A) A
B) B
C) both are equally consistent
D) None of the above
Correct Answer: A
Solution :
Given that, mean and variance of players A and B are \[\overline{{{X}_{1}}}=50\sigma _{1}^{2}=36\,\,or\,\,{{\sigma }_{1}}=6\] and \[\overline{{{X}_{2}}}=60\sigma _{2}^{2}=81\,\,or\,\,{{\sigma }_{2}}=9\] From the following formula, we can check consistency coefficient of variation \[=\frac{SD}{Mean}\times 100\] Therefore, coefficient of variation of player A \[=\frac{{{\sigma }_{1}}}{{{X}_{2}}}\times 100=\frac{6}{50}\times 100=12%\] and coefficient of variation of player B \[=\frac{{{\sigma }_{2}}}{{{X}_{2}}}=\frac{9}{60}\times 100=15%\] Since, coefficient of variation of player A is less, hence it is more consistent in comparison to player B.
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