5(iii)
Using step-deviation method, find mean for the following frequency distribution
| Class | Frequency |
|---|---|
| 0-15 | 3 |
| 15-30 | 4 |
| 30-45 | 7 |
| 45-60 | 6 |
| 60-75 | 8 |
| 75-90 | 2 |
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In the given table i is the class interval which is equal to 15.
| Class | Class mark (x) | d = (x - A) | u = d/i | Frequency (f) | fu |
|---|---|---|---|---|---|
| 0-15 | 7.5 | -45 | -3 | 3 | -9 |
| 15-30 | 22.5 | -30 | -2 | 4 | -8 |
| 30-45 | 37.5 | -15 | -1 | 7 | -7 |
| 45-60 | A = 52.5 | 0 | 0 | 6 | 0 |
| 60-75 | 67.5 | 15 | 1 | 8 | 8 |
| 75-90 | 82.5 | 30 | 2 | 2 | 4 |
| Total | Σf = 30 | Σfu = -12 |
Mean = A + $\frac{Σfu}{Σf} \times i = 52.5 + \frac{-12}{30} \times 15$
= $52.5 - \frac{180}{30}$
= 52.5 - 6
= 46.50
Hence, mean = 46.50