નીચે આપેલ માહિતી પરથી બતાવો કે $A$ અને $B$ માંથી કયા સમૂહમાં વધારે ચલન છે?

ગુણ	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
સમૂહ  $A$	$9$	$17$	$32$	$33$	$40$	$10$	$9$
સમૂહ $B$	$10$	$20$	$30$	$25$	$43$	$15$	$7$

Firstly, the standard deviation of group $A$ is calculated as follows.

Marks	Group $A$ ${f_i}$	Mid-point ${x_i}$	${y_i} = \frac{{{x_i} - 45}}{{10}}$	${y_i}^2$	${f_i}{y_i}$	${f_i}{y_i}^2$
$10-20$	$9$	$15$	$-3$	$9$	$-27$	$81$
$20-30$	$17$	$25$	$-2$	$4$	$-34$	$68$
$30-40$	$32$	$35$	$-1$	$1$	$-32$	$32$
$40-50$	$33$	$45$	$0$	$0$	$0$	$0$
$50-60$	$40$	$55$	$1$	$1$	$40$	$40$
$60-70$	$10$	$65$	$2$	$4$	$20$	$40$
$70-80$	$9$	$75$	$3$	$9$	$27$	$81$
	$150$				$-6$	$342$

Here, $h =10, N =150, A =45$

Mean $ = A + \frac{{\sum\limits_{i = 1}^7 {{x_i}} }}{N} \times h$

$ = 45 + \frac{{\left( { - 6} \right) \times 10}}{{150}} \times 45 - 0.4 = 44.6$

$\sigma _1^2 = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^7 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^7 {{f_i}{y_i}} } \right)}^2}} } \right]$

$=\frac{100}{22500}\left[150 \times 342-(-6)^{2}\right]$

$=\frac{1}{225}(51264)$

$=227.84$

$\therefore$ Standard deviation $\left(\sigma_{1}\right)=\sqrt{227.84}=15.09$

The standard deviation of group $B$ is calculated as follows.

Marks	Group $A$ ${f_i}$	Mid-point ${x_i}$	${y_i} = \frac{{{x_i} - 45}}{{10}}$	${y_i}^2$	${f_i}{y_i}$	${f_i}{y_i}^2$
$10-20$	$9$	$15$	$-3$	$9$	$9$	$-30$
$20-30$	$17$	$25$	$-2$	$4$	$4$	$-40$
$30-40$	$32$	$35$	$-1$	$1$	$1$	$-30$
$40-50$	$33$	$45$	$0$	$0$	$0$	$0$
$50-60$	$40$	$55$	$1$	$1$	$1$	$43$
$60-70$	$10$	$65$	$2$	$4$	$4$	$30$
$70-80$	$9$	$75$	$3$	$9$	$9$	$21$
	$150$					$-6$

Mean  $ = A + \frac{{\sum\limits_{i = 1}^7 {{f_i}{y_i}} }}{N} \times h$

$ = 45 + \frac{{\left( { - 6} \right) \times 10}}{{150}} \times 45 - 0.4 = 44.6$

$\sigma _2^2 = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^7 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^7 {{f_i}{y_i}} } \right)}^2}} } \right]$

$=\frac{100}{22500}\left[150 \times 366-(-6)^{2}\right]$

$=\frac{1}{225}(54864)=243.84$

$\therefore$ Standard deviation $\left(\sigma_{1}\right)=\sqrt{243.84}=15.61$

Since the mean of both the groups is same, the group with greater standard deviation will be more variable.

Thus, group $B$ has more variability in the marks.