...分布列,试用概率平均的方法计算x的数学期望和方差?
发布网友
发布时间：2024-08-18 14:14
共1个回答
热心网友
时间：2024-08-28 11:43
P(X=0) = (0.8)^5 =0.32768
P(X=1) = C(5,1)(0.2)(0.8)^4 =0.4096
P(X=2) = C(5,2)(0.2)^2.(0.8)^3 =0.2048
P(X=3) = C(5,3)(0.2)^3.(0.8)^2 =0.0512
P(X=4) = C(5,4)(0.2)^4.(0.8) =0.0064
P(X=5) = (0.2)^5 =0.00032
E(X)
=(0)(0.32768)+(1)(0.4096)+2(0.2048)+3(0.0512)+4(0.0064)+5(0.00032)
=1
E(X^2)
=(0^2)(0.32768)+(1^2)(0.4096)+(2^2)(0.2048)+(3^2)(0.0512)
+(4^2)(0.0064)+(5^2)(0.00032)
=1.8
D(X)
=E(X^2) -[E(X)]^2
=1.8-1
=0.8
