Question

Let T denote the time in minutes for a customer service representative to respond to 10 telephone inquiries. T is uniformly distributed on the interval with endpoints 8 minutes and 12 minutes. Let R denote the average rate, in customers per minute, at which the representative responds to inquiries. What is the density function for the random variable R on the interval 10/12 <= r <= 10/8

Answer 1

Answer:

The answer is "$f_{R} (r) =\frac{5}{2r^2}; \frac{10}{12} \leq r \leq \frac{10}{8}$"

Step-by-step explanation:

They have the distributional likelihood function of T: $f_{\Gamma } \ (t)= \frac{1}{12-8}= \frac{1}{4}; 8 \leq t \leq 12$

This is the following PDF of transformation $R =\frac{10}{T}$

They know that PDF is the Y=g(X) transformation

$f_{y} (y)=f_{x} (g^{-1} (y))|\frac{dg^{-1} (y)}{dy}$

Using theformula, the PDF of  $R =\frac{10}{T}$ is

$f_{R} (\Gamma)=f_{(\Gamma)} |\frac{d(\frac{10}{r})}{dr}| \\\\f_{R}(r) =\frac{1}{4}| -\frac{20}{r^2}|\\\\f_{R} (r) =\frac{5}{2r^2}; \frac{10}{12} \leq r \leq \frac{10}{8}$