GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.6

Gujarat Board Statistics Class 11 GSEB Solutions Chapter 4 Measures of Dispersion Ex 4.6 Textbook Exercise Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.6

Question 1.
The information regarding marks of the students of two classes of a school is given below. Find the combined standard deviation of the marks obtained by the students.

	Division A	Division B
No. of Students	50	60
Mean marks	60	48
Standard deviation	10	12

Answer:
Class A:
n₁ = No. of students = 50
x̄₁ = Average marks = 60
s₁ = Standard deviation = 10

Class B:
n₂ = No. of students = 60
x̄₂ = Average marks = 48
s₂ = Standard deviation = 12

Combined mean of marks of students of Class A and Class B:
x̄_c = \(\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}\)
= \(\frac{(50 \times 60)+(60 \times 48)}{50+60}\)
= \(\frac{3000+2880}{110}=\frac{5880}{110}\) = 53.45 marks

Combined standard deviation of marks of students of Class A and Class B:
d₁ = x̄₁ – x̄_c = 60 – 53.45 = 6.55
∴ d₁² = 42.9025

d₂ = x̄₂ – x̄_c = 48 – 53.45 = 5.45
∴d₂¹ = 29.7025
s₁ = 10 ∴ s₁² = 100: s₂ = 12 ∴ s₂² = 144

Question 2.
The following information is available for two sections of a factory. Obtain the combined standard deviation of the production time.

	Section A	Section B
No. of workers	10	40
Mean production time per unit (minute)	25	20
Variance	16	25

Answer:
Section A:
n₁ = No. of workers =10
x̄₁ = Average production time per unit = 25
s₁¹ = Variance =16

Section B :
n₂ = No. of workers = 40
x̄₂ = Average production time per unit = 20
s₂² = Variance = 25

Combine mean of production time per unit of Section A and Section B:
x̄_c = \(\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}\)
= \(\frac{(10 \times 25)+(40 \times 20)}{10+40}\)
= \(\frac{250+800}{50}=\frac{1050}{50}\) = 21 minutes

Combined standard deviation of production time per unit of Section A and Section B :
d₁ = x̄₁ – x̄_c = 25 – 21 = 4
∴ d₁² = 16

d₂ = x̄₂ – x̄_c = 20 – 21 = -1
∴d₂¹ = 1

s₁ = 4 ∴ s₁² = 16: s₂ = 5 ∴ s₂² = 25