Central Limit Theorem: Finding the probability that a sample proportion will differ from the population by more than a given amount Suppose 48% of the registered voters in a county are Republican. If a sample of 586 voters is selected, what is the probability that the sample proportion of Republicans will be greater than 42%?

Progress: I was trying to solve this using the central limit theorem, the difference from the population proportion by more than a given amount equation. But every time I plugged it into my calculator I got 0 or 1 and I know those aren't right. 
 A: Every time you pick one voter and count the number $X$ of Republicans in that sample of $1$, you get either $0$ or $1$.  You have
$$
\operatorname{E}(X) = 0\cdot\Pr(X=0)+1\cdot\Pr(X=1) = 0.48.
$$
$$
\operatorname{var}(X) = (0-0.48)^2\Pr(X=0)+(1-0.48)^2\Pr(X=1)
$$
and with a bit of algebra this simplifies to $(0.48)(1-0.48) = 0.2496$.
Let $X_1,\ldots,X_{586}$ be the observations, each either $0$ or $1$, and let
$$
\bar X = \frac{X_1+\cdots+X_{586}}{586}
$$
be their average.  Then
\begin{align}
& \operatorname{E}\left( \bar X \right) = \operatorname{E} \left( \frac{X_1+\cdots+X_{586}}{586} \right) = \frac 1 {586} \operatorname{E}(X_1+\cdots+X_{586}) \\[10pt]
= {} & \frac 1 {586}(\operatorname{E}(X_1)+\cdots+\operatorname{E}(X_{586})) = \frac 1 {586}(0.48+\cdots+0.48) = 0.48.
\end{align}
and
\begin{align}
& \operatorname{var}\left( \bar X \right) = \operatorname{var} \left( \frac{X_1+\cdots+X_{586}}{586} \right) = \frac 1 {586^2} \operatorname{var}(X_1+\cdots+X_{586}) \\[10pt]
= {} & \frac 1 {586^2}(\operatorname{var}(X_1)+\cdots+\operatorname{var}(X_{586})) = \frac 1 {586^2} (0.2496+\cdots+0.2496) = \frac 1 {586} 0.2496.
\end{align}
so the standard deviation of $\bar X$ is $\dfrac{\sqrt{0.2496}}{\sqrt{586}}$.
By the central limit theorem, it follows that
$$
\frac{\bar X - 0.48}{\sqrt{0.2496/586}} \approx N(0,1),
$$
and you seek the probability that that exceeds
$$
\frac{0.42 - 0.48}{\sqrt{0.2496/586}}.
$$
