Question

An insurance company offers policies on cars and light trucks, with cars comprising 70% of the company’s accident claims and light trucks 30%. Suppose the damages for cars are normally distributed with mean 2000 and standard deviation 1000, while those for light trucks are normally distributed with mean 4000 and standard deviation 2000. Given a random vehicle insured by the company that has sustained at least 1000 in damages, what is the probability it is a car?

Answer 1

P(X>1000)=P(Car and X>1000)+P(light truck and X>1000)

=0.7*P(Z>(1000-2000)/1000)+0.3*P(Z>(1000-4000)/2000)

=0.7*P(Z>-1)+0.3*P(Z>-1.5)

-0.7*0.8413+0.3*0.9332=0.8689

hence P(Car given X>1000)=P(Car and X>1000)/P(X>1000)

                                       =0.7*0.8413/0.8689=0.6778