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

The answer is "
`f_(R) (r) =(5)/(2r^2); (10)/(12) \leq r \leq (10)/(8)`"

Explanation:

They have the distributional likelihood function of T:
`f_(\Gamma ) \ (t)= (1)/(12-8)= (1)/(4); 8 \leq t \leq 12`

This is the following PDF of transformation
`R =(10)/(T)`

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

`f_(y) (y)=f_(x) (g^(-1) (y))|(dg^(-1) (y))/(dy)`

Using theformula, the PDF of
`R =(10)/(T)` is

`f_(R) (\Gamma)=f_((\Gamma)) |(d((10)/(r)))/(dr)| \\\\f_(R)(r) =(1)/(4)| -(20)/(r^2)|\\\\f_(R) (r) =(5)/(2r^2); (10)/(12) \leq r \leq (10)/(8)\\\\`