NCERT Solutions Class 11th Maths Chapter – 15 – Statistics Exercise 15.3

NCERT Solutions Class 11th Maths Chapter – 15 – Statistics Exercise 15.3

NCERT Solutions Class 11th Maths Chapter – 15 – Statistics Exercise 15.3

?Chapter – 15?

✍Statistics✍

?Exercise 15.3?

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

?‍♂️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

mean   x̅ = A + ∑^a_{i = 1}f_iy_i/N x h

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

the variance σ²= h²/h²[N∑f_iy²_i – (∑f_iy_i)²]

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.

x̅ = A + ∑^a_{i = 1}f_iy_i/N x h

Where A = 45,
h = 10
So, x̅ = 45 + ((-6/150) × 10)
= 45 – 0.4
= 44.6

σ² = h²/h²[N∑f_iy²_i – (∑f_iy_i)²]

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:

?‍♂️Solution: From the given data,

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

= (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/n²[N∑yi² – (∑y_i)²]

= (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:

(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:

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.

For Team A:

Mean of rounds scored = ∑f_ix_i/N = 50/25 = 2

Standard Deviation = 1/N √N∑f_ix_i² – (∑f_ix_i)²

= 1/25√25×130-50×50

= 2/25√130-100

= 1/5√30

= 1.095

Coefficient of Variance = σ/x×100

= 1.095/2×100

= 54.75

For Team B:

mean x̅ = 2

Standard Deviation = 1.25

Coefficient of Variance = σ/x×100

= 1.25/2×100

= 62.5

Team A has less coefficient of variance than Team B.

So team A has more stability than team B.

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,

First we have to calculate the mean for length x,

n = 50, ∑x_i⁵⁰_{i = 1}

x̅ =212/50

= 4.24

σ² = 1/n√[ N ∑x²_i – (∑x_i)²]

= 1/50√50×902.8-(212)²

= 1/50√45140-44944

= √196/50

= 14/50

= 0.28

Coefficient of Variance,

C.V. = σ / x̅ × 100

= 0.28/4.24×100

= 6.60

n = 50, ∑y_i⁵⁰_{i = 1} = 261  ∑y²_i⁵⁰_{i = 1} = 1457.6

x̅ =∑y_i/n

= 261/50

= 5.22

σ² = 1/n√[ N ∑y²_i – (∑y_i)²]

= 1/50√50×1457.6-(261)²

For load:

= 1/50√72880-68121

= √4759/50

= 68.9855/50

= 1.38

Coefficient of Variance, C.V. = σ / x̅ ×100

= 1.38/5.22×100

= 26.44

The coefficient of variance of the load is greater than the coefficient of variance of the length.

Hence there is more variance in the distribution of weights.