# Standard Deviation of a Proportion

It's easy to explain why the standard deviation of a data set is computed the way it is. If you can compute an average, it is, in some sense simply an average deviation of each score from the mean for all scores (yes, I know that average deviation and standard deviation are not the same thing). The standard deviation of a proportion is harder to understand intuitively. Why would you multiply p times (1-p)? What's up with that? Is it analogous to the idea of average deviation, that is so apparent in the standard deviation formula used for a set of scores? If so, how?

The intuitive explanation for $p$ and $(1-p)$ may not be best explained using an analogy with the arithmetic mean (not impossible, just saying it may not be the most fruitful path).
If we accept that there is some value for the variance (standard deviation) when we sample proportions, then we are saying there is some measure of spread for $p$ successes. But, for every $p$ successes, there are $q=1-p$ failures. Thus, the measure of spread for the two should be the same. (For every observed value $\hat{p}$ from $p$, the distance between the two will be the same as the distance between $\hat{q}$ from $q$....thus, the spread of the values should be the same.)
So, if there is no difference in the measure of spread for $p$ and $1-p$, the formula should be symmetric. (Also, a visual display of the histograms for sampled $p$ and $q$ would show this symmetry, too.)