Abakery 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.

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:

P(X > x) = 0.95

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

P(X > x)  

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

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

4+n=5

step-by-step explanation:

this is because the four is four hours and the n represents the 1 extra hour over the weekend.

b= i'm srry i don't know that one.

4+n=5 n=1 because if vernon plays 5 hours a week then 4+1=5 hours.

hope this : )