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.

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