设离散型随机变量X的分布律为 X-100.512 P0.10.50.10.10.2 求E(X)，E(X2)，D(X)．

设离散型随机变量X的分布律为
X-100.512
P0.10.50.10.10.2
求E(X)，E(X²)，D(X)．
【正确答案】：E(X)=-1×0.1+0×0.5+0.5×0.1+1×0.1+2×0.2=0.45 E(X²)=1²×0．1+0²×0.5+0.5²×0.1+1²×0.1+2²×0.2=1.025 ∴D(X)=E(²)-[E(X)]²=1.025-0.45²=0.8225