Calculating sample size with unknown standard deviation I'm having troubles with understanding a problem which is basically asking to calculate a sample size (in order to estimate what portion of population has income more than 40K) with 95% confidence level and margin of error (E) of 1%. The standard deviation is not given and it says that I should take a maximum possible value for that. In addition, it says that 2 should be used as the quantile of the normal distribution of 0.975.
Two things confuse me here: how to find maximum standard deviation and how quantile is different from a z-score?
If quantile and a z-score are the same thing, then I thought that it should be 0.475 => 1.96 (and not 0.975 => 2)? (Since the confidence interval is 95% the alpha value is going to be 0.05 and the tails will have an area of 0.025 each. Having this the critical z-value is going to be 1.96 (z of 0.475 (0.5 - 0.025)), right?)
Any feedback is appreciated, thanks in advance
 A: Your assertion that the correct value of the quantile should be 1.96 (if we assume the normal approximation is accurate) is completely correct. 
However, the suggested value of 2 is a common approximation; it's only 2% larger, and that small additional margin by rounding the value up may be a good idea since several approximations are involved in doing the calculations.
That is, while you know how to compute 1.96 correctly, just use 2 anyway, like it says.
Secondly, if I told you the population proportion, could you compute the standard deviation? 
Edit after chat discussion with OP:
As you figured out, the standard error is maximized when $p=0.5$. By changing the scale on the y-axis (a simple monotonic transformation), it's perhaps easier to see:

What remains is to figure out the margin or error in terms of the standard deviation.
Your question says to take it to be twice the standard error, so all you have left to do is find the smallest value of $n$ that has $2\sqrt{0.25/n}\leq 0.01$. Even if you're not able to do the algebraic manipulation to solve for $n$, you can find this by trial and error. 
[n=1 will be too wide. What happens at n=10? 100? 1000? etc ... if you go past it, you can then try the middle of the interval until you hit it exactly or you get two consecutive numbers that give a margin of error either side of the right answer]
