Question

Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of 7 per hour. If it takes approximately 10 minutes to serve each customer, find the mean and variance of the total service time for customers arriving during a 1-hour period. Assume that a sufficient number of servers are available so that no customer must wait for service. What is the probability that the total service time will exceed 2.5 hours? (Use at least four digits after the decimal if rounding.)

Answer 1

the mean service time = 70 minutes

the variance of the total service time for customers arriving during a 1-hour period is 700

the probability that the total service time will exceed 2.5 hours is 0.0012

Explanation:

Given that:

Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of 7 per hour.

If it takes approximately 10 minutes to serve each customer

From the given information:

The mean total service time=
`10 \ minutes * \lambda` (i.e the average number to turn up during the hour) .

where;

`\lambda =` the average rate per hour i.e 7

Thus; the mean service time = 10 × 7 = 70 minutes

For a Poisson Distribution;

The variance of the the total service time for customers arriving during a 1-hour period can be expressed by the relation:

`\sigma(aX) = a* \sigma(X). \\ \\ where ; \sigma = {√(S^2)} \\ \\ \sigma(service \ time) = 10 * \sigma(arrival \ time) \\ \\S^2 (service \ time) = (10* \sigma(arrival \ time))^2 \\ \\ S^2(service \ time) = 100 * S^2(arrival \ time) \\ \\ = 100*\lambda \\ \\ = 100*7 = 700`

where S² = variance

What is the probability that the total service time will exceed 2.5 hours?

If we convert 2.5 hours to minutes ; we have

2.5 × 60 minutes =150 minutes

The probability that the total service time will exceed 2.5 hours is as follows:

`Z = \frac{S-70}{\sqrt {700}} \ \ \ \ follow \ N(0,1) \\ \\ P(S > 150) = (S-70)/((√(700)>(150-70))/(√(700))) \\ \\ =P(Z > -2.5512) \\ \\ =0.0012`