Question

Willow Brook National Bank operates a drive-up teller window that allows customers to complete bank transactions without getting out of their cars. On weekday mornings, arrivals to the drive-up teller window occur at random, with an arrival rate of 30 customers per hour or 0.5 customers per minute.

(a)

What is the mean or expected number of customers that will arrive in a four-minute period?

(b)

Assume that the Poisson probability distribution can be used to describe the arrival process. Use the arrival rate in part (a) and compute the probabilities that exactly 0, 1, 2, and 3 customers will arrive during a four-minute period. (Round your answers to four decimal places.)

x	P(x)
0
1
2
3

(c)

Delays are expected if more than three customers arrive during any four-minute period. What is the probability that delays will occur? (Round your answer to four decimal places.)

Answer 1

Given,

On weekday mornings, arrivals to the drive-up teller window occur at random, with an arrival rate of 30 customers per hour or 0.5 customers per minute.

(a)

Mean or expected number of customers that will arrive in a four-minute period = 4 x Customer arrival rate per minute = 4 x 0.5 =2

(b)

Probability mass function of Poisson Distribution with mean

X : Number of customer arrive in 4-minute period;

Mean or expected number of customers that will arrive in a four-minute period : =2

Therefore,

X follows Poisson distribution with =2; Probability mass function is given by

x	P(x)
0	0.1353
1	0.2707
2	0.2707
3	0.1804

(c)

Delays are expected if more than three customers arrive during any four-minute period. What is the probability that delays will occur?

Probability that delays will occur

= probability that more than three customers arrive during any four-minute period

= P(X>3)=1-P(X 3)

= 1 - [P(X=0)+P(X=1)+P(X=2)+P(X=3)] (Substituting the values of P(X=0),P(X=1),P(X=2) and P(X=3) From (b))

= 1- [0.1353+0.2707+0.2707+0.1804]

= 1- 0.8571

= 0.1429

Probability that delays will occur = 0.1429