The lifetime of a printer costing 200 is exponentially distributed with mean 2 years. The manufacturer agrees to pay a full refund to a buyer if the printer fails during the first year following its purchase, a one-half refund if it fails during the second year, and no refund for failure after the second year.

Calculate the expected total amount of refunds from the sale of 100 printers.

• 6,321
• 7,358
• 7,869
• 10,256
• 12,642

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

Solution: D

Let [math]T[/math] denote printer lifetime. Then [math]f(t) =\frac{1}{2} e^{–t/2}, 0 ≤ t ≤ \infty [/math]. Note that

[[math]] \begin{align*} P[T \leq 1] = \int_0^1 \frac{1}{2} e^{-t/2} dt = e^{-t/2} \Big |_0^1 = 1 - e^{-1/2} = 0.393. \\ P[1 ≤ T ≤ 2] = \int_1^2 \frac{1}{2} e^{-t/2} dt = e^{-t/2} \Big |_1^2 = e^{-1/2} - e^{-1} = 0.239. \end{align*} [[/math]]

Next, denote refunds for the 100 printers sold by independent and identically distributed random variables [math]Y_1, . . . , Y_{100}[/math] where

[[math]] Y_i = \begin{cases} 200 \quad \textrm{with probability } 0.393 \\ 100 \quad \textrm{with probability } 0.239 \\ 0 \quad \textrm{with probability } 0.368 \, \end{cases} [[/math]]

Now [math]\operatorname{E}[Y_i] = 200(0.393) + 100(0.239) = 102.56 [/math]. Therefore the expected refunds equals

[[math]] \sum_{i=1}^{100} \operatorname{E}[Y_i] = 100(102.56) = 10,256. [[/math]]

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