A Wendy’s fast-food restaurant sells hamburgers and chicken sandwiches. On a typical weekday, the demand for hamburgers is norma
lly distributed with a mean of 450 and standard deviation of 80 and the demand for chicken sandwiches is normally distributed with a mean of 120 and standard deviation of 30. How many hamburgers must the restaurant stock to be 99% sure of not running out on a given day?
Our given: Hamburgers sold daily are normally distributed and have a mean of 450 and standard deviation of 80 which is denoted by X ~ N (450,80)
What we do not know: How many hamburgers must the restaurant stock so that they won't run out. So let that be X: Let x: the number of required hamburgers the restaurant needs
Solution: So our X should be greater than 0.99
P(X>0.99) = 0.99 means P(Z>(X-450) divided by 80) > 0.99
From the normal distribution table, .99 Z score is = 2.3263
