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
Answer: 636 hamburgers should be stocked.
