Determine the mean of the following frequency distribution by short-cut (assumed mean) method:
| Class Interval | $10 - 16$ | $16 - 22$ | $22 - 28$ | $28 - 34$ | $34 - 40$ |
|---|---|---|---|---|---|
| Frequency | $1$ | $10$ | $5$ | $3$ | $6$ |
| Class Interval | Frequency $f_i$ | Mid value $x_i$ | Assumed Mean $A = 25; \;\; d_i = x_i - A$ | $f_i d_i$ |
|---|---|---|---|---|
| $10 - 16$ | $1$ | $13$ | $-12$ | $-12$ |
| $16 - 22$ | $10$ | $19$ | $-6$ | $-60$ |
| $22 - 28$ | $5$ | $25$ | $0$ | $0$ |
| $28 - 34$ | $3$ | $31$ | $6$ | $18$ |
| $34 - 40$ | $6$ | $37$ | $12$ | $72$ |
$\Sigma f_i = N = 25$, $\;\;$ $\Sigma f_i d_i = 18$
Mean $= A + \dfrac{\Sigma f_i d_i}{\Sigma f_i} = 25 + \dfrac{18}{25} = 25 + 0.72 = 25.72$