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Chapter 15 Statistics Ex-15.3 Interview Questions Answers

Question 1 : From the data given below state which group is more variable, A or B?

Marks

10 – 20

20 – 30

30 – 40

40 – 50

50 – 60

60 – 70

70 – 80

Group A

9

17

32

33

40

10

9

Group B

10

20

30

25

43

15

7

Answer 1 :

each series. Theseries having greater C.V. is said to be more variable than the other. Theseries having lesser C.V. is said to be more consistent than the other.

Co-efficient ofvariation (C.V.) = (σ/ x̅) × 100

Where, σ = standarddeviation, x̅ = mean

For Group A

Marks

Group A

fi

Mid-point

Xi

Yi = (xi – A)/h

(Yi)2

fiyi

fi(yi)2

10 – 20

9

15

((15 – 45)/10) = -3

(-3)2

= 9

– 27

81

20 – 30

17

25

((25 – 45)/10) = -2

(-2)2

= 4

– 34

68

30 – 40

32

35

((35 – 45)/10) = – 1

(-1)2

= 1

– 32

32

40 – 50

33

45

((45 – 45)/10) = 0

02

0

0

50 – 60

40

55

((55 – 45)/10) = 1

12

= 1

40

40

60 – 70

10

65

((65 – 45)/10) = 2

22

= 4

20

40

70 – 80

9

75

((75 – 45)/10) = 3

32

= 9

27

81

Total

150

-6

342

Where A = 45,

and yi =(xi – A)/h

Here h = class size =20 – 10

h = 10

So, x̅ = 45 +((-6/150) × 10)

= 45 – 0.4

= 44.6

σ2 =(102/1502) [150(342) – (-6)2]

= (100/22500) [51,300– 36]

= (100/22500) × 51264

= 227.84

Hence, standarddeviation = σ = √227.84

= 15.09

 C.V for group A= (σ/ x̅) × 100

= (15.09/44.6) × 100

= 33.83

Now, for group B.

Marks

Group B

fi

Mid-point

Xi

Yi = (xi – A)/h

(Yi)2

fiyi

fi(yi)2

10 – 20

10

15

((15 – 45)/10) = -3

(-3)2

= 9

– 30

90

20 – 30

20

25

((25 – 45)/10) = -2

(-2)2

= 4

– 40

80

30 – 40

30

35

((35 – 45)/10) = – 1

(-1)2

= 1

– 30

30

40 – 50

25

45

((45 – 45)/10) = 0

02

0

0

50 – 60

43

55

((55 – 45)/10) = 1

12

= 1

43

43

60 – 70

15

65

((65 – 45)/10) = 2

22

= 4

30

60

70 – 80

7

75

((75 – 45)/10) = 3

32

= 9

21

63

Total

150

-6

366

Where A = 45,

h = 10

So, x̅ = 45 +((-6/150) × 10)

= 45 – 0.4

= 44.6

σ2 =(102/1502) [150(366) – (-6)2]

= (100/22500) [54,900– 36]

= (100/22500) × 54,864

= 243.84

Hence, standarddeviation = σ = √243.84

= 15.61

 C.V for group B= (σ/ x̅) × 100

= (15.61/44.6) × 100

= 35

By comparing C.V. ofgroup A and group B.

C.V of Group B >C.V. of Group A

So, Group B is morevariable.

Question 2 : From the prices of shares X and Y below, find out which is more stable in value:

X

35

54

52

53

56

58

52

50

51

49

Y

108

107

105

105

106

107

104

103

104

101

Answer 2 :

From the given data,

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

X (xi)

Y (yi)

Xi2

Yi2

35

108

1225

11664

54

107

2916

11449

52

105

2704

11025

53

105

2809

11025

56

106

8136

11236

58

107

3364

11449

52

104

2704

10816

50

103

2500

10609

51

104

2601

10816

49

101

2401

10201

Total = 510

1050

26360

110290

We have to calculateMean for x,

Mean x̅ = ∑xi/n

Where, n = number ofterms

= 510/10

= 51

Then, Variance for x =

= (1/102)[(10× 26360) – 5102]

= (1/100) (263600 –260100)

= 3500/100

= 35

WKT Standard deviation= √variance

= √35

= 5.91

So, co-efficient ofvariation = (σ/ x̅) × 100

= (5.91/51) × 100

= 11.58

Now, we have tocalculate Mean for y,

Mean ȳ = ∑yi/n

Where, n = number ofterms

= 1050/10

= 105

Then, Variance for y =

= (1/102)[(10× 110290) – 10502]

= (1/100) (1102900 –1102500)

= 400/100

= 4

WKT Standard deviation= √variance

= √4

= 2

So, co-efficient ofvariation = (σ/ x̅) × 100

= (2/105) × 100

= 1.904

By comparing C.V. of Xand Y.

C.V of X > C.V. ofY

So, Y is more stablethan X.

Question 3 :

An analysis of monthly wages paid to workers in two firms A and B,belonging to the same industry, gives the following results:

Firm A

Firm B

No. of wages earners

586

648

Mean of monthly wages

Rs 5253

Rs 5253

Variance of the distribution of wages

100

121

(i) Which firm A or B pays larger amount as monthly wages?

(ii) Which firm, A or B, shows greater variability in individual wages?

Answer 3 :

(i) From the giventable,

Mean monthly wages offirm A = Rs 5253

and Number of wageearners = 586

Then,

Total amount paid =586 × 5253

= Rs 3078258

Mean monthly wages offirm B = Rs 5253

Number of wage earners= 648

Then,

Total amount paid =648 × 5253

= Rs 34,03,944

So, firm B pays largeramount as monthly wages.

(ii) Variance of firmA = 100

We know that, standarddeviation (σ)= √100

=10

Variance of firm B =121

Then,

Standard deviation(σ)=√(121 )

=11

Hence the standarddeviation is more in case of Firm B that means in firm B there is greatervariability in individual wages.

Question 4 :
The following is the record of goals scored by team A in a football session:

No. of goals scored

0

1

2

3

4

No. of matches

1

9

7

5

3

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?

Answer 4 :

From the given data,

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

Number of goals scored xi

Number of matches fi

fixi

Xi2

fixi2

0

1

0

0

0

1

9

9

1

9

2

7

14

4

28

3

5

15

9

45

4

3

12

16

48

Total

25

50

130

Since C.V. of firm Bis greater

 Team A is moreconsistent.

Question 5 :
The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
 
Which is more varying, the length or weight?

Answer 5 :

First we have tocalculate Mean for Length x,


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Chapter 15 Statistics Ex-15.3 Contributors

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