Question

Lee works in an insurance call centre where she handles claim enquiries from customers whose surnames
start with A to E. Studies have revealed that the volume of calls arriving in Lee’s queue follows a Poisson
distribution with an average of 18 per hour. If Lee’s queue receives more than 2 calls in 5 minutes, Lee
won’t be able to keep up and customers will abandon their call.
a) Determine the probability that Lee’s queue will have abandoned calls during a randomly chosen
interval of five minutes.
P (At least 2 calls in 5 minutes in queue)
F(0) = 18^0*e^-18/0! = 1.5229(10^-8)
F(1) = 18

Answer 1

Just want to know if I'm on the right track here...
P (At least 2 calls in 5 minutes in queue)
F(0) = 18^0*e^-18/0! = 1.5229(10^-8)
F(1) = 18^1*e^-18/1! = 2.7414(10^-7)
F(2) = 18^2*e^-18/2! = 2.4673(10^-6)

P(>2) = 1 – {1.5229(10^-8) + 2.7414(10^-7) + 2.4673(10^-6)}
= .9999972 or 99.99972% chance of abandoned calls in random 5 minute period