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

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

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

${\displaystyle Y=\frac{X-\mu }{\sigma };}$

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 $\mathrm{Prob}[X\le 400.5$] 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 $Y$.