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

Answer 1

Answer:

The value is  $P(X = 3) = 0.1804$  

Step-by-step explanation:

From the question we are told that

   The number of customer that enter the tellers every 5 minutes is  $\lambda = 2$

Generally the  probability that exactly three customers enter the queue in a randomly selected five-minute period is mathematically represented as

      $P(X = 3) = \frac{e^{- \lambda} * \lambda^{3}}{3!}$

=>   $P(X = 3) = \frac{e^{- 2} * \lambda^{3}}{3!}$

=>   $P(X = 3) = 0.1804$