confidence intervals with a mean How many people must be surveyed to estimate $90\%$ confidence for the mean, if the standard deviation is known to be $21.0$ and the allowable margin of error is $5$?
What I have so far is:
$\frac{1}{2}(1-0.90)=0.05$,
$Z= 1.645$
$(1.645\cdot 21.0/5)^2= 47.73$
Is that right?

Yes, what you have is correct.  Confidence intervals are represented as 
$$\bar{X}\pm E$$
Where $E$ is the maximum error of estimate and 
$$E=z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right)$$
Solving the last equation for $n$ yields
$$n=\left(\frac{z_{\alpha/2}\cdot\sigma}{E}\right)^2$$
So for your values
$$n=\left(\frac{1.645\cdot 21}{5}\right)^2\approx47.73$$
And so we round to $48$ and you are correct.
