设X,Y相互独立,且X~B(3,),Y~N(0,1),令U=max(X,Y),求P{1<U≤1.96}(其中Ф(1)=0.841,Ф(1.96)=0.975).

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问题 设X,Y相互独立,且X~B(3,),Y~N(0,1),令U=max(X,Y),求P{1<U≤1.96}(其中Ф(1)=0.841,Ф(1.96)=0.975).

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答案P(U≤u)=P{max(X,Y)≤u)=P{X≤u,Y≤u}=P(X≤u)P(Y≤u), P(U≤1.96)=P(X≤1.96)P(Y≤1.96)=[P(X=0)+P(X=1)]P(Y≤1.96) [*]Ф(1.96)=0.4875 P(U≤1)=P(X≤1)P(Y≤1)=[*]×Ф(1)=0.4205, 则P(1<U≤1.96)=P(U≤1.96)-P(U≤1)=0.067.

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