Question

your turn
40% of the students at a university live off - campus. when sampling from this population, consider
\success\ to be students who live off - campus.

1. for random samples of 80 students, what interval captures 68% of all sample proportions under the normal

curve?

Explanation:

Step1: Calculate the standard deviation of the sampling distribution

The formula for the standard deviation of the sampling distribution of the sample proportion $\hat{p}$ is $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$, where $p = 0.4$ (population proportion) and $n=80$ (sample size).
$$\sigma_{\hat{p}}=\sqrt{\frac{0.4\times(1 - 0.4)}{80}}=\sqrt{\frac{0.4\times0.6}{80}}=\sqrt{\frac{0.24}{80}}=\sqrt{0.003}\approx0.055$$

Step2: Use the empirical rule

The empirical rule for a normal distribution states that approximately 68% of the data lies within $1$ standard deviation of the mean. The mean of the sampling distribution of the sample proportion is $\mu_{\hat{p}}=p = 0.4$.
The lower bound is $\mu_{\hat{p}}-\sigma_{\hat{p}}=0.4 - 0.055=0.345$
The upper bound is $\mu_{\hat{p}}+\sigma_{\hat{p}}=0.4+0.055 = 0.455$

Answer:

The interval is from $0.345$ to $0.455$.