Question

A bank manager estimates that an average of two customers enter the tellers’ queue every five minutes. Assume that the number of customers that enter the tellers’ queue is Poisson distributed. What is the probability that exactly three customers enter the queue in a randomly selected five-minute period?

a. 0.2707
b. 0.0902
c. 0.1804
d. 0.2240

Answer 1

Option C) 0.1804

Explanation:

We are given the following information in the question:

Mean customers enter the tellers’ queue every five minutes = 2

`\lambda = 2`

Thus, the number of customers that enter the tellers’ queue is Poisson distributed

Formula:

`P(X =k) = \displaystyle(\lambda^k e^(-\lambda))/(k!)\\\\ \lambda \text{ is the mean of the distribution}`

We have to evaluate:

`P( x = 2)\\\\= \displaystyle(2^3 e^(-2))/(3!)\\\\= 0.1804`

0.1804 is the probability that exactly three customers enter the queue in a randomly selected five-minute period.

Thus, the correct answer is

Option C) 0.1804