Question

On Monday mornings, a CIBC branch has only one teller window open for deposits and withdrawals. Experience has shown that the average number of arriving customers in a four-minute interval on Monday mornings is 2.6, and each teller can serve more than that number efficiently. The random arrivals at this bank on Monday mornings are Poisson distributed.

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

(b) When demand cannot be met during any given interval, a second window is opened. What percentage of the time will a second window have to be opened?

(c) What is the probability that exactly three people will arrive at the bank during a two-minute period on Monday mornings to make a deposit or a withdrawal?

(g) What is the probability that five or more customers will arrive during an eight-minute period?

Answer 1

Answer:

Given,

P(x) = e^-*^x/x!

= 2.6

a)

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

P(x <= 4) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)

= e^-2.6*2.6^0/0! + e^-2.6*2.6^1/1! + e^-2.6*2.6^2/2! + e^-2.6*2.6^3/3! + e^-2.6*2.6^4/4!

= 0.0743 + 0.1932 + 0.2511 + 0.2176 + 0.1415

= 0.8777

Now to give the probability that the teller will be able to meet the demand

Probability = P(x >= 4)

= 1 - P(x <= 4)

= 1 - 0.8777

So the probability that the teller will be able to meet the demand = 0.1223

b)

To give the percentage of the time will a second window have to be opened

i.e.,

= 0.1223*100

= 12.23%

c)

To give the probability that exactly three people will arrive at the bank during a two-minute period on Monday mornings to make a deposit or a withdrawal

Here = 2.6 for the 4 minutes

now for 2 minutes interval ,

= 2.6/2

= 1.3

P(x = 3) = e^-1.3*1.3^3/3!

= 0.0998

g)

To give the probability that five or more customers will arrive during an eight-minute period

Here = 2.6 for the 4 minutes

now for the 8 min interval,

= 2.6*2

= 5.2

consider,

P(x >= 5) = 1 - P(x < 4)

= 1 - [P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)]

= 1 - [ e^-5.2*5.2^0/0! + e^-5.2*5.2^1/1! + e^-5.2*5.2^2/2! + e^-5.2*5.2^3/3! + e^-5.2*5.2^4/4! ]

= 1 - [0.0055 + 0.0287+ 0.0746 + 0.1289 + 0.1676]

= 1 - 0.4053

= 0.5947

Hence the probability that five or more customers will arrive during an eight-minute period is 0.5947