Suppose there is an ordinary deductible of $40 and the maximum payment by the insurer is $130. Determine the standard deviation of annual losses incurred by the automobile owner before the insurance reimbursement.

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.

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

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