First,calculate the mean $\bar{x}$:
| $x_i$ | $f_i$ | $f_i x_i$ |
| $5$ | $7$ | $35$ |
| $10$ | $4$ | $40$ |
| $15$ | $6$ | $90$ |
| $20$ | $3$ | $60$ |
| $25$ | $5$ | $125$ |
| Total | $N = 25$ | $\sum f_i x_i = 350$ |
$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{350}{25} = 14$
Now,calculate the mean deviation about the mean:
| $x_i$ | $f_i$ | $|x_i - \bar{x}|$ | $f_i |x_i - \bar{x}|$ |
| $5$ | $7$ | $|5-14| = 9$ | $63$ |
| $10$ | $4$ | $|10-14| = 4$ | $16$ |
| $15$ | $6$ | $|15-14| = 1$ | $6$ |
| $20$ | $3$ | $|20-14| = 6$ | $18$ |
| $25$ | $5$ | $|25-14| = 11$ | $55$ |
| Total | $25$ | - | $158$ |
$M.D.(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{158}{25} = 6.32$