From the data given below state which group is more variable, $A$ or $B$ ?

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

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

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.

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.