设X~B(3,0.4),求下列随机变量的数学期望
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解决时间 2021-01-24 04:17
- 提问者网友：送舟行
- 2021-01-23 19:32
设X~B(3,0.4),求下列随机变量的数学期望
最佳答案
- 五星知识达人网友：怀裏藏嬌
- 2021-01-23 21:10
已知X～B(3, 0.4),则X的概率分布为
X 0 1 2 3
pk 0.216 0.432 0.288 0.064
∴
E(X1)=E(X^2)=0×0.216+1×0.432+4×0.288+9×0.064=2.16 .
E(X2)=E(X^2-2X)=0×0.216+(-1)×0.432+0×0.288+3×0.064=-0.24 .追问第三问咩~追答X 0 1 2 3
X(3-X)/2) 0 1 1 0
pk 0.216 0.432 0.288 0.064
E(X3)=E(X(3-X)/2)=0×0.216+1×0.432+1×0.288+0×0.064=0.72X 0 1 2 3
X(3-X)/2) 0 1 1 0
pk 0.216 0.432 0.288 0.064
E(X3)=E(X(3-X)/2)=0×0.216+1×0.432+1×0.288+0×0.064=0.72X 0 1 2 3
X(3-X)/2) 0 1 1 0
pk 0.216 0.432 0.288 0.064
E(X3)=E(X(3-X)/2)=0×0.216+1×0.432+1×0.288+0×0.064=0.72X 0 1 2 3
X(3-X)/2) 0 1 1 0
pk 0.216 0.432 0.288 0.064
E(X3)=E(X(3-X)/2)=0×0.216+1×0.432+1×0.288+0×0.064=0.72X 0 1 2 3
X(3-X)/2) 0 1 1 0
pk 0.216 0.432 0.288 0.064
E(X3)=E(X(3-X)/2)=0×0.216+1×0.432+1×0.288+0×0.064=0.72
X 0 1 2 3
pk 0.216 0.432 0.288 0.064
∴
E(X1)=E(X^2)=0×0.216+1×0.432+4×0.288+9×0.064=2.16 .
E(X2)=E(X^2-2X)=0×0.216+(-1)×0.432+0×0.288+3×0.064=-0.24 .追问第三问咩~追答X 0 1 2 3
X(3-X)/2) 0 1 1 0
pk 0.216 0.432 0.288 0.064
E(X3)=E(X(3-X)/2)=0×0.216+1×0.432+1×0.288+0×0.064=0.72X 0 1 2 3
X(3-X)/2) 0 1 1 0
pk 0.216 0.432 0.288 0.064
E(X3)=E(X(3-X)/2)=0×0.216+1×0.432+1×0.288+0×0.064=0.72X 0 1 2 3
X(3-X)/2) 0 1 1 0
pk 0.216 0.432 0.288 0.064
E(X3)=E(X(3-X)/2)=0×0.216+1×0.432+1×0.288+0×0.064=0.72X 0 1 2 3
X(3-X)/2) 0 1 1 0
pk 0.216 0.432 0.288 0.064
E(X3)=E(X(3-X)/2)=0×0.216+1×0.432+1×0.288+0×0.064=0.72X 0 1 2 3
X(3-X)/2) 0 1 1 0
pk 0.216 0.432 0.288 0.064
E(X3)=E(X(3-X)/2)=0×0.216+1×0.432+1×0.288+0×0.064=0.72
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