stat 7 hw - Stat 101-002 Rahmioglu Heller 37479511 Cengiz...

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Stat 101-002 Cengiz Rahmioglu Heller Penn ID: 37479511 STAT 101 HW 1. (a) E(X)=∑xP(x)=0*0.4+1*0.6= 0.6 Var(X) = ∑(x-μ) 2 P(x)= (0-0.6) 2 *(0.4)+ (1-0.6) 2 *(0.6)= 0.144 + 0.096=0.24 (b) E(Y)= ∑yP(y)=0*0.3+1*0.7= 0.7 Var(Y)= ∑(y-μ) 2 P(y)= (0-0.7) 2 *(0.3)+ (1-0.7) 2 *(0.7)= 0.147 + 0.063=0.207 (c) E(X Y)= ∑(xy)P(x,y)= (0*0)(0.1)+ (0*1)(0.3) +(1*0)(0.2) +(1*1)(0.4)= 0.4 E(X Y) gives the expectation of the joint probability function of X and Y. In this case, it is the probability of having a customer who buys both car life and auto insurance. (d) Cov(X,Y)=E(X Y) -μ x μ y = 0.4-0.6*0.7=-0.02 Corr(X, Y)=Cov(X, Y)/σ x σ y = -0.02/(0.24 1/2 *0.207 1/2 )= -0.089 (e) The correlation is found to be -0.089 which means that there is almost no correlation between getting a life insurance and auto insurance. In other words, it can be assumed that whether someone buys one policy cannot be used to predict if they will buy the other. (f) E(750X+300Y)=750E(X)+300E(Y)=$660 Var(750X+300Y)=750 2 Var(X)+300 2 Var(Y)+2*750*300*Cov(X, Y)=153630- 450000(0.02)= 144630dollar 2 Standard deviation=√ Var(750X+300Y)=$380.66 2. (a) The utility should model X and Y as dependent because we expect to
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