NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.3

NCERT Solutions for Class 11 Maths Chapter 15 Statistics, contains solutions for all exercise 15.3 questions. These questions are solved by subject experts using step by step approach. By practising NCERT solutions, students will be able to clear their doubts on statistics and solve problems at their own pace. Download the NCERT Solutions of Class 11 maths now and start practising to score well in exams.

Download PDF of NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.3

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Access other exercise solutions of Class 11 Maths Chapter 15 Statistics

Exercise 15.1 Solutions : 12 Questions

Exercise 15.2 Solutions : 10 Questions

Miscellaneous Exercise Solutions : 7 Questions

Access Solutions for Class 11 Maths Chapter 15.3 Exercise

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

Solution:-

For comparing the variability or dispersion of two series, we calculate the coefficient of variance for each series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other.

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

Where, σ = standard deviation, 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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 37

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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 38

σ2 = (102/1502) [150(342) – (-6)2]

= (100/22500) [51,300 – 36]

= (100/22500) × 51264

= 227.84

Hence, standard deviation = 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

160

70 – 80

7

75

((75 – 45)/10) = 3

32

= 9

21

189

Total

150

-6

592

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 39

Where A = 45,

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 40

σ2 = (102/1502) [150(592) – (-6)2]

= (100/22500) [88,800 – 36]

= (100/22500) × 88,764

= 394.50

Hence, standard deviation = σ = √394.50

= 19.86

∴ C.V for group B = (σ/ x̅) × 100

= (19.86/44.6) × 100

= 44.53

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

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

So, Group B is more variable.

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

Solution:-

From the given data,

Let us make the table of 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 calculate Mean for x,

Mean x̅ = ∑xi/n

Where, n = number of terms

= 510/10

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 41= 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 of variation = (σ/ x̅) × 100

= (5.91/51) × 100

= 11.58

Now, we have to calculate Mean for y,

Mean ȳ = ∑yi/n

Where, n = number of terms

= 1050/10

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 42= 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 of variation = (σ/ x̅) × 100

= (2/105) × 100

= 1.904

By comparing C.V. of X and Y.

C.V of X > C.V. of Y

So, Y is more stable than X.

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?

Solution:-

From the given table,

Mean monthly wages of firm A = Rs 5253

and Number of wage earners = 586

Then,

Total amount paid = 586 × 5253

= Rs 3078258

Mean monthly wages of firm B = Rs 5253

Number of wage earners = 648

Then,

Total amount paid = 648 × 5253

= Rs 34,03,944

(i) So, firm B pays larger amount as monthly wages.

(ii) Variance of firm A = 100

We know that, standard deviation (σ)= √100

=10

Variance of firm B = 121

Then,

Standard deviation (σ)=√(121 )

=11

Hence the standard deviation is more in case of Firm B that means in firm B there is greater variability in individual wages.

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?

Solution:-

From the given data,

Let us make the table of 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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 43

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 44

Since C.V. of firm B is greater

∴ Team A is more consistent.

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:

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 45

Which is more varying, the length or weight?

Solution:-

First we have to calculate Mean for Length x,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 46

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 47


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