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:
Find the mean deviation of the heights of the soldiers.
Answer:
Here, n = 10
| Height (in cm) | | x – xÌ„ | xÌ„Â = 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 – xÌ„| = 50.0 |
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.
Answer:
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
Question 3.
Find the mean deviation and the coefficient of mean deviation of the distribution of talk time (in minutes) per call.
Answer:
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:
Answer:
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
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:
Answer:
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