为自己加油

设X的分布列为：
X-1023
p1/81/43/81/4
求：(1)E(X)；
(2)E(X²)；
(3)E(-2X+1)．

(1)E(X)=∑⁴_i=1x_iP_i=[(-1)•1/8]+[0•(1/4)]+[2•(3/8)]+[3•(1/4)] =11/8 (2)E(X²)=∑⁴_i=1x_²iP_i =[(-1)²•(1/8)]+[0²•(1/4)]+[2²•(3/8)]+[3²•(1/4)]=31/8． (3)E(-2X+1)=∑(-2x_i+1)•p_i=[3•(1/8)]+[1•(1/4)]-[3•(3/8)]-[5•(1/4)]=-7/4．