【正确答案】(Ⅰ)EZ = E(X+2Y)=EX+2EY =∫
—∞+∞xf(x)dx+2∫
—∞+∞y.2f(—2y)dy
=∫
—∞+∞xf(x)dx+∫
—∞+∞(一2y)f(一2y)d(一2y)

∫
—∞+∞xf(x)dx+tf(t)dt=0,
由此可知,EZ=0,EY=

EX.又DY=EY
2一(EY)
2,而
EY
2=∫
—∞+∞y
2.2f(一2y)dy=

∫
—∞+∞(一2y)
2f(一2y)d(一2y)

所以 DY=EY
2一(EY)
2=

DZ = D(X+2Y) = DX+4DY+4cov(X,Y)
= DX+4DY+

(Ⅱ)由切比雪夫不等式
P{|Z|≥4} = P{|Z—EZ|≥4}≤
