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

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

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) = (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) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 0) = (e^(-2)*2^(0))/((0)!) = 0.1353`

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