store manager is doing a study on how to staff his franchise. He knows that the average number of pizzas ordered is 12 per hour with a standard deviation of 1.4 pizzas. Staffed with 3 employees, he can assemble up to 15 pizzas per hour. If his store is open for 10 hours each day, what is the probability that he will receive more orders than he can assemble in a given hour?

Question

store manager is doing a study on how to staff his franchise.  He knows that the average number of pizzas ordered is 12 per hour with a standard deviation of 1.4 pizzas.  Staffed with 3 employees, he can assemble up to 15 pizzas per hour.  If his store is open for 10 hours each day, what is the probability that he will receive more orders than he can assemble in a given hour?

PsyDAG · Answer

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n(hours)

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.