Binomial Probability Distribution

Question

A small airline has a policy of booking as many as 58 persons on an airplane that can seat only 51. (Past studies have revealed that only 78% of the booked passengers actually arrive for the flight.)

Find the probability that if Air-USA books 58 persons, not enough seats will be available. 

Answer 1

use normal approximation to binomial distribution

p(> 51 seats arrive out of 58) when probability arrive is 78%

mean is 58 * 0.78 = 45.24

std dev = sqrt(58 * 0.78 * 0.22) = 3.15

z = (51-45.25)/3.15 = 1.83

P(z > 1.83) = 1 - P(z < 1.83) = 1 = 0.9664 = 0.0336