Question

The scores of two batsmen A and B in five innings during a certain match are :

$\begin{array}{cccccc}\text{A}& 32& 28& 47& 63& 71\\ \text{B}& 19& 31& 48& 53& 67\end{array}$

Identify the better player. Who is more consistent?

Answer 1

$\therefore \overline{X}=\frac{\sum X}{N}=\frac{241}{5}=48.2$

$\sigma =\sqrt{\frac{\sum d^2}{N}-{\left(\frac{\sum d}{N}\right)}^2}=\sqrt{\frac{1418}{5}-{\left(\frac{6}{5}\right)}^2}$

$=\sqrt{283.6-14.44}=16.79$

$C.V_A=\frac{\sigma }{\overline{X}}\times 100=\frac{16.79}{48.2}\times 100=34.83\%$

$\therefore \overline{Y}=\frac{\sum Y}{N}=\frac{218}{5}=43.6$

$\sigma =\sqrt{\frac{\sum d^2}{N}-{\left(\frac{\sum d}{N}\right)}^2}=\sqrt{\frac{1516}{5}-{\left(\frac{-22}{5}\right)}^2}$

$=\sqrt{303.2-19.36}=16.84$

$C.V_B=\frac{\sigma }{\overline{Y}}\times 100=\frac{16.84}{43.6}\times 100=38.62\%$

Since, $\overline{X}>\overline{Y}$

A is a better player.

$C.V_A<C.V_B$

Hence, A is more consistent than B.