View source for Sampling Distribution of the Sample Proportion as an Example of the Sampling Distribution of the Sample Mean

Proportion problems can be likened to a coin toss, where we think of a "heads" as 1, and a "tails" as 0. Proportions are related to percentages and probabilities, but proportions are (like probabilities) always given as numbers between 0 and 1. Examples include the proportion of 
* Americans in favor of the death penalty, 
* water samples that turn up E.Coli, 
* paint samples that turn up lead, 
etc.

If all samples of paint turn up lead, then the proportion of contaminated samples is 1; if no samples turn up lead, then the proportion is 0; and generally, the truth lies somewhere between!

You might wonder how one could ever know whether a coin is fair (that is, the chance of a head is the same as the chance of a tail). The truth of the matter is, you can't. All conditions might point to a fair coin -- it may be perfectly symmetric, etc. -- but you'll never know for sure. This points to the existence of a '''parameter''', which we call <math>\left.\pi\right.</math>, and which indicates the true underlying proportion of heads (say). Note well: this <math>\left.\pi\right.</math> is not the same as the <math>\left.\pi\right.</math> that plays such an important role in the study of circles. 

From a "frequentist's" perspective, the only way to understand <math>\left.\pi\right.</math> is to toss the coin forever and see what happens to the ratio of heads to tosses. That's how we find the underlying parameters of the coin: 
<center>
<math>\pi = \lim_{n \rightarrow \infty}\frac{\Sigma_{i=1}^nx_i}{n}
</math></center>

Then the variance of that coin is 
<center>
<math>\sigma^2 = \lim_{n \rightarrow \infty}\frac{\Sigma_{i=1}^n(x_i-\pi)^2}{n} = \lim_{n \rightarrow \infty}\frac{\Sigma_{i=1}^n(x_i^2-2x_i\pi + \pi^2)}{n}
</math>
</center>
or
<center>
<math>\sigma^2 = \lim_{n \rightarrow \infty}\frac{\Sigma_{i=1}^nx_i}{n}
- 2\pi\lim_{n \rightarrow \infty}\frac{\Sigma_{i=1}^nx_i}{n}
+ \lim_{n \rightarrow \infty}\frac{\Sigma_{i=1}^n\pi^2}{n}
</math>
</center>
(because <math>x_i^2=x_i</math> for the coin toss); or
<center>
<math>\sigma^2= \pi - 2\pi^2 - \pi^2 = \pi - \pi^2 = \left(1-\pi\right)\pi
</math></center>

Therefore
<center><math>\sigma = \sqrt{(1-\pi)\pi}</math></center>

Therefore, according to '''the theory of the sampling distribution of the sample mean''', the parameters of the distribution of the sample mean are
<center> <math>\mu_{\overline{x}}=\pi</math>,</center>
and 
<center> <math>\sigma_{\overline{x}} = \sqrt{\frac{(1-\pi)\pi}{n}}</math>.</center>

The proportion problem is interesting for (at least) two reasons:
# once the mean is given, the standard deviation is known (somewhat unusual); and
# Our rule for when the normal distribution assumption is valid casts a shadow on the old lie that <math>\left.n=30\right.</math> is magic: for the proportion problem the rule we use is: <center> <math>0 < \pi-3\sigma_{\overline{x}} < \pi+3\sigma_{\overline{x}}<1</math></center> That is: <math>\left.n=30\right.</math> won't save you: it's a rule of thumb, but it depends entirely on the underlying distribution of ''x''....