Question 2
Consider the following distribution of daily wages of 50 workers of a factory.
| Daily wages (in Rs.) | 500-520 | 520-540 | 540- 560 | 560-580 | 580-600 |
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Find the midpoint of the given interval using the formula.
| Midpoint $(x_i)$ = $\dfrac{(upper limit + lower limit)}{2}$ |
In this case, the value of Mid-point $(x_i)$ is very large, so let us assume the mean value, a = 550.
Class interval (h) = 20
So, $u_i = \dfrac{(x_i – a)}{h}$
$u_i$ = $\dfrac{(x_i – 550)}{20}$
Substitute and find the values as follows:
| Daily wages(Class interval) | Number of workers (frequency $(f_i)$) | Mid-point $(x_i)$ | $u_i$ = $\dfrac{(x_i – 550)}{20}$ | $f_i u_i$ |
|---|---|---|---|---|
| 500- 520 | 12 | 510 | -2 | -24 |
| 520-540 | 14 | 530 | -1 | - 14 |
| 540-560 | 8 | 550 = a | 0 | 0 |
| 560- 580 | 6 | 570 | 1 | 6 |
| 580- 600 | 10 | 590 | 2 | 20 |
| Total | Sum $f_i$ = 50 | Sum $f_iu_i$ = -12 |
So, the formula to find out the mean is:
| Mean = $\overline{x} = a + h(\dfrac{∑f_iu_i }{∑f_i} )$ |
= 550 + [20 × $(-\dfrac{12}{50})$] = 550 – 4.8 = 545.20
Thus, mean daily wage of the workers = Rs. 545.20