A batsman is to be selected for a cricket team. The choice is between X and Y on the basis of their five previous scores which are:
X25854080120Y5070654580
(a) Calculate coefficient of standard deviation, variance and coefficient of variation.
(b) Which batsman should be selected if we want:
(i) A high run scorer,
(ii) A more reliable batsman in the team.
BatsmanX−¯¯¯¯¯X(x−¯¯¯¯¯X)2BatsmanX−¯¯¯¯Y(X−¯¯¯¯Y)2Score Xxx2Yyy225−45202550−1214485152257086440−309006539801010045−172891205025008018324∑X=350∑x2=5750∑Y=310∑y2=830
Batsman XBatsman YArithmetic mean¯¯¯¯¯X=∑XN¯¯¯¯Y=∑YN∑X=350,N=5∑Y=310N=5¯¯¯¯¯X=3505=70¯¯¯¯Y=3105=62Average score = 70 RunsAverage Score = 62 RunsStandard Deviationσx=√∑x2Nσy=√∑y2N∑x2=5750 and N=5∑y2=830 and N=5σX=√57505=√1150σY=√8305=√166σX=33.91σY=12.88Coefficient of S.DCoefficient of σX=Coefficient of x¯¯¯¯XCoefficient of σY=Coefficient of y¯¯¯¯Y=33.9170=0.484=12.8862=0.207Varianceσ2X=∑x2N=57505=1150σ2Y=∑y2N=8305=166Coefficient of variationC.V.=σX¯¯¯¯X×100=33.9170×100C.V.=σY¯¯¯¯Y×100=12.8862×100=48.44%=20.77%
(b) (i) Batsman X should be selected as a higher run scorer as his average score (70 runs) is greater than the average score of Y (62 runs).
(ii) Batsman Y is a more reliable batsman in the team because his coefficient of variation (C.V = 20.77%) is less than that of batsman X (C.V = 48.44%).