 # 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 a 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        ### 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 Exercise 15.3

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 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, 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 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, standard deviation = σ = √243.84

= 15.61

∴ C.V for group B = (σ/ x̅) × 100

= (15.61/44.6) × 100

= 35

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

= 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

= 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:-

(i) 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

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  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: Which is more varying, the length or weight?

Solution:-

First we have to calculate Mean for Length x,  