Question

4. From past experience, the owner of a restaurant knows that, on average, 5% of the groups that make reservations never show, and 7% of the groups that make reservations show up late (the other 88% 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”?

Answer 1

Answer:

a)

Here 5% of the gatherings that reserve a spot never demonstrate the

100 - 5 = 95% who reserve a spot appear.

In this way the likelihood that he will show is 0.95 on the off chance that he makes just a single reservation.

Also in the event that he reserves two spot, at that point the opportunity to appear in both the shows is 0.95*0.95 = 0.9025

Rehash the procedure until you arrive at number of individuals with likelihood under .80

That is 0.95*0.95*0.95*0.95*0.95*0.95 = 0.735

The necessary answer is 6 reservations.

b)

The subsequent inquiry is nearly equivalent to the primary issue. Be that as it may, when you discover the appropriate response, you simply add one to is. This extra one records for the individual who could possibly appear. As a result, this last individual doesn't make a difference in the real likelihood.

(0.95)^n >= 0.90

n = 3

So the response to this part is 4 reservations