Question

The following are the scores made by two batsmen A and B in a series of innings:

A	:	12	115	6	73	7	19	119	36	84	29
B	:	47	12	76	42	4	51	37	48	13	0

Who is better as a run-getter ? Who is more consistent?

Answer 1

For Batsman A

S. No.	Score (XA)	Deviation from mean $\left(x_{\mathrm{A}}=X_{\mathit{A}}\mathit{-}{\overline{\mathit{X}}}_{\mathit{A}}\right)$	$x_{\mathrm{A}}^2$
1 2 3 4 5 6 7 8 9 10	12 115 6 73 7 19 119 36 84 29	−38 65 −44 23 −43 −31 69 −14 34 −21	144 4225 1936 529 1849 961 4761 196 1156 441
N = 10	ΣxA = 500		$\mathrm{\Sigma }{x}_{\mathrm{A}}^2=17498$

$$\begin{gathered} \overline{X}=\frac{\Sigma X}{N}=\frac{500}{10}=50 \\ SD\left({\sigma }_A\right)=\sqrt{\frac{\mathrm{\Sigma }{x}_{\mathrm{A}}^2}{\mathrm{N}}}=\sqrt{\frac{17498}{10}}=41.83 \\ \mathrm{Coefficient}\mathrm{of}\mathrm{Variation}=\frac{{\sigma }_{\mathrm{A}}}{{\overline{\mathit{X}}}_{\mathrm{A}}}\times 100=\frac{41.83}{50}\times 100=83.66\% \end{gathered}$$

For Batsman B

S. No.	Score XB	Deviation from Mean $\left(x_{\mathrm{B}}=X_{\mathrm{B}}-{\overline{\mathit{X}}}_{\mathrm{B}}\right)$	$x_{\mathrm{B}}^2$
1 2 3 4 5 6 7 8 9 10	47 12 76 42 4 51 37 48 13 0	14 −21 43 9 −29 18 4 15 −20 −33	196 441 1849 81 841 324 16 225 400 1089
N = 10	ΣxB = 330		$\mathrm{\Sigma }{x}_{\mathrm{B}}^2=5462$

$$\begin{gathered} {\overline{\mathit{X}}}_{\mathrm{B}}=\frac{\Sigma X}{N}=\frac{330}{10}=33 \\ SD\left({\sigma }_B\right)=\sqrt{\frac{\mathrm{\Sigma }{x}_{\mathrm{B}}^2}{N}}=\sqrt{\frac{5462}{10}}=\sqrt{546.2}=23.37 \\ \mathrm{Coefficient}\mathrm{of}\mathrm{Variation}=\frac{{\sigma }_B}{{\overline{\mathit{X}}}_{\mathrm{B}}}\times 100=\frac{23.37}{33}\times 100=70.82\% \end{gathered}$$
Since average score of A is more than B, he is better as run-getter but since, coefficient of variance is less for B, therefore B is more consistent.