Based on long experience, an airline found that about 5% of the people making reservations on a flight from Miami to Denver do not show up for the flight. Suppose the airline overbooks this flight by selling 266 ticket reservations for an airplane with only 255 seats. (Round your answers to four decimal places.)
(a) What is the probability that a person holding a reservation will show up for the flight?
(b) Let n = 269 represent the number of ticket reservations. Let r represent the number of people with reservations who show up for the flight. What expression represents the probability that a seat will be available for everyone who shows up holding a reservation?
P(r ≥ 269)
P(r ≤ 255)
P(r ≥ 255)
P(r ≤ 269)
(c) Use the normal approximation to the binomial distribution and part (b) to answer the following question: What is the probability that a seat will be available for every person who shows up holding a reservation?

a) 0.95

b) P(r ≤ 255)

c) P(r ≤ 255) = 0.4404

Step-by-step explanation:

There are 266 ticket reservations and 255 seats.

5% of people making reservations don't show up, hence the probability of making a reservation and not showing up = 0.05

a) The probability that a person holding a reservation will show up for the flight = 1 - 0.05 = 0.95

b) Let n = 269 represent the number of ticket reservations. Let r represent the number of people with reservations who show up for the flight. What expression represents the probability that a seat will be available for everyone who shows up holding a reservation?

For everyone that shows up with a reservation to gets seat, the number of people that show up must be at most 255 because the maximum number of seats available is 255.

So, the probability that everyone that shows up with a reservation gets a seat is

P(r ≤ 255)

c) And the probability, can be calculated using the normal distribution table or the binomial approximation. But the question asks us to use the normal distribution method.

Using normal distribution table

Mean = np = xbar = 269 × 0.95 = 255.55

Standard deviation = σ = √(np(1-p) = √(269×0.95×0.05) = 3.58

To obtain P(r ≤ 255)

We standardize 255

z = (r - xbar)/σ = (255-255.55)/3.58

z = -0.15

P(r ≤ 255) = P(z ≤ -0.15) = 0.4404

Hope this Helps!!!

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step-by-step explanation:

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