GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3

   

Gujarat Board Statistics Class 11 GSEB Solutions Chapter 4 Measures of Dispersion Ex 4.3 Textbook Exercise Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3

Question 1.
The measurements of height (in centimeters) of 10 soldiers are given below:
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3 1
Find the mean deviation of the heights of the soldiers.
Answer:
Here, n = 10

Height (in cm) | x – x̄ |
x̄  = 167.4
160 7.4
175 7.6
158 9.4
165 2.4
170 2.6
166 1.4
173 5.6
176 8.6
163 4.4
168 0.6
Σx = 1674 Σ |x – x̄| = 50.0

GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3 2

Question 2.
The distribution of number of ball bearings used in machines of a factory is given below. Calculate the mean deviation and coefficient of mean deviation of number of ball bearings per machine.
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3 3
Answer:
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3 4
Mean:
x̄ = \(\frac{\Sigma f x}{n}=\frac{160}{20}\) = 8 ball bearing

Mean deviation of numbers of ball bearing:
MD = \(\frac{\Sigma f|x-\bar{x}|}{n}\)
= \(\frac{56}{20}\) = 28 ball bearing

Coefficient Mean deviation of ball bearing:
Coefficient of mean deviation = \(\frac{\mathrm{MD}}{\bar{x}}\)
= \(\frac{2.8}{8}\) = 0.35

GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3

Question 3.
Find the mean deviation and the coefficient of mean deviation of the distribution of talk time (in minutes) per call.
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3 5
Answer:
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3 6
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3 7
Coefficient of Mean deviation of talk time : Coefficient of Mean deviation = \(\frac{\mathrm{MD}}{\bar{x}}=\frac{3.33}{7.3}\) = 0.46

Question 4.
Find the mean deviation and coefficient of mean deviation of number of TV sets using the following frequency distribution of TV sets sold in last 16 months in a town:
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3 8
Answer:
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3 9
Mean:
x̄ = \(\frac{\Sigma f x}{n}=\frac{960}{16}\) = 60 TV sets

Mean deviation of monthly sales of TV sets:
MD = \(\frac{\Sigma f|x-\bar{x}|}{n}=\frac{240}{16}\) = 15 TV sets

Coefficient of Mean deviation of monthly sales of TV sets:
Coefficient of mean deviation = \(\frac{M D}{\bar{x}}=\frac{15}{60}\) = 0.25

GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3

Question 5.
There are 50 boxes containing different number of units of an item in a factory. Find the mean deviation of number of units per box using the following distribution of the units:
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3 11
Answer:
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3 12

Mean:
x̄ = \(\frac{\Sigma f x}{n}=\frac{1670}{50}\) = 33.4 units

Mean deviation of number of units per box:
MD = \(\frac{\Sigma f|x-\bar{x}|}{n}=\frac{659.2}{50}\) = 13.18 units

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