Awendy’s fast-food restaurant sells hamburgers and chicken sandwiches. on a typical weekday, the demand for hamburgers is normally 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

solve for X by:

2.3263 = (X-450) divided by 80
X-450= 80*2.3263 = 186.1
X= 450+186.1=636.1

636 hamburgers should be stocked.

(6,0) i hope it makes enough sense