Using step deviation method, calculate the mean for the following data:
| Height (in cm) | Number of boys |
|---|---|
| $135 - 140$ | $4$ |
| $140 - 145$ | $9$ |
| $145 - 150$ | $18$ |
| $150 - 155$ | $28$ |
| $155 - 160$ | $24$ |
| $160 - 165$ | $10$ |
| $165 - 170$ | $5$ |
| $170 - 175$ | $2$ |
Class size $= $ Upper class limit $-$ Lower class limit $= h = 5$
Let assumed mean $= A = 152.5$
| Height (in cm) | Number of boys (frequency) $\left(f_i\right)$ | Mid-value $\left(x_i\right)$ | deviation $= x_i - A $ | $ t_i = \dfrac{x_i - A}{h} $ | $ f_i \times t_i $ |
|---|---|---|---|---|---|
| $135 - 140$ | $4$ | $137.5$ | $-15$ | $-3$ | $-12$ |
| $140 - 145$ | $9$ | $142.5$ | $-10$ | $-2$ | $-18$ |
| $145 - 150$ | $18$ | $147.5$ | $-5$ | $-1$ | $-18$ |
| $150 - 155$ | $28$ | $152.5$ | $0$ | $0$ | $0$ |
| $155 - 160$ | $24$ | $157.5$ | $5$ | $1$ | $24$ |
| $160 - 165$ | $10$ | $162.5$ | $10$ | $2$ | $20$ |
| $165 - 170$ | $5$ | $167.5$ | $15$ | $3$ | $15$ |
| $170 - 175$ | $2$ | $172.5$ | $20$ | $4$ | $8$ |
$\Sigma f_i = 100$, $\;$ $\Sigma f_i \times t_i = 19$
Mean $= A + \dfrac{\Sigma f_i \times t_i}{\Sigma f_i} \times h$
i.e. $\;$ Mean $= 152.5 + \dfrac{19}{100} \times 5 = 153.45$