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

P(3,7) = 0.95

He should prepare between 3 and 7 doughnuts    

Step-by-step explanation:

We need determine C such that

P(μ-c, μ+c) = 0.95

               Where μ: Mean of doughnuts demands

For calculating C, we may use the table of the normal standard distribution

If P(μ-c, μ+c) = 0.95 by symetria P(μ+c) = 0.975, then

              (μ+c-μ)/δ

                                Where δ: Standard deviation

How     1.96

             1.96 =

             1.96 = c

Thus

P(3.04, 6.96) = 0.95

P(3,7) = 0.95

He should prepare between 3 and 7 doughnuts    

answer: a. 13/44

step-by-step explanation: judging by the information given to us, we are looking for experimental probability. we can see that 13 males have chosen chocolate ice cream. there are 44 ice creams in total (remember that the total is always the denominator). so our answer is 13/44.

step-by-step explanation: