设随机变量X与Y相互独立,且X~N(2,1),Y~N(-2,4),Z=3X-2Y+4,求:D(Z) 与 P{Z<=9}
2个回答
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Z还是正态分布,利用公式
E(aX+bY)=aE(X)+bE(Y)
D(aX+bY)=a²D(X)+b²D(Y)
由已知得E(X) = 2, D(X) = 1,E(Y) = -2, D(Y) = 4,则:
E(Z) = E(3X-2Y+4) = 3E(X) - 2E(Y) + E(4) = 14
D(Z) = D(3X-2Y+4) = 9D(X) + 4D(Y) + 0 = 25
所以 Z~N(14,25)
P{Z<=9} = P{(z-14)/5 <= (9-14)/5} = φ(-1) = 1-φ(1) = 0.16
E(aX+bY)=aE(X)+bE(Y)
D(aX+bY)=a²D(X)+b²D(Y)
由已知得E(X) = 2, D(X) = 1,E(Y) = -2, D(Y) = 4,则:
E(Z) = E(3X-2Y+4) = 3E(X) - 2E(Y) + E(4) = 14
D(Z) = D(3X-2Y+4) = 9D(X) + 4D(Y) + 0 = 25
所以 Z~N(14,25)
P{Z<=9} = P{(z-14)/5 <= (9-14)/5} = φ(-1) = 1-φ(1) = 0.16
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