88.4k views
0 votes
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.

User Prashant G
by
7.4k points

2 Answers

1 vote

Final answer:

To determine how many trays of doughnuts the bakery should prepare each day to meet a service level of at least 95%, the owner can use the normal distribution and calculate the z-score corresponding to the desired service level. The z-score can then be used to determine the number of trays to be prepared.

Step-by-step explanation:

To determine how many trays of doughnuts the bakery should prepare each day, the owner can use the concept of service levels. The service level represents the probability of meeting or exceeding the demand. In this case, the owner wants a service level of at least 95%. Using the normal distribution, the owner can calculate the z-score corresponding to a service level of 0.95. A z-score of 1.645 corresponds to a service level of 0.95, and it represents the number of standard deviations above the mean that will satisfy the desired service level.

Next, the owner can use the z-score formula to calculate the number of trays that should be prepared:

Z = (X - mean) / standard deviation

where Z is the z-score, X is the number of trays to be prepared, mean is the mean demand (5 trays), and standard deviation is the standard deviation of demand (1 tray).

Substituting the values into the formula:

1.645 = (X - 5) / 1

Solving for X, we get:

X = 6.645

Rounding to the nearest whole tray, the owner should prepare 7 trays each day to meet the desired service level of at least 95%.

User Kit Sunde
by
6.7k points
1 vote

Answer:

7 trays

Step-by-step explanation:

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


z=(x-\mu)/(\sigma)

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

#Substitute our z value and solve for x:


z=(x-\mu)/(\sigma)\\\\1.645=(x-5)/(1)\\\\1.645+5=x\\\\x=6.645\approx7

Hence, the bakery must prepare 7 trays of doughnuts .

User Jintin
by
8.0k points