An automobile owner has 40% chance of having exactly one accident in a year and 60% chance of having no accidents in a year. There is no chance that the automobile owner incurs more than one accident in a year.
If there is an accident, the loss amount has the following distribution:
Loss amount & Prob
30 0.3
60 0.15
100 0.45
180 0.1
Suppose there is an ordinary deductible of $40 and the maximum payment by the insurer is $130.
Determine the standard deviation of the payment made by the insurer to the automobile owner.
Answer:
The answer is 32.437
Explanation:
Let X be the loss and let Y be the payment by the insurer.
Hence, we have Y = o if x<= 40,
x-40 if 40 < x < 130
130 if x > = 170
Therefore, Var (y) = E (y^2) - [E(y)]^2
E(y) = 0.4[130 (0.1)+60 (0.45) +20 (0.15)] = 17.2
E(^2) = 0.4 [130^2(0.1) + 60^2(0.45) +20^2(0.15)] = 1348
Finally, Var(y) = 1348 -17.2^2 = 1052.16
Sqrt(1052.16) = 32.437
Therefore, the standard deviation of the payment made by the insurer to the automobile owner is 32.437
