Question

6. Suppose you just bought a new sports car. Based on historical data of the city you live in, there is a 5% chance that a car gets severely damaged within the first five years of purchase. If your car gets severely damaged, your estimated loss is $50,000. Now you have the option of buying a premium insurance plan that can cover your loss if your car gets severely damaged within the next five years. The insurance plan costs $2,000 total and it covers your loss with 95% probability (i.e., if you severely damage your car within the first five years, with 95% chance, the insurance company covers your loss). a. (5 points) What is your expected loss if you don't get the insurance? What is the standard deviation of the loss? b. (5 points) What is your expected loss if you get the insurance? What is the standard deviation of the loss?

Answer 1

Answer :- Suppose you just bought a new sports car. Based on historical data of the city you live in, there is 5℅ chance that a car get severally damage within the first five years of purchase....

a) Let 'L' be the random variable of loss if insurance isn't bought. The expected value of 'L' will be calculate as :

=> E(L) = 0 × 0.95 + 50000 × 0.05

= 0 + 2500

= 2500

=> E(L^2) = 0^2 × 0.95 + 50000^2 × 0.05

= 0 × 0.95 + 2500000000 × 0.05

= 0 + 125000000

= 125000000

=> Var(L^2) = - [E(L)]^2

= 125000000 - 2500^2

= 125000000 - 6250000

= 118750000

So, standard deviation = Var(L)^0.5

= 118750000^0.5

= 10897

b) Let 'X' be the random variable of the loss if insurance is bought. The expected value of 'X' will be calculate as :

=> E(X) = 2000 × 0.95 + (50000 × 0.05 + 2000) × 0.05

= 1900 + (2500 + 2000) × 0.05

= 1900 + 4500 × 0.05

= 1900 + 225

= 2125

=> E(X^2) = 2000^2 × 0.95 + (50000^2 × 0.05 × 0.05 + 2000^2) × 0.05

= 4000000 × 0.95 + (2500000000 × 0.05 × 0.05 +4000000) × 0.05

= 10250000

So, Var(X) = E(X^2) - [E(X)]^2

= 10250000 - 2125^2

= 10250000 - 4515625

= 5734375

So, standard deviation = Var(X)^0.5

= 5734375^0.5

= 2394.66