Answer to Question #179046 in Statistics and Probability for robin

Question

The arrival rates of telephone calls at telephone booth are according to poisson distribution with

an average time of 12 minutes between arrival of two consecutive calls. The length of telephone

calls is assumed to be exponentially distributed with mean 4 minutes. (1) Determine the

probability that the person arriving at the booth will have to wait. (2) Find the average queue

length that is formed from time to time. (3) The telephone company will install the second booth

when convinced that an arrival would expect to wait atleast 5 minutes for the phone. Find the

increase in flows of arrivals which will justify the second booth. (4) What is the probability that

an arrival will have to wait for more than 15 minutes before the phone is free?

Answer 1

Solution:

Arrival rate, $\lambda=\frac1{12}$ per minute

service rate, $\mu=\frac1{4}$ per minute

(1). $p=\frac{\lambda}{\mu}=\frac1{3}$

(2). $L=\mu/(\mu-\lambda)=\frac1{4} /(\frac1{4}-\frac1{12})=1.5$

(3). average waiting time in the queue $\mathrm{W}=\lambda/(\mu(\mu-\lambda))$

The waiting time to install a second booth is 5 minutes

$5=\dfrac{\lambda}{(1 / 4)((1 / 4)-\lambda)}$

$\Rightarrow \lambda=\frac5{36}$ arrivals/minute

increase in flow of arrivals $=\frac5{36}-\frac1{12}=\frac1{18}$ per minute

(4). $P(W>15)=(\lambda / \mu) \int_{15}^{\infty}(\mu-\lambda) e^{-(\mu-\lambda){t}} d t$

$=(1 / 3 )(1/6)\int_{15}^{\infty}e^{-t(1/6)}dt$

$=(1 /18)/(-1/6)\times[e^{-\infty(1/6)}-e^{-15(1/6)}]$

$=(-1/3)\times[0-0.082084]=0.027361$