Question

Exercise 2.3 The time between phone calls to a call center is exponentially distributed with mean 60 seconds. (a) What is the probability that exactly 4 calls arrive in the next 2 minutes? (6) What is the probability that at least 2 calls arrive in the next 2 minutes? (c) What is the probability that no buses arrive in the next 2 minutes? (d) Given that a call has just arrived, what is the probability that the next call arrives between .5 and 1.5 minutes later?

Answer 1

a)

expected number of calls in 2 minutes =2

therefore P(exactly 4 calls in 2 minutes )=P(X=4) =e^-2*2⁴/4! =0.0902

b)

P(X>=2) =1-P(X<=1) =1-(e^-2*2⁰/0! +e^-2*2¹/1! )=1-(0.1353+0.2707)=0.5940

c)

P(No call ) =P(X=0) =e^-2*2⁰/0! =0.1353

d)

since intervals are indepeddnent:

P(neext call arrive between 0.5 and 1.5 minute) =1-P(no call in next 0.5 and 1.5 later (in 1 minute))

=1-(e^-11⁰/0!) =1-0.3679 =0.6321