ટૂંકી રીતનો ઉપયોગ કરીને મધ્યક, વિચરણ અને પ્રમાણિત વિચલન શોધો.

ઊંચાઈ  સેમીમાં	$70-75$	$75-80$	$80-85$	$85-90$	$90-95$	$95-100$	$100-105$	$105-110$	$110-115$
બાળકોની  સંખ્યા	$3$	$4$	$7$	$7$	$15$	$9$	$6$	$6$	$3$

Class Interval	Frequency ${f_i}$	Mid-point ${f_i}$	${y_i} = \frac{{{x_i} - 92.5}}{5}$	${y_i}^2$	${f_i}{y_i}$	${f_i}{y_i}^2$
$70-7$	$3$	$72.5$	$-4$	$16$	$-12$	$48$
$75-80$	$4$	$77.5$	$-3$	$9$	$-12$	$36$
$80-85$	$7$	$82.5$	$-2$	$4$	$-14$	$28$
$85-90$	$7$	$87.5$	$-1$	$1$	$-7$	$7$
$90-95$	$15$	$92.5$	$0$	$0$	$0$	$0$
$95-100$	$9$	$97.5$	$1$	$1$	$9$	$9$
$100-105$	$6$	$102.5$	$2$	$4$	$12$	$24$
$105-110$	$6$	$107.5$	$3$	$9$	$18$	$54$
$110-115$	$3$	$112.5$	$4$	$16$	$12$	$48$
	$60$				$6$	$254$

Mean, $\bar x = A + \frac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{N} \times h$

$ = 92.5 + \frac{6}{{60}} \times 5 = 92.5 + 0.5 = 93$

Variance,  $\left( {{\sigma ^2}} \right) = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^9 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)}^2}} } \right]$

$=\frac{(5)^{2}}{(60)^{2}}\left[60 \times 254-(6)^{2}\right]$

$=\frac{25}{3600}(15204)=105.58$

$\therefore$ Standard deviation $(\sigma)=\sqrt{105.58}=10.27$