Binomial Probability Distribution

Question

A small airline has a policy of booking as many as 60 persons on an airplane that can seat only 52. (Past studies have revealed that only 76% of the booked passengers actually arrive for the flight.)

Find the probability that if Air-USA books 60 persons, not enough seats will be available. 

Answer 1

P(number of passengers arrive out of 60 > 52, when probability arrive is 0.76.

Can use normal approximation to binomial distribution.

mean = np = 60 * 0.76 = 45.6

standard deviation = sqrt(npq) = 3.31

z = (x - mean)/standard deviation where x = number that arrive

compute z for x = 52 and above mean and standard deviation:

z = (52 - 45.6)/3.31 = 1.93

P(z > 1.93) = 1 - P(z < 1.93) = 1 - 0.9732 -= 0.0268