Answer to Question #222309 in Statistics and Probability for ANESTO

Question

Let X denote the amount of time (in minutes) a postal clerk spends with his/her customer. The time is known to have an exponential distribution with an average amount of time equal to 4 minutes.

i)Find the probability that a clerk spends four to five minutes with a randomly selected customer.

ii)Half of all the customers are finished within how long? (Find the median)

iii)Which is larger, the mean or the median?

Answer 1

$X$ is a continuous random variable since time is measured. It is given that $\mu=4$ minutes. 

$X\sim Exp(\lambda)$

$$\lambda=\dfrac{1}{\mu}=\dfrac{1}{4}=0.25$$

i)

$$P(4<X<5)=P(X<5)-P(X<4)$$

$$=1-e^{-\lambda(5)}-(1-e^{-\lambda(4)})$$

$$=e^{-0.25(4)}-e^{-0.25(5)}=e^{-1}-e^{-1.25}$$

$$\approx0.081375$$

ii)

$$P(X<T)=0.5$$

$$1-e^{-0.25(T)}=0.5$$

$$e^{-0.25(T)}=0.5$$

$$0.25(T)=\ln 2$$

$$T=4\ln 2$$

$$T\approx2.77\ minutes$$

iii) The theoretical mean is  $4$ minutes. We have just calculated the median (or 50th percentile) as  $2.77$ minutes. So the mean is larger.