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

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

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(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(x - 5)/(1))=0.95`

`= 1 -P( z \leq \displaystyle(x - 5)/(1))=0.95`

`=P( z \leq \displaystyle(x - 5)/(1))=0.05`

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

`\displaystyle(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%.