# Chapter 3 and 4 60 113 random variables recap adding

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Chapter 3 and 460 / 113
Random variablesRecapAdding or multiplying?A company has 5 Lincoln Town Cars in its fleet. Historical data showthat annual maintenance cost for each car is on average \$2,154 witha standard deviation of \$132.What is the mean and the standarddeviation of the total annual maintenance cost for this fleet?Note that we have 5 cars each with the given annual maintenancecost(X1+X2+X3+X4+X5), not one car that had 5 times the givenannual maintenance cost(5X).Chapter 3 and 460 / 113
Random variablesRecapAdding or multiplying?A company has 5 Lincoln Town Cars in its fleet. Historical data showthat annual maintenance cost for each car is on average \$2,154 witha standard deviation of \$132.What is the mean and the standarddeviation of the total annual maintenance cost for this fleet?Note that we have 5 cars each with the given annual maintenancecost(X1+X2+X3+X4+X5), not one car that had 5 times the givenannual maintenance cost(5X).E(X1+X2+X3+X4+X5)=E(X1)+E(X2)+E(X3)+E(X4)+E(X5)Chapter 3 and 460 / 113
Random variablesRecapAdding or multiplying?A company has 5 Lincoln Town Cars in its fleet. Historical data showthat annual maintenance cost for each car is on average \$2,154 witha standard deviation of \$132.What is the mean and the standarddeviation of the total annual maintenance cost for this fleet?Note that we have 5 cars each with the given annual maintenancecost(X1+X2+X3+X4+X5), not one car that had 5 times the givenannual maintenance cost(5X).E(X1+X2+X3+X4+X5)=E(X1)+E(X2)+E(X3)+E(X4)+E(X5)=5×E(X)=5×2,154=\$10,770Chapter 3 and 460 / 113
Random variablesRecapAdding or multiplying?A company has 5 Lincoln Town Cars in its fleet. Historical data showthat annual maintenance cost for each car is on average \$2,154 witha standard deviation of \$132.What is the mean and the standarddeviation of the total annual maintenance cost for this fleet?Note that we have 5 cars each with the given annual maintenancecost(X1+X2+X3+X4+X5), not one car that had 5 times the givenannual maintenance cost(5X).E(X1+X2+X3+X4+X5)=E(X1)+E(X2)+E(X3)+E(X4)+E(X5)=5×E(X)=5×2,154=\$10,770Var(X1+X2+X3+X4+X5)=Var(X1)+Var(X2)+Var(X3)+Var(X4)+Var(X5)Chapter 3 and 460 / 113
Random variablesRecapAdding or multiplying?A company has 5 Lincoln Town Cars in its fleet. Historical data showthat annual maintenance cost for each car is on average \$2,154 witha standard deviation of \$132.What is the mean and the standarddeviation of the total annual maintenance cost for this fleet?Note that we have 5 cars each with the given annual maintenancecost(X1+X2+X3+X4+X5), not one car that had 5 times the givenannual maintenance cost(5X).E(X1+X2+X3+X4+X5)=E(X1)+E(X2)+E(X3)+E(X4)+E(X5)=5×E(X)=5×2,154=\$10,770Var(X1+X2+X3+X4+X5)=Var(X1)+Var(X2)+Var(X3)+Var(X4)+Var(X5)=5×V(X)=5×1322=\$287,120Chapter 3 and 460 / 113
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