Let [math]T[/math] denote the time in minutes for a customer service representative to respond to 10 telephone inquiries. [math]T[/math] is uniformly distributed on the interval with endpoints 8 minutes and 12 minutes. Let [math]R[/math] denote the average rate, in customers per minute, at which the representative responds to inquiries, and let [math]f(r)[/math] be the density function for [math]R[/math].

Determine [math]f(r)[/math], for [math]\frac{10}{12} \lt r \lt \frac{10}{8}[/math].

• 12/5
• 3-5/2r
• [math]3r-5\ln(r)/2[/math]
• [math]\frac{10}{r^2}[/math]
• [math]\frac{5}{2r^2}[/math]

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

Solution: E

First note R = 10/T . Then

[[math]] F_R(r) = \operatorname{P}[R ≤ r] = \operatorname{P}[\frac{10}{T} \leq r ] = \operatorname{P}[T \geq \frac{10}{r}] = 1 - F_T(\frac{10}{r}). [[/math]]

Differentiating with respect to

[[math]] \begin{align*} r f_R(r) = F^{'}_R(r) = d/dr (1 - F_T(\frac{10}{r})) = -(\frac{d}{dt}F_T(t)) (\frac{-10}{r^2}) \\ \frac{d}{dt}F_T(t) = f_T(t) = \frac{1}{4} \end{align*} [[/math]]

since [math]T[/math] is uniformly distributed on [8,12]. Therefore

[[math]] f_R(r) = \frac{-1}{4} (\frac{-10}{r^2}) = \frac{5}{2r^2}. [[/math]]

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