Public Two-Year Colleges From the data in CollegeScores2yr we know that about $66 \%$ of two-year colleges in the US are public institutions. Suppose that we select samples of $n=90$ schools at a time from the population of all two-year colleges and find the proportion of public schools in each sample. (a) Consider the distribution of these sample proportions when $n=90$ and $p=0.66$. What is its form? Center $=$ Standard error $=$ (round to three decimal places) (b) What proportion of these samples will have more than $75 \%$ public schools? Round to three decimal places. eTextbook and Media

Step 1 ：The distribution of these sample proportions is approximately normal because the sample size is large enough (n=90). This is due to the Central Limit Theorem.

Step 2 ：The center of the distribution is equal to the population proportion, p. So, Center = p = 0.66.

Step 3 ：The standard error (SE) of the proportion is given by the formula: \(SE = \sqrt{\frac{p(1 - p)}{n}}\)

Step 4 ：Substituting the given values: \(SE = \sqrt{\frac{0.66(1 - 0.66)}{90}} = \sqrt{\frac{0.66*0.34}{90}} = \sqrt{\frac{0.2244}{90}} = \sqrt{0.002493} = 0.050\). So, the standard error is approximately 0.050 (rounded to three decimal places).

Step 5 ：To find the proportion of these samples that will have more than 75% public schools, we first need to standardize the proportion using the z-score formula: \(z = \frac{x - \mu}{\sigma}\) where x is the sample proportion, \(\mu\) is the population proportion, and \(\sigma\) is the standard error.

Step 6 ：Substituting the given values: \(z = \frac{0.75 - 0.66}{0.050} = 1.8\)

Step 7 ：We then look up this z-score in the standard normal distribution table or use a calculator to find the area to the right of this z-score (since we want the proportion that is more than 75%). The area to the right of z = 1.8 is approximately 0.036.

Step 8 ：\(\boxed{0.036}\) or \(\boxed{3.6\%}\) of these samples will have more than 75% public schools.