Question 960750: A Wendy’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?
Answer by mathmate(429)   (Show Source):
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Given:
Hamburgers sold daily are normally distributed and have a mean of 450 and standard deviation of 80, i.e. N(450,80).
Require:
Daily stock of hamburgers necessary in order not to run out 99 times out of 100.
Let X=number required, then we need
P(x>X)=0.99 which means
P(Z>(X-450)/80)>0.99
From normal distribution tables, Z=2.327,
solve for X
2.3263=(X-450)/80
X-450=80*2.3263=186.1
X=450+186.2=636.1
Answer:
637 hamburgers should be stocked.
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