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 $\mu =\mathrm{Tp}$ and variance ${\sigma }^2=\mathrm{Tp}(1-p)$.

If $T=400$ and $p=0.95$ then $\mu =380$. 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

${\displaystyle N(\mu ,{\sigma }^2).}$

Thus we now assume that $X$ has distribution $N(\mu ,{\sigma }^2)$. Put

${\displaystyle Y=\frac{X-\mathrm{Tp}}{\sqrt{\mathrm{Tp}(1-p)}};}$

then $Y$ has distribution $N(0,1)$.

The probability $P$ that all passengers who arrive for the flight can actually fly is $\mathrm{Prob}\{X\le S+0.5\}$ (because $X=400$ is fine, but $X=401$ is not). Thus

${\displaystyle P=\mathrm{Prob}\{X\le S+0.5\}=\mathrm{Prob}\{Y\le \frac{S+0.5-\mathrm{Tp}}{\sqrt{\mathrm{Tp}(1-p)}}\}.}$

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%