Questions: According to the U.S. Census Bureau, 26.1% of Floridians have earned a bachelor's degree. Let p̂ be the proportion of people that have earned a bachelor's degree in a random sample of 807 Floridians. Determine the following probabilities. Round solutions to four decimal places, if necessary. Find the probability that in a random sample of 807 Floridians, less than 21.9% have earned a bachelor's degree. P(p̂<0.219)= Find the probability that in a random sample of 807 Floridians, between 20.9% and 30.6% have earned a bachelor's degree. P(0.209<p̂<0.306)=

Solution Steps

Given the population proportion \( p = 0.261 \) and the sample size \( n = 807 \), we calculate the standard deviation of the sample proportion \( \sigma_{\widehat{p}} \) using the formula:

\[ \sigma_{\widehat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.261(1-0.261)}{807}} \approx 0.0175 \]

We need to find the probability \( P(\widehat{p} < 0.219) \). The Z-score for the upper limit is calculated as follows:

\[ Z_{end} = \frac{0.219 - 0.261}{0.0175} \approx -2.4 \]

Using the standard normal distribution, we find:

\[ P(\widehat{p} < 0.219) = \Phi(Z_{end}) \approx \Phi(-2.4) \approx 0.0 \]

Next, we calculate the probability \( P(0.209 < \widehat{p} < 0.306) \). The Z-scores for the limits are:

\[ Z_{start} = \frac{0.209 - 0.261}{0.0175} \approx -2.97 \] \[ Z_{end} = \frac{0.306 - 0.261}{0.0175} \approx 2.57 \]

Using the standard normal distribution, we find:

\[ P(0.209 < \widehat{p} < 0.306) = \Phi(Z_{end}) - \Phi(Z_{start}) \approx 1.0 - 0.0 = 1.0 \]

Final Answer