91Ó°ÊÓ

Start by noting that the sample size (n) is 75, population size (N) is 10,000, and population proportion (p) is 0.8.The mean of the sampling distribution of \hat{p} (the sample proportion) is given by \mu_{\hat{p}}=p. Thus, \mu_{\hat{p}}=0.8.The standard deviation of the sampling distribution of \hat{p} is calculated using the formula \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \times \sqrt{\frac{N-n}{N-1}}.First, calculate the \sqrt{\frac{p(1-p)}{n}} term:\[\sqrt{\frac{0.8(1-0.8)}{75}} = \sqrt{\frac{0.8(0.2)}{75}} = \sqrt{\frac{0.16}{75}} = \sqrt{0.0021333} \approx 0.0462 \]Then, adjust for the finite population using the finite population correction factor:\[\sqrt{\frac{10,000 - 75}{10,000 - 1}} = \sqrt{\frac{9,925}{9,999}} \approx \sqrt{0.9925} \approx 0.9962 \]Multiply these two results together to find \sigma_{\hat{p}}:\[0.0462 \times 0.9962 \approx 0.046 \]Thus, the standard deviation \sigma_{\hat{p}} \approx 0.046So, the sampling distribution of \hat{p} is approximately normal with mean 0.8 and standard deviation 0.046.

First, convert the sample proportion \hat{p} = 0.84 into a z-score:\[z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.84 - 0.8}{0.046} = \frac{0.04}{0.046} \approx 0.87 \]Using the standard normal distribution table, find the probability corresponding to z = 0.87.\[P(Z \geq 0.87) = 1 - P(Z < 0.87) \]From the standard normal table, P(Z < 0.87) \approx 0.8078.Thus, the probability is:\[P(Z \geq 0.87) = 1 - 0.8078 = 0.1922 \]So, \[P(\hat{p} \geq 0.84) \approx 0.1922 \]

Convert the sample proportion \hat{p} = 0.68 into a z-score:\[z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.68 - 0.8}{0.046} = \frac{-0.12}{0.046} \approx -2.61 \]Using the standard normal distribution table, find the probability corresponding to z = -2.61.\[P(Z \leq -2.61) \]From the standard normal table, P(Z \leq -2.61) \approx 0.0045.So, \[P(\hat{p} \leq 0.68) \approx 0.0045 \]