For the following probability distribution:

The mean, variance and standard deviation respectively are:

3.8, 0.76 and 0.87

The correct answer is Option (4) → 3.8, 0.76 and 0.87

$E(X)=∑(X.P(X))$

$=(3×0.5)+(4×0.2)+(5×0.3)$

$=1.5+0.8+1.5=3.8$

$Var(X)=E(X^2)-(E(X))^2$

$E(X^2)=∑(X^2.P(X))$

$=(3^2×0.5)+(4^2×0.2)+(5^2×0.3)$

$=4.5+3.2+7.5=15.2$

$Var(X)=15.2-(3.8)^2$

$=15.2-14.44$

$=0.76$

Standard Deviation = $SD(X)$

$SD(X)=\sqrt{Var(X)}$

$=\sqrt{0.76}≈0.8718$