The warranty on a machine specifies that it will be replaced at failure or age 4, whichever occurs first. The machine’s age at failure, [math]X[/math], has density function

[[math]] f(x) = \begin{cases} \frac{1}{5}, \, 0\lt x \lt 5 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Let [math]Y[/math] be the age of the machine at the time of replacement. Calculate the variance of [math]Y[/math].

• 1.3
• 1.4
• 1.7
• 2.1
• 7.5

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

Solution: C

Note

[[math]] Y = \begin{cases} X, \quad 0 \leq X \leq 4 \\ 4, \quad 4 \lt X \leq 5 \end{cases} [[/math]]

Therefore,

[[math]] \begin{align*} \operatorname{E}[Y] = \int_0^4 \frac{1}{5} x dx + \int_4^5 \frac{4}{5} dx &= \frac{1}{10} x^2 \Big |_0^4 + \frac{4}{5} x \Big |_4^5 \\ &= \frac{16}{10} + \frac{20}{5} - \frac{16}{5} \\ &= \frac{8}{5} + \frac{4}{5} = \frac{12}{5} \\ &= \frac{112}{15} \end{align*} [[/math]]

[[math]] \operatorname{Var}[Y] = \operatorname{E}[Y^2] - (\operatorname{E}[Y])^2 = \frac{112}{15} - \left( \frac{12}{5} \right)^2 = 1.707. [[/math]]

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