Let S be the number of seats in the aircraft;
T be the number of tickets sold;
p be the probability that any given passenger arrives for the flight.

Let X be the number of passengers that arrrive for a given flight.

Then X has the binomial distribution for T trials with the probability of success 0.95. This distribution has mean m = Tp and variance s2=Tp(1-p).

If T=400 and p=0.95 then m = 380. This means that if the airline sold 400 tickets it would expect to have on average 20 empty seats.

We approximate this Binomial distribution by a Normal distribution with the same mean and variance, i.e. by
N(m,s2).
Thus we now assume that X has distribution N(m,s2). Put
Y = X-Tp
  ______
ÖTp(1-p)
 ;
then Y has distribution N(0,1).

The probability P that all passengers who arrive for the flight can actually fly is Prob{X £ S+0.5} (because X=400 is fine, but X=401 is not). Thus
P = Prob{X £ S+0.5} = Prob ì
ï
í
ï
î 		Y £ S+0.5-Tp
  ______
ÖTp(1-p)
 ü
ï
ý
ï
þ 		.
If we insert actual values for S, T and p, this can now be found from tables on the Normal distribution.
S T p Y
Prob all can fly
(from Normal tables)
Prob of more than 400
passengers arriving
400 410 0.95 2.4926 0.99365 0.00635 or 0.6%
411 2.2746 0.98855 0.01145 or 1.1%
412 2.0571 0.98017 0.01983 or 2.0%
413 1.8401 0.9671 0.0329 or 3.3%
414 1.6236 0.9477 0.0523 or 5.2%
415 1.4077 0.9203 0.0797 or 8.0%

The airline can sell 412 tickets but not more if it wants to have to refuse passengers with
tickets in no more than 2% of the flights.