In this problem $n=400$ and $p=0.95$.

So, $E(X)=380$ and $\sigma =4.36$ and the expected number of empty seats is $20$.

It is known that, if the value of $n$ is large, the variable $X$ could be considered to have a probability distribution that approximates to the standard normal distribution, with the same mean and variance.

To verify that the normal distribution could be, in the conditions of the problem, a good approximation for the binomial distribution, I have to verify that both the mean $\mu =\mathrm{np}$ and the variance ${\sigma }^2=n(1-p)$ are greater than 5. Here $\mathrm{np}=380$ and $n(1-p)=20$. So, the use of the normal distribution is acceptable.

Using the applet at http://davidmlane.com/hyperstat/z_table.html , I tried to find the number of tickets, $x$, that the airline should sell to satisfy the conditions of the problem.

Let $x$ be the number of tickets sold, which, as explained before, could be considered to have a normal distribution $N(\mu ,{\sigma }^2)$. The mean of the distribution is $x\times 0.95$, and the standard deviation is $\sqrt{x\times 0.95\times 0.05}$. The area under the curve and below 400 is 98 per cent or 0.98 and the area above 400 is 2 per cent or 0.02 (the probability that too many passengers will turn up for the flight).

Trying for some values of $x$ I obtained the number of tickets that the airline must sell. Put

${\displaystyle y=\frac{x-\mathrm{np}}{\sqrt{\mathrm{np}(1-p)}};}$

then $y$ has distribution $N(0,1)$. The probability that all passengers who arrive for the flight can actually get a seat is $\mathrm{Prob}\{x\le 400.5\}$ (because $x=400$ is fine, but $x=401$ is not). Thus

${\displaystyle \mathrm{Prob}\{x\le 400.5\}=\mathrm{Prob}\{y\le \frac{400.5-\mathrm{np}}{\sqrt{\mathrm{np}(1-p)}}\}}$

and this can now be found from tables of the normal distribution.

We find that if 411 tickets are sold then the probability of too many passengers arriving is less than 2 percent but for 412 it is more than 2 percent so the ideal number of tickets to be sold is 411.