| Height in cms | Number of boys |
| $95-105$ | $9$ |
| $105-115$ | $13$ |
| $115-125$ | $26$ |
| $125-135$ | $30$ |
| $135-145$ | $12$ |
| $145-155$ | $10$ |
| height in cms | Number of boys ${f_i}$ | Mid-point ${x_i}$ | ${f_i}{x_i}$ | $\left| {{x_i} - \bar x} \right|$ | ${f_i}\left| {{x_i} - \bar x} \right|$ |
| $95-105$ | $9$ | $100$ | $900$ | $25.3$ | $227.7$ |
| $105-115$ | $13$ | $110$ | $1430$ | $15.3$ | $198.9$ |
| $115-125$ | $26$ | $120$ | $3120$ | $5.3$ | $137.8$ |
| $125-135$ | $30$ | $130$ | $3900$ | $4.7$ | $141$ |
| $135-145$ | $12$ | $140$ | $1680$ | $14.7$ | $176.4$ |
| $145-155$ | $10$ | $150$ | $1500$ | $24.7$ | $247$ |
Here, $N = \sum\limits_{i = 1}^6 {{f_i}} = 100,\sum\limits_{i = 1}^6 {{f_i}{x_i}} = 12530$
$\therefore \bar x = \frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}{x_i}} = \frac{1}{{100}} \times 12530 = 125.3$
$M.D.\left( {\bar x} \right) = \frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - \bar x} \right|} = \frac{1}{{100}} \times 1128.8 = 11.28$
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