Question

At the Fidelity Credit Union, a mean of 3.5 customers arrive hourly at the drive-through window. What is the probability that, in any hour, more than 5 customers will arrive? Round your answer to four decimal places.

Answer 1

0.1423 = 14.23% probability that, in any hour, more than 5 customers will arrive.

Explanation:

We have the mean, which means that the Poisson distribution is used to solve this question.

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 interval.

A mean of 3.5 customers arrive hourly at the drive-through window.

This means that
`\mu = 3.5`

What is the probability that, in any hour, more than 5 customers will arrive?

This is:

`P(X > 5) = 1 - P(X \leq 5)`

In which

`P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)`

Then

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

`P(X = 0) = (e^(-3.5)*3.5^(0))/((0)!) = 0.0302`

`P(X = 1) = (e^(-3.5)*3.5^(1))/((1)!) = 0.1057`

`P(X = 2) = (e^(-3.5)*3.5^(2))/((2)!) = 0.1850`

`P(X = 3) = (e^(-3.5)*3.5^(3))/((3)!) = 0.2158`

`P(X = 4) = (e^(-3.5)*3.5^(4))/((4)!) = 0.1888`

`P(X = 5) = (e^(-3.5)*3.5^(5))/((5)!) = 0.1322`

Finally

`P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0302 + 0.1057 + 0.1850 + 0.2158 + 0.1888 + 0.1322 = 0.8577`

`P(X > 5) = 1 - P(X \leq 5) = 1 - 0.8577 = 0.1423`

0.1423 = 14.23% probability that, in any hour, more than 5 customers will arrive.