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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

Respuesta :

Answer:

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

Step-by-step explanation:

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

This is the following PDF of transformation [tex]R =\frac{10}{T}[/tex]

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

[tex]f_{y} (y)=f_{x} (g^{-1} (y))|\frac{dg^{-1} (y)}{dy}[/tex]

Using theformula, the PDF of  [tex]R =\frac{10}{T}[/tex] is

[tex]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}\\\\[/tex]