Unformatted text preview: f) (5 points) Find the variance of Y? a) E(X)=0.8*0+1*0.1+2*0.05+3*0.03+4*0.02=0.37 b) E(X 2 )=0.8*0+0.1*1 2 +0.05*2 2 +0.03*3 2 +0.02*4 2 =0.89 Var(X)= E(X 2 )E(X)E(X)=0.890. 37 2 =0.7531 c) E((XE(X)) 3 )=0.8(00.37) 3 +0.1(10.37) 3 +0.05(20.37) 3 +0.03(30.37) 3 +0.02(40.37) 3 = =1.703 skewness=1.703\(0.7531) 3/2 =2.6058 d) MX(t)=0.8+0.1e 1t +0.05e 2t +0.03e 3t +0.02 e 4t e) X=0 => Y=0, X=1 => Y=0, X=2 => Y=10, X=3 =>Y=36, X=4=> Y=84 So E(Y)= 0.8*0+0*0.1+10*0.05+36*0.03+84*0.02=3.26 f) E(Y 2 )=0.8*0+0.1*0+0.05*100+0.03*1296+7056*0.02=185 So Var(Y)=1853.26*3.66=174.37...
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 Fall '08
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 Probability, Variance, Probability theory, Notes exam Exercise

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