Question

Arrivals of customers at a telephone booth follow poisson distribution, with an average time of 10 minutes between one arrival and the next. The length of the phone call is assumed to be distributed exponentially with a mean of 3 minutes. Find a. The average number of persons waiting and making telephone calls b. The average length of the queue that is formed from time to time. c. Probability that a customer arrive and find telephone booth is busy. d. Probability that a customer arrive and find telephone booth is empty. e. The average time spent by a customer in telephone booth

Answer 1

Answer:

Step-by-step explanation:

Given λ= 1/10 = 0.10 person per minute.

μ = 1/3 = 0.33 person per minute.

(a) Probability that a person arriving at the booth will have to wait,

P (w > 0) = 1 – P0

= 1 – (1 - λ / μ) = λ / μ

= 0.10/0.33 = 0.3

(b) The installation of second booth will be justified if the arrival rate is more than the waiting time. Expected waiting time in the queue will be,            Wq = l/ m (m-l)                                                                                        Where, E(w) = 3 and λ = λ (say ) for second booth.                                         λ = 0.16                                                                                                   Hence the increase in arrival rate is, 0.16-0.10 = 0.06 arrivals per minute.

(c) Average number of units in the system is given by,                                Ls= ρ/1- ρ                                                                                                  =0.3/1-0.3                                                                                                          = 0.43 customers

(d)) Probability of waiting for 10 minutes or more is given by

3 percent of the arrivals on an average will have to wait for 10 minutes or more before they can use the phone.

Answer 2

Concept:

Poisson distribution is a discrete probability distribution which is used to express the probability of events occurring in a fixed interval of time or space.

Given:

Average time between arrivals=10 minutes

Mean of phone call length= 3 minutes.

Find:

The average number of persons who are waiting to make telephone calls The average length of the queue. Probability that a  booth is busy. Probability that a telephone booth is empty. The average time of a customer in telephone booth.

Solution:

a)The average number of persons waiting and making telephone calls :

Let a be the service rate and b be the arrival rate.

The probability that a person will wait and make a call = b/a=0.1/0.33

Probability=0.3

b)The average length of the queue that is formed from time to time:

Length=a/(a-b)=0.33/(0.33-0.1)

L=0.33/(0.23)=33/23=1.435 (approx.)

c)Probability that a customer arrive and find telephone booth is busy:

Probability=0.1/0.33=0.3

d)Probability that a customer arrive and find telephone booth is empty:

Probability=1-0.3=0.7

e)The average time spent by a customer in telephone booth:

The average time will be:

0.33/0.1=3.3.

Therefore, we get the probability of the different aspects of the Poisson table.

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