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 of the random variable R on the interval [10/12, 10/8]?
-----
I started by saying, ok, so f(t) = 1/(12-8) = 1/4
Now, the wording confuses me. So T is the time in minutes to respond to 10 inquiries, or 10 customers. R is the average customers per minute. So, R should be T/10, correct? So that means
F_t(T) = Pr(T<=t), and F_r(R) = Pr(R<=r) --> Pr(T/10<=r) -> Pr(T<=10r) = F_t(10r)
Well, F_t(t) = (t-8)/(12-8) = (t-8)/4 ---> F_t(10r) = (10r-8)/4
Therefore f(t) = d/dt [F_t(t)] = 1/4 * r * (5r-8) which is not even in the same realm as the answer
-----
I started by saying, ok, so f(t) = 1/(12-8) = 1/4
Now, the wording confuses me. So T is the time in minutes to respond to 10 inquiries, or 10 customers. R is the average customers per minute. So, R should be T/10, correct? So that means
F_t(T) = Pr(T<=t), and F_r(R) = Pr(R<=r) --> Pr(T/10<=r) -> Pr(T<=10r) = F_t(10r)
Well, F_t(t) = (t-8)/(12-8) = (t-8)/4 ---> F_t(10r) = (10r-8)/4
Therefore f(t) = d/dt [F_t(t)] = 1/4 * r * (5r-8) which is not even in the same realm as the answer
May 2000 Course 1 Exam Problem 4 - Transformation
