Question

A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 5 trays and standard deviation of 1 tray. If the owner wants a service level of at least 95%, how many trays should he prepare (rounded to the nearest whole tray)? Assume doughnuts have no salvage value after the day is complete.

Answer 1

Answer:

7 trays

Step-by-step explanation:

Given the $\mu=5, \sigma=1$, the z score is obtained using the formula:

$z=\frac{x-\mu}{\sigma}$

The z score value for a 95% interval is 1.645.

#Substitute our z value and solve for x:

$z=\frac{x-\mu}{\sigma}\\\\1.645=\frac{x-5}{1}\\\\1.645+5=x\\\\x=6.645\approx7$

Hence, the bakery must prepare 7 trays of  doughnuts .