Chapter 3 - Section 3-4 - Mean and Variance of a Discrete Random Variable - Exercises - Page 77: 3-58

μ=1.3333, σ²=1.139

Given, $P(x=0)=2/6,P(x=1.5)=2/6,P(x=2)=1/6,P(x=3)=1/6$, then, $μ=E(x)=∑xP(x)=0×\frac{2}{6}+1.5×\frac{2}{6}+2×\frac{1}{6}+3×\frac{1}{6}=4/3\approx1.3333$, $σ²=V(x)=∑x²P(x)-μ²=0²×\frac{2}{6}+1.5²×\frac{2}{6}+2²×\frac{1}{6}+3²×\frac{1}{6}-\frac{4}{3}^{2}=41/36\approx1.139$