A bank manager estimates that an average of two customers enters the tellers' queue every five minutes. Assume that the number of customers that enters the tellers' queue is Poisson distributed. What is the probability that at least two customers enter the queue in a randomly selected five-minute period?
A) 0.1353
B) 0.2707
C) 0.4060
D) 0.5940

Question

Quizplus · Accepted Answer

The parameter for the Poisson distribution is λ = 2 (since on average, 2 customers enter the queue in a five-minute period). To find the probability that at least two customers enter the queue, we need to find P(X ≥ 2), where X is the number of customers entering the queue in a five-minute period. 
Using the Poisson distribution, we have: 
P(X ≥ 2) = 1 - P(X < 2) 
P(X < 2) = P(X = 0) + P(X = 1) 
P(X = 0) = e^-2 * (2^0 / 0!) = 0.1353 
P(X = 1) = e^-2 * (2^1 / 1!) = 0.2707 
P(X < 2) = 0.1353 + 0.2707 = 0.406 
P(X ≥ 2) = 1 - 0.406 = 0.594. Therefore, the answer is D.

A Bank Manager Estimates That an Average of Two Customers