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设离散型随机变量X的分布律为
X-100.512
P0.10.50.10.10.2
求E(X),E(X
2
),D(X).
分类:
概率论与数理统计(经管类)(04183)
发表:2024年09月12日 09时09分54秒
作者:
admin
阅读:
(12)
设离散型随机变量X的分布律为
X-100.512
P0.10.50.10.10.2
求E(X),E(X
2
),D(X).
【正确答案】:E(X)=-1×0.1+0×0.5+0.5×0.1+1×0.1+2×0.2=0.4 5, E(X
2
)=1
2
×0.1+0
2
×0.5+0.5
2
×0.1+1
2
×0.1+2
2
×0.2 =1.025, D(X)=E(X
2
)-[E(X)]
2
=1.025-0.45
2
=0.8225.
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