Question

Fidelity credit union, a mean of five customers arrives hourly at the drive-through window what is the probability that in any hour, less than one customer will rise on your answer to 4 Decimal Pl. Pl.

Answer 1

This scenario follows a Poisson distribution since the arrivals of customers can be considered random and occur at a constant average rate. The formula for the Poisson distribution is:

P(X = x) = (e^(-λ) * λ^x) / x!

where P(X = x) is the probability of x events occurring in a given time period, λ is the mean number of events per time period, and x is the number of events.

In this case, λ = 5 since the mean number of customers arriving hourly is 5.

To find the probability of less than one customer arriving in an hour, we need to calculate P(X < 1).

P(X < 1) = P(X = 0) + P(X = 1)

P(X = 0) = (e^(-5) * 5^0) / 0! = e^(-5) = 0.0067 (rounded to 4 decimal places)

P(X = 1) = (e^(-5) * 5^1) / 1! = 0.0337 (rounded to 4 decimal places)

Therefore,

P(X < 1) = 0.0067 + 0.0337 = 0.0404 (rounded to 4 decimal places)

So, the probability of less than one customer arriving in an hour at Fidelity credit union is 0.0404.