From past experience, the owner of a restaurant knows that, on average, 4% of th

Question

From past experience, the owner of a restaurant knows that, on average, 4% of the groups that make reservations never show and 7% of the groups that make reservations show up late (the other 89% percent show up on time).

a) How many reservations can the owner accept and still be at least 80% sure that all parties that make a reservation will show?

b) How many reservations can the owner accept and still be at least 90% sure that there will be at most one “no-show”?

c) What is the expected number of late groups?

Show your work and justify your answer.

Explanation / Answer

a)since 4% of the groups that make reservations never show the 100-4=96% who make reservations show up.
Thus the probability that he will show is 0.96 if he makes only one reservation.
Similarly if he makes two reservations, then the chance to show up in both the shows is 0.96*0.96 = 0.922.

Repeat the process until you reach number of people with probability less than .80

That is 0.96*0.96*0.96*0.96*0.96*0.96=0.782
The required answer is 6 reservations.

b) The second question is almost the same as the first problem. However, once you find the answer, you just add one to is. This additional one accounts for the person who may or may not show. In effect, this last person does not matter in the actual probability.

(.96)^n >= 0.90
n = 3

So the answer to this part is 4 reservations

c) the expected number of late groups is 0.07

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