Question

A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and standard deviation of 5 tray. If the owner wants a service level a at least 95%?

Answer 1

The number of trays that should be prepared if the owner wants a service level of at least 95% is; 7 trays

How to utilize z-score statistics?

We are given;

Mean; μ = 15

Standard Deviation; σ = 5

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

Formula for z-score is;

z = (x' - μ)/σ

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

P(X > x) = P(z > (x - 15)/5) = 0.95

⇒ 1 -  P(z ≤ (x - 15)/5) = 0.95

Thus;

P(z ≤ (x - 15)/5) = 1 - 0.95

P(z ≤ (x - 15)/5) = 0.05

The value of z from the z-table of 0.05 is -1.645

Thus;

(x - 15)/5 = -1.645

x ≈ 7

Complete Question is;

A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and standard deviation of 5 trays. 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.

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