Question

Weekly demand at a convenience store for a brand of breakfast cereal is normally distributed with a mean of 8 boxes and a standard deviation of 2 boxes. How many boxes should the store order for a week to have at most a 2.5 percent chance of running short of this brand of cereal during the week?

Answer 1

Let N be the number of boxes the store should order to have at most 2.5% chance of running short of this brand of cereal during the week.

P(X < N) = P(Z < (N - mean)/standard deviation)

Here, mean = 8

standard deviation = 2

P(X > N) = 0.025

P(X < N) = 1 - 0.025 = 0.975

P(Z < (N - 8)/2 = 0.975

Take Z value corresponding to 0.975 from standard normal distribution table

(N - 8)/2 = 1.96

N = 1.96x2 + 8 = 11.92

So, the store should order 12 boxes of cereal.