设随机变量X,Y,Z相互独立,且X~N(4,5),Y~N(-2,9),Z~N(2,2),则P{0≤X+Y-Z≤3}=_______.(Ф()=0.7734)

admin2019-01-05  32

问题 设随机变量X,Y,Z相互独立,且X~N(4,5),Y~N(-2,9),Z~N(2,2),则P{0≤X+Y-Z≤3}=_______.(Ф()=0.7734)

选项

答案0.2734

解析 E(X+Y-Z)=EX+EY-EZ=4-2-2=0,D(X+Y-Z)=DX+DY+DZ=5+9+2=16,
    ∴X+Y-Z~N(0,16),故
    P{0≤X+Y-Z≤3}=-Ф(0)=0.7734-0.5=0.2734.
转载请注明原文地址:https://kaotiyun.com/show/9YW4777K
0

随机试题
最新回复(0)