5(i)
Shown below is a table illustrating the monthly income distribution in a company with 100 employees.
| Monthly income (in ₹ 10,000) | Number of employees |
|---|---|
| 0-4 | 55 |
| 4-8 | 15 |
| 8-12 | 06 |
| 12-16 | 08 |
| 16-20 | 12 |
| 20-24 | 4 |
Using step-deviation method, find the mean monthly income of an employee.
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In the given table,
Class size (i) = 4.
| Monthly income | No.of employees (f) | Class mark | d = x - A | u = d/i | fu |
|---|---|---|---|---|---|
| 0-4 | 55 | 2 | -4 | -1 | -55 |
| 4-8 | 15 | A = 6 | 0 | 0 | 0 |
| 8-12 | 06 | 10 | 4 | 1 | 06 |
| 12-16 | 08 | 14 | 8 | 2 | 16 |
| 16-20 | 12 | 18 | 12 | 3 | 36 |
| 20-24 | 4 | 22 | 16 | 4 | 16 |
| Total | Σf = 100 | Σfu = 19 |
By formula,
Mean = A + $\frac{Σfu}{Σf} \times i$
= 6 + $\frac{19}{100} \times 4$
= 6 + $\frac{76}{100}$
= 6 + 0.76
= 6.76
Hence, mean = 6.76