企源知识库

设离散型随机蛮量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 所以 D(X)=E(X²)-[E(X)]²=1.025-0.45²=0．8225