Question

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.

Answer 1

The standard deviation of annual losses can be determined using the formula σ = √(∑((x - μ)^2) / n). In this case, the mean loss is $1410 and the standard deviation is approximately $13,967.61.

Step-by-step explanation:

The standard deviation of annual losses incurred by the automobile owner before insurance reimbursement can be determined using the formula:

Standard Deviation formula:

σ = √(∑((x - μ)^2) / n)

Where:

• σ is the standard deviation
• x is the value of each loss
• μ is the mean loss
• n is the number of losses

In this case, the losses are $100, $1000, and $15000. The mean loss can be calculated as:

μ = (60 * $100 + 30 * $1000 + 10 * $15000) / 100 = $1410

Using this mean loss, the standard deviation can be calculated as:

σ = √(((60 * ($100 - $1410)^2) + (30 * ($1000 - $1410)^2) + (10 * ($15000 - $1410)^2)) / 100)

σ = √(((60 * (-$1310)^2) + (30 * (-$410)^2) + (10 * ($13590)^2)) / 100)

σ = √(((60 * $1716100) + (30 * $168100) + (10 * $184259100)) / 100)

σ = √((102966600 + 5043000 + 1842591000) / 100)

σ = √(1957474600 / 100) ≈ $13,967.61

Answer 2

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

Step-by-step 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