Class 11 NCERT Solutions- Chapter 15 Statistics – Exercise 15.3

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

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 y_i = (x_i – A)/h

Here h = class size = 20 – 10

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

Then, variance σ² = 

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

Mean, 

Where A = 45,

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

Then, Variance σ² = 

σ² = (10²/150²) [150(366) – (-6)²]

= (100/22500) [54,900 – 36]

= (100/22500) × 54,864

= 243.84

Hence, standard deviation = σ = 

= 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.

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

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	26360	110290
Total = 510	1050	26360	110290

We have to calculate Mean for x,

Mean x̅ = 

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 = 

= √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 

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 = 

= √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.

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?

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.

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?

Solution:

From the given data,

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

Number of goals 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
Total	25	50		130

First we have to calculate Mean for Team A,

Mean = 

Then,

Variance = 

We know that, standard deviation σ = 

Hence co-efficient of variation of team A,

For Team B

Given, x̅  = 2

Standard deviation σ = 1.25

So, coefficient of variation of Team B,

Since C.V. of firm B is greater

∴ Team A is more consistent.

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?

Solution:

First we have to calculate Mean for Length x,

Mean = 

Then,

Variance = 

We know that, standard deviation σ = 

Hence co-efficient of variation of team A,

Now we have to calculate mean of weight y

Then,

Variance = 

When know that, standard deviation σ = 

So, co-efficient of variance of Team B,

Since C.V. of firm weight y is greater

∴ Weight is more varying.