An airline is interested in determining the proportion of its customers who are flying for reasons of business. If they want to be 90 percent certain that their estimate will be correct to within two percent, how large a random sample should they select?
This involves using the Z-distribution to find the 90% confidence interval of the actual value of the proportion. But for this to be done, there needs to be an estimate of the proportion, which is not given. How do we find this out?ie
(p)*(1-p)*za/2⁄n=0.0004 How do we find the value of p?
Take the most "pessimistic" estimate of $p(1-p)$, that is, the largest possible variance. Note that the maximum value of $p(1-p)$ is $\frac{1}{4}$, attained at $p=\frac{1}{2}$.
The standard deviation of the sample mean is therefore $\le \frac{1/2}{\sqrt{n}}$. So we use $\frac{1/2}{\sqrt{n}}$.
Remarks: $1$. There may be a mistake, or at least a typo, in your equation. Probably $z_{\alpha/2}^2$ is intended.
$2$. After the experiment has been performed, if $\hat{p}$ turns out to be not far from $0$ or not far from $1$, one may conclude that one has obtained a better estimate than the one planned for. However, $p(1-p)$ stays remarkably close to $\frac{1}{4}$ for $p$ not too far from $\frac{1}{2}$.