The sum of the squares deviations for 10 observations taken from their mean 50 is 250. The coefficient of variation is
10%
We have :
¯¯¯¯¯X=50,n=10
∑10i=1(xi−¯¯¯¯¯X)2=250
∴SD=√Variance of X =√∑10i=1(xi−¯¯¯¯X)2n=√25010=5
Using CV =σ¯¯¯¯X×100⇒CV=550×100=10%