An airline finds that 5% of the persons who make reservations on a certain flight do not show up for the flight. if the airline sells 160 tickets for a flight with only 155 seats, what is the probability that a seat will be available for every person holding a reservation and planning to fly?

Probability of success (showing up) = 1-0.05=0.95 is constant and known.

Trials are Bernoulli (show or no show).

Trials are independent and random (assumed from context)

Number of trials is known, n=160.

All being satisfied, we can then model with binomial distribution, where

P (x) = C (n, x) p^x * (1-p) ^ (n-x)

where C (n, x) = n! / (x! (n-x) !)

Here we look for

P (X<=155) = P (X=0) + P (X=1) + P (X=2) + ... + P (X=155)

=0.9061461 (using technology, or add up 156 values calculated, or read from binomial distribution table).

Alternatively, the normal approximation can be used, when n is large.

mean=np=160*0.95=152

standard deviation=sqrt (np (1-p)) = 2.75681

Apply continuity correction, x=155.5

Z = (155.5-152) / 2.75681=1.26958

P (z<=Z) = 0.89788 (read from normal distribution tables)

Error = (0.89788-0.9061461) * 100%=-0.83%

The approximation is considered good considering p=0.95 is quite skewed, but compensated by n>>50.