How do I find if the probability of the sample proportion is greater than something?

Question

I have this problem and I have no clue how to solve it. In 2012, 31% of the adult population in the US had earned a bachelor’s degree or higher. One hundred people are randomly sampled from the population. What is the probability that the sample proportion p-hat is greater than 0.40?

Answer 1

The mean of $\hat{p}$ is equal to p, which is 0.31, and the standard deviation of $\hat{p}$ is:

$$\sqrt{p(1-p)/n} = (0.31*0.69/100)^{1/2} = 0.0462$$

So

$$P(\hat{p} > 0.4) = P(z > (0.4 - 0.31)/0.0462)$$

$$= P(z > 1.948) = 1 - P(z < 1.948) = 1 - .9744 = .0256$$

Answer 2

DeepSea's answer is a pretty standard way to do this. But it's important to remember, that treating $\hat p$ as Normally distributed is an approximation (albeit a quite good one most of the time).

If you wanted to have a more "exact" answer, you could use the Binomial distribution.

Let $Y$ be the number of people in the sample who have earned a Bachelors or higher. Then $Y \sim Binom(100, 0.31)$ and $\hat P = \frac{Y}{n}$. Therefore:

\begin{equation} P(\hat p > .40) = P(Y > 40) = 1 - F_Y(40) = 0.0218 \end{equation}

Notice that DeepSea's answer gives a good approximation. But this is answer is more "exact".