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

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 $B$	$10$	$20$	$30$	$25$	$43$	$15$	$7$

Finding Mean and Variance for Group $A$
Let, assumed mean $\left(a\right)=45$
$h=\text{class size}=10$
$d_i=$ step deviation
For Marks $(10-20),f_i=9$,
$x_i=\frac{10+20}{2}=15,$
$d_i=\frac{x_i-a}{h}=\frac{15-45}{10}=-3$
$d_i^2=(-3{)}^2=9$
$f_i{d}_i=9\times -3=-27$
$f_i{d}_i^2=9\times 9=81$

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

Mean $\left(\overline{x_A}\right)=\text{assumed mean}+\frac{\sum f_i{d}_i}{\sum f_i}\times h$
Mean $\left(\overline{x_A}\right)=45-\frac{6}{150}\times 10$
Mean $\left(\overline{x_A}\right)=45-\frac{2}{5}$
Mean $\left(\overline{x_A}\right)=44.6$
Variance $\left({\sigma }_A^2\right)=\frac{h^2}{(\sum f_i{)}^2}[(\sum f_i).(\sum f_i{d}_i^2)-{(\sum f_i{d}_i)}^2]$
$\left({\sigma }_A^2\right)=\frac{(10{)}^2}{(150{)}^2}[150\times (342)-(-6{)}^2]$
$\left({\sigma }_A^2\right)=\frac{100}{150\times 150}[51300-36]$
$\left({\sigma }_A^2\right)=\frac{1}{225}\times 51264=227.84$

Finding Standard deviation for $\text{Group}A$
Standard deviation $\left({\sigma }_A\right)$
$\left({\sigma }_A\right)=\sqrt{\text{Variance}}$
$\left({\sigma }_A\right)=\sqrt{227.84}=15.09$

Finding Coefficient of Variation for $\text{Group}A$
Coefficient of Variation
$(C.V{.}_A)=\frac{{\sigma }_A}{{\overline{x}}_A}\times 100$
$(C.V{.}_A)=\frac{15.09}{44.6}\times 100$
$(C.V{.}_A)=33.83$


Finding Mean and Variance for Group $B$
Let,
$a=\text{assumed mean}=45$
$h=\text{width of class-intervals}=10$
$d_i=$ step deviation

For class $(10-20),x_i=\frac{10+20}{2}=15,$
$d_i=\frac{x_i-a}{h}=\frac{15-45}{10}=-3$
$d_i^2=(-3{)}^2=9$
$f_i{d}_i=10\times (-3)=-30$
$f_i{d}_i^2=10\times (-3{)}^2=90$

Marks	Group $B\left(f_i\right)$	Mid point $(x_i$)	$d_i=\frac{x_i-a}{h}$	$d_i^2$	$f_i{d}_i$	$f_i{d}_i^2$
$10-20$	$10$	$15$	$-3$	$9$	$-30$	$90$
$20-30$	$20$	$25$	$-2$	$4$	$-40$	$80$
$30-40$	$30$	$35$	$-1$	$1$	$-30$	$30$
$40-50$	$25$	$45$	$0$	$0$	$0$	$0$
$50-60$	$43$	$55$	$1$	$1$	$43$	$43$
$60-70$	$15$	$65$	$2$	$4$	$30$	$60$
$70-80$	$7$	$75$	$3$	$9$	$21$	$63$
	$\sum f_i=150$				$\sum f_i{d}_i=-6$	$\sum f_i{d}_i^2=366$

Mean $\left(\overline{x_B}\right)=\text{assumed mean}+\frac{\sum f_i{d}_i}{\sum f_i}\times h$
Mean $\left(\overline{x_B}\right)=45-\frac{6}{150}\times 10$
Mean $\left(\overline{x_B}\right)=45-\frac{2}{5}$
Mean $\left(\overline{x_B}\right)=44.6$
Variance $\left({\sigma }_B^2\right)=\frac{h^2}{(\sum f_i{)}^2}[(\sum f_i).(\sum f_i{d}_i^2)-{(\sum f_i{d}_i)}^2]$
$\left({\sigma }_B^2\right)=\frac{(10{)}^2}{(150{)}^2}[150\times (366)-(-6{)}^2]$
$\left({\sigma }_B^2\right)=\frac{100}{150\times 150}[54900-36]$
$\left({\sigma }_B^2\right)=\frac{1}{225}\times 54864=243.84$

Finding Standard deviation for $\text{Group}B$
Standard deviation $\left(\sigma B\right)$
$\left({\sigma }_B\right)=\sqrt{\text{Variance}}$
$\left({\sigma }_B\right)=\sqrt{243.84}$
$\left({\sigma }_B\right)=15.62$

Finding Coefficient of Variation for $\text{Group}B$
Coefficient of Variation
$(C.V{.}_B)=\frac{{\sigma }_B}{{\overline{x}}_B}\times 100$
$=\frac{15.62}{44.6}\times 100$
$=35.02$
$\because C.V{.}_B=35.02\text{and}(C.V{.}_A)=33.83$
$\Rightarrow C.V{.}_B>C.V{.}_A$
$\therefore \text{Group}B$ is more variable.