Question

The following is the record of goals scored by team $A$ in a football session:

No.of goals scored	$0$	$1$	$2$	$3$	$4$
No.of matches	$1$	$9$	$7$	$5$	$3$

For the team $B$, mean number of goals scored per match was $2$ with a standard deviation $1.25$ goals. Find which team may be considered more consistent?

Answer 1

Finding mean and variance of team $A$

No.of goals scored $\left(x_i\right)$	No.of matches $\left(f_i\right)$	$x_i{f}_i$	$x_i^2$	$f_i{x}_i^2$
$0$	$1$	$0$	$0$	$0$
$1$	$9$	$9$	$1$	$9$
$2$	$7$	$14$	$4$	$28$
$3$	$5$	$15$	$9$	$45$
$4$	$3$	$12$	$16$	$48$
	$\sum f_i=25$	$\sum x_i{f}_i=50$		$\sum f_i{x}_i^2=130$

Mean $\left({\overline{x}}_A\right)=\frac{\sum f_1{x}_1}{\sum f_i}=\frac{50}{25}=2$
Variance $\left({\sigma }_A^2\right)$
$=\frac{1}{(\sum f_i{)}^2}[(\sum f_i).(\sum f_i{x}_i^2)-{(\sum f_i{x}_i)}^2]$
$\left({\sigma }_A^2\right)=\frac{1}{{25}^2}[25\times 130-(50{)}^2]$
$\left({\sigma }_A^2\right)=\frac{1}{625}[3250-2500]$
$\left({\sigma }_A^2\right)=\frac{1}{625}\times 750=1.2$

Finding standard deviation for team $A$
Standard deviation $\left({\sigma }_A\right)=\sqrt{\text{Variance}}$
$\left({\sigma }_A\right)=\sqrt{1.2}$
$\left({\sigma }_A\right)=1.09$

Finding coefficient of variance for team $A$
$C.V_A=\frac{{\sigma }_A}{{\overline{x}}_A}\times 100$
$C.V_A=\frac{1.09}{2}\times 100$
$C.V_A=54.5$

Finding coefficient of variance for team $B$
Mean $\left({\overline{x}}_B\right)=2$ (Given)
Standard deviation $\left({\sigma }_B\right)=1.25$ (Given)
$C.V_B=\frac{{\sigma }_B}{{\overline{x}}_B}\times 100$
$C.V_B=\frac{1.25}{2}\times 100$
$C.V_B=62.5$
$\therefore C.V_B>C.V_A$
Team $A$ is more consistent than $B$.