Answer
$P(4)=0.345$ and $P(6)=0.23$
$\mu= 3.485$
$\sigma^2= 3.82$
$\sigma=1.95$
Work Step by Step
Given $P(6)=\frac{2}{3}P(4)$, as $P(X)$ forms a probability distribution,
we have $\sum P(X)=1$. Thus
$0.23+0.18+P(4)+\frac{2}{3}P(4)+0.015=1$,
we can find that $P(4)=0.345$ and $P(6)=0.23$
The mean, variance, and standard deviation can be calculated based on formulas as the following
$\mu= (0.23×1+0.18×2+0.345×4+0.23×6+0.015×9) / (0.23+0.18+0.345+0.23+0.015) = 3.485$
$\sigma^2=(0.23×(1-3.485)²+0.18×(2-3.485)²+0.345×(4-3.485)²+0.23×(6-3.485)²+0.015×(9-3.485)²) / (0.23+0.18+0.345+0.23+0.015) = 3.82$
$\sigma=\sqrt {\sigma^2}=1.95$
