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)．
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
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)．
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
