statistic question

Question

Theaters nationwide accept online reservations for movie shows. Of immediate concern is the number of online reservations they will accept relative to how many of those customers will actually show up to watch the movie. Suppose the capacity of a movie theater is 150 persons. The theater knows that 75% of the online reservation made will purchase tickets. For a typical movie showing, the theater earmarks 100 tickets to be sold to walk-in customers. The remaining 50 tickets are set aside for those making online reservation. How many online reservation should the movie theater accept in order to be 90% certain that they do not exceed the 50 tickets mark?

Answer 1

I think Michael misinterpreted the question.

 

You are looking for a (1-sided) 90% confidence interval where the upper bound is 50 (the maximum number of tickets bought through the online reservations.)

 

Let n= number online reservations to be made.

 

since p=.75 and q=.25; μ = np =.75n and σ =√[(0.75)(0.25))n]

 

50 = .75n + 1.28√[(0.75)(0.25))n]

 

Solve this equation for n (and round down if necessary to an integer.)

 

 

Any questions, comment back.

 

 

 

 

 

 

Answer 2

Hi Lejin,

 

Use a normal distribution approximation

n = 50, p = 0.75, q =1-p = 0.25

μ = np = 50(0.75) = 37.5,  σ =√(50(0.75)(0.25))=3.0619

X = probability of not exceeding 50 ticket mark 

P(X ≤ x) = 0.90 

Z = (x - μ)/σ  = (x-37.5)/3.0619

P(X ≤ x) = P(Z ≤ (x - 37.5)/3.0619)

P(Z ≤ (x - 37.5)/3.0619) = 0.90

Z ≈ 1.28

(x - 37.5)/3.0619 = 1.28  

x - 37.5 = 1.28*3.0619

x = 37.5 + 1.28*3.0619

x = 41

41 online reservation should be made in order to be 90% certain not to exceed the 50 tickets mark.