Independent solution

How to solve this Sums of Independent Random Variables question

Setup

Setup

Translate each coefficient of variation into a standard deviation using the common positive mean.

E[X]=E[Y]=μE[X]=E[Y]=\mu
SD⁡(X)=3μ,SD⁡(Y)=4μ\operatorname{SD}(X)=3\mu,\qquad \operatorname{SD}(Y)=4\mu

Model

Model

Find the mean and variance of the average, using independence for the variance sum.

E[X+Y2]=μE\left[\frac{X+Y}{2}\right]=\mu
Var⁡(X+Y2)=9μ2+16μ24=25μ24\operatorname{Var}\left(\frac{X+Y}{2}\right)=\frac{9\mu^2+16\mu^2}{4}=\frac{25\mu^2}{4}

Compute

Compute

Take the square root and divide by the mean.

SD⁡(X+Y2)=5μ2\operatorname{SD}\left(\frac{X+Y}{2}\right)=\frac{5\mu}{2}
CV⁡(X+Y2)=5μ/2μ=52\operatorname{CV}\left(\frac{X+Y}{2}\right)=\frac{5\mu/2}{\mu}=\frac52

Answer

Answer

The coefficient of variation is 2.5.

52(C)\boxed{\frac52\quad\text{(C)}}