 16.5 = -10.5

110.25

9

9 – 16.5 = -7.5

56.25

12

12 – 16.5 = -4.5

20.25

15

15 – 16.5 = -1.5

2.25

18

18 – 16.5 = 1.5

2.25

21

21 – 16.5 = – 4.5

20.25

24

24 – 16.5 = 7.5

56.25

27

27 – 16.5 = 10.5

110.25

30

30 – 16.5 = 13.5

182.25

742.5

Then, Variance

= (1/10) × 742.5

= 74.25

Mean = 16.5 andVariance = 74.25

Question 4 : Find the mean and variance for the data 

xi

6

10

14

18

24

28

30

fi

2

4

7

12

8

4

3

Answer 4 :

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

Xi

fi

fixi

Deviations from mean

(xi – x̅)

(xi – x̅)2

fi(xi – x̅)2

6

2

12

6 – 19 = 13

169

338

10

4

40

10 – 19 = -9

81

324

14

7

98

14 – 19 = -5

25

175

18

12

216

18 – 19 = -1

1

12

24

8

192

24 – 19 = 5

25

200

28

4

112

28 – 19 = 9

81

324

30

3

90

30 – 19 = 11

121

363

N = 40

760

1736

Question 5 : Find the mean and variance for the data 

xi

92

93

97

98

102

104

109

fi

3

2

3

2

6

3

3

Answer 5 :

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

Xi

fi

fixi

Deviations from mean

(xi – x̅)

(xi – x̅)2

fi(xi – x̅)2

92

3

276

92 – 100 = -8

64

192

93

2

186

93 – 100 = -7

49

98

97

3

291

97 – 100 = -3

9

27

98

2

196

98 – 100 = -2

4

8

102

6

612

102 – 100 = 2

4

24

104

3

312

104 – 100 = 4

16

48

109

3

327

109 – 100 = 9

81

243

N = 22

2200

640

Question 6 : Find the mean and standard deviation using short-cut method.

xi

60

61

62

63

64

65

66

67

68

fi

2

1

12

29

25

12

10

4

5

Answer 6 :

Let the assumed mean A= 64. Here h = 1

We obtain thefollowing table from the given data.

Xi

Frequency fi

Yi = (xi – A)/h

Yi2

fiyi

fiyi2

60

2

-4

16

-8

32

61

1

-3

9

-3

9

62

12

-2

4

-24

48

63

29

-1

1

-29

29

64

25

0

0

0

0

65

12

1

1

12

12

66

10

2

4

20

40

67

4

3

9

12

36

68

5

4

16

20

80

0

286

Mean,

Where A = 64, h = 1

So, x̅ = 64 + ((0/100)× 1)

= 64 + 0

= 64

Then, variance,

σ2 =(12/1002) [100(286) – 02]

= (1/10000) [28600 –0]

= 28600/10000

= 2.86

Hence, standarddeviation = σ = √2.886

= 1.691

 Mean = 64 andStandard Deviation = 1.691

Question 7 : Find the mean and variance for the following frequency distributions in Exercises 7 and 8.

Classes

0 – 30

30 – 60

60 – 90

90 – 120

120 – 150

150 – 180

180 – 210

Frequencies

2

3

5

10

3

5

2

Answer 7 :

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

Classes

Frequency fi

Mid – points

xi

fixi

(xi – x̅)

(xi – x̅)2

fi(xi – x̅)2

0 – 30

2

15

30

-92

8464

16928

30 – 60

3

45

135

-62

3844

11532

60 – 90

5

75

375

-32

1024

5120

90 – 120

10

105

1050

-2

4

40

120 – 150

3

135

405

28

784

2352

150 – 180

5

165

825

58

3364

16820

180 – 210

2

195

390

88

7744

15488

N = 30

3210

68280

Question 8 : Find the mean and variance for the following frequency distributions in Exercises 7 and 8.

Classes

0 – 10

10 – 20

20 – 30

30 – 40

40 –50

Frequencies

5

8

15

16

6

Answer 8 :

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

Classes

Frequency fi

Mid – points

xi

fixi

(xi – x̅)

(xi – x̅)2

fi(xi – x̅)2

0 – 10

5

5

25

-22

484

2420

10 – 20

8

15

120

-12

144

1152

20 – 30

15

25

375

-2

4

60

30 – 40

16

35

560

8

64

1024

40 –50

6

45

270

18

324

1944

N = 50

1350

6600

Question 9 : Find the mean, variance and standard deviation using short-cut method

Height in cms

70 – 75

75 – 80

80 – 85

85 – 90

90 – 95

95 – 100

100 – 105

105 – 110

110 – 115

Frequencies

3

4

7

7

15

9

6

6

3

Answer 9 :

Let the assumed mean,A = 92.5 and h = 5

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

Height (class)

Number of children

Frequency fi

Midpoint

Xi

Yi = (xi – A)/h

Yi2

fiyi

fiyi2

70 – 75

2

72.5

-4

16

-12

48

75 – 80

1

77.5

-3

9

-12

36

80 – 85

12

82.5

-2

4

-14

28

85 – 90

29

87.5

-1

1

-7

7

90 – 95

25

92.5

0

0

0

0

95 – 100

12

97.5

1

1

9

9

100 – 105

10

102.5

2

4

12

24

105 – 110

4

107.5

3

9

18

54

110 – 115

5

112.5

4

16

12

48

N = 60

6

254

Mean,

Where, A = 92.5, h = 5

So, x̅ = 92.5 +((6/60) × 5)

= 92.5 + ½

= 92.5 + 0.5

= 93

Then, Variance,

σ2 =(52/602) [60(254) – 62]

= (1/144) [15240 – 36]

= 15204/144

= 1267/12

= 105.583

Hence, standarddeviation = σ = √105.583

= 10.275

 Mean = 93,variance = 105.583 and Standard Deviation = 10.275

Question 10 :
The diameters of circles (in mm) drawn in a design are given below:

Diameters

33 – 36

37 – 40

41 – 44

45 – 48

49 – 52

No. of circles

15

17

21

22

25

Calculate the standard deviation and mean diameter of the circles.
[Hint first make the data continuous by making the classes as 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5 – 48.5, 48.5 – 52.5 and then proceed.]

Answer 10 :

Let the assumed mean,A = 42.5 and h = 4

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

Height (class)

Number of children

(Frequency fi)

Midpoint

Xi

Yi = (xi – A)/h

Yi2

fiyi

fiyi2

32.5 – 36.5

15

34.5

-2

4

-30

60

36.5 – 40.5

17

38.5

-1

1

-17

17

40.5 – 44.5

21

42.5

0

0

0

0

44.5 – 48.5

22

46.5

1

1

22

22

48.5 – 52.5

25

50.5

2

4

50

100

N = 100

25

199

Mean,

Where, A = 42.5, h = 4

So, x̅ = 42.5 +(25/100) × 4

= 42.5 + 1

= 43.5

Then, Variance,

σ2 =(42/1002)[100(199) – 252]

= (1/625) [19900 –625]

= 19275/625

= 771/25

= 30.84

Hence, standarddeviation = σ = √30.84

= 5.553

 Mean = 43.5,variance = 30.84 and Standard Deviation = 5.553.


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Chapter 15 Statistics Ex-15.2 Contributors

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