Question

A telephone switchboard handles on the average 4 calls per minute. If the calls follow a Poisson distribution, what is the probability of:

(a) exactly 8 calls per minute, more than 10 calls per minute?

(b) a call comes in with the next 30 seconds?

(c) assuming no call comes in during a 30 second period, what is the probability that a call comes in the following 30 seconds?

(d) Using the Central Limit Theorem, how many calls can we expect in a day? What is the standard deviation of this number? Using the normal approximation, what is the bar for the number of calls for a day to be considered 10% of the busiest?

Answer 1

a)

here for poisson distribtuion: =4

P(exaclty 8 calls) =P(X=8)=e^-4*4⁸/8! =0.0298

P(more than 10 calls)=P(X>10)=1-P(X<=10)=1- =1-0.9972 =0.0028

b)

for 30 seconds expected calls =4*30/60=2

P(call comes in with the next 30 seconds)=1-P(no calls in 30 seconds) =1-(e^-2*2⁰/0!) =1-0.1353=0.8647

c)

for poisson process ; probability of outcme of an event is independent from interval to interval

therefore  probability that a call comes in the following 30 seconds given no call in fist 30 seconds

=  probability that a call comes in the following 30 seconds =0.8647

d)

expected number of calls in a day (1440 minutes) =4*1440=5760 =

std deviation=sqrt() =sqrt(4*1440)=141.986

for top 10 percentile ; critial z score =1.28

hence corresponding bar =mean+z*std deviation =5760+1.28*141.986=5941.74