Over-booking

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 $0.95$. You can calculate the mean $\mu$ and variance $\sigma^2$ for this distribution.

In order to avoid lengthy calculations in the discrete case we approximate the Binomial distribution by a Normal distribution with the same mean and variance, i.e. by $$N(\mu,\sigma^2)$$

Thus we now assume that $X$ has distribution $N(\mu,\sigma^2)$. In order to use the standard Normal probability tables $N(0,1)$ we have to put $$Y = {X-\mu\over \sigma}$$ then $Y$ has distribution $N(0,1)$.

So as to allow for the approximation to the discrete data by the continuous Normal distribution, we want to find $\text{Prob}[X\leq 400.5$] and look up the probability for the corresponding value of $Y$ in the Normal table.

If you use a Normal distribution table you need to check to see if it gives the area $\Phi(Y)$ under the Normal curve to the left of $Y$, that is the probability that the variable is less than $Y$, or to the right of $Y$.