stat 7 hw

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)
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This note was uploaded on 06/22/2008 for the course STAT 101 taught by Professor Heller during the Fall '08 term at UPenn.

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

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