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. 6 5 4 7 unable to determine with the above information.

Answer 1

Answer:

4 trays should he prepared, if the owner wants a service level of at least 95%.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 5

Standard Deviation, σ = 1

We are given that the distribution of demand score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(X > x) = 0.95

We have to find the value of x such that the probability is 0.95

P(X > x)  

$P( X > x) = P( z > \displaystyle\frac{x - 5}{1})=0.95$  

$= 1 -P( z \leq \displaystyle\frac{x - 5}{1})=0.95$  

$=P( z \leq \displaystyle\frac{x - 5}{1})=0.05$  

Calculation the value from standard normal z table, we have,  

$\displaystyle\frac{x - 5}{1} = -1.645\\x = 3.355 \approx 4$  

Hence,  4 trays should he prepared, if the owner wants a service level of at least 95%.