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 |
By formula,
Mean = A + $\frac{Σfu}{Σf} \times i$
= 52.5 + $\frac{-12}{30}$ × 15
= = 52.5 - $\frac{180}{30}$
= 52.5 - 6
= 46.50
Hence, mean = 46.50.