Question

a small commuter airline is concerned about reservation no shows and, how much they should overbook flights to compensate. assume their commuter planes will hold 19 people. research indicates that 32% of people making a reservation will not show up for the flight. if the airline decided to book 27, what is the probability that at least 1 person does not get a seat?

Answer 1

P(a person showing up), p = 1 - 0.32 = 0.68

q = 0.32

Sample size, n = 27

Normal approximation for binomial distribution: P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = np

= 27 x 0.68

= 18.36

Standard deviation =

=

= 2.424

P(at least one person does not get a seat) = 1 - P(all people get a seat)

= 1 - P(X 19)

= 1 - P(Z < (19.5 - 18.36)/2.424) (with continuity correction)

= 1 - P(Z < 0.47)

= 1 - 0.6808

= 0.3192