The owner of an automobile insures it against damage by purchasing an insurance policy with a deductible of 250. In the event that the automobile is damaged, repair costs can be modeled by a uniform random variable on the interval (0, 1500).

Calculate the standard deviation of the insurance payment in the event that the automobile is damaged.

• 361
• 403
• 433
• 464
• 521

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: B

Let [math]X[/math] and [math]Y[/math] denote repair cost and insurance payment, respectively, in the event the auto is damaged. Then

[[math]] Y = \begin{cases} 0, \quad x \leq 250 \\ x-250, \quad x \gt 250 \end{cases} [[/math]]

and

[[math]] \operatorname{E}[Y] = \int_{250}^{1500} \frac{1}{1500} (x-250) dx = \frac{1}{3000} (x-250)^2 \Big |_{250}^{1500} = \frac{1250^2}{3000} = 521 [[/math]]

[[math]] \operatorname{E}[Y^2] = \int_{250}^{1500} \frac{1}{1500} (x-250)^2 dx = \frac{1}{3000} (x-250)^3 \Big |_{250}^{1500} = \frac{1250^3}{4500} = 434,028 [[/math]]

[[math]] \operatorname{Var}[Y] = \operatorname{E}[Y^2] - (\operatorname{E}[Y])^2 = 434028 - (521)^2 [[/math]]

[[math]] \sqrt{\operatorname{Var}[Y]} = 403. [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.