Question

The arrival of customers at a service desk follows a Poisson distribution. If they arrive at a rate of two every five minutes, what is the probability that no customers arrive in a five-minute period? ​

Answer 1

Answer:

13.53% probability that no customers arrive in a five-minute period

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

They arrive at a rate of two every five minutes

This means that $\mu = 2$

What is the probability that no customers arrive in a five-minute period?

This is P(X = 0).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

13.53% probability that no customers arrive in a five-minute period