For a probability distribution of a random variable X, mean is 2.05 and variance is 1.44. Then values of x $(\frac{X}{2})$ and $var (\frac{X}{2})$ respectively are :

1.025, 0.36

The correct answer is Option (4) → 1.025, 0.36

Mean of $X=2.05$

Variance of $X=1.44$

Mean, $E(y)=E(\frac{X}{2})=\frac{1}{2}E(X)$

$=\frac{1}{2}×2.05=1.025$

Variance,

$Var(\frac{X}{2})=(\frac{1}{2})^2Var(X)$

$=\frac{1}{4}×1.44=0.36$