Question

Data shows that the average number of customers vising a bank branch in a four-minute interval on Friday mornings is 2.6. The data also shows that available tellers can serve more than that number efficiently.

Suppose you work in the Operations Management division of the bank and are tasked to find the following.

1. What is the probability that five or more customers will arrive during an eight-minute period?

2. Suppose that tellers can serve no more than four customers in any 4-minute interval at this window on a Friday morning. What is the probability that, during any given four-minute interval, tellers will be unable to meet the demand?

(I assume it is Poisson distributed)

Answer 1

a)

expected number of customers in 8 minute period=2.6*8/4=5.2

hence probability that five or more customers will arrive during an eight-minute period =P(X>=5)=1-P(X<=4)

=1- =1-0.406128 =0.593872

b)

probability that, during any given four-minute interval, tellers will be unable to meet the demand

=P(more then 4 customers arrive in a 4 minute period)=P(X>4)=1-P(X<=4)=1- =1-0.877423 =0.122577