My Notes Based on long experience, an airline found that about 7% of the people

Question

My Notes Based on long experience, an airline found that about 7% of the people making reservations on a night from Miami to Denver do not show up for the night. suppose the airline overbooks this night by selling 268 ticket reservations for an airplane with only 255 seats. (Round your answers to four decimal places.) What is the probability that a person holding a reservation will show up for the flight? 93 Let n = 268 represent the number of ticket reservations 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 reservations? P(r lessthanorequalto 268) P(r lessthanorequalto 255) P(r greaterthanorequalto 268) P(r greaterthanorequalto 255) Use the normal approximation to the binomial distribution and part (b) to answer the for question: What is the probability that a seat will be available for every person who shows up holding the reservation? _____

Explanation / Answer

ans=

1) 1-.07=.93.
2) let X=# who show up for flight; X is approx. normal with mean=268(.93)=249.24 and
std dev = sqrt[268(.93)(1-.93)] =sqrt[174.468]

=13.208.
Using the 0.5 normal correction, Pr(X > 255.5) = Pr((X-249.24)/13.208 > (255.5 - 249.24/13.208)=
Pr(Z > 0.473)=.3121 from normal table.

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