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In this question, we are given that 63% of U.S. adults drink coffee daily, and a random sample of 275 U.S. adults is selected. We are asked to find the mean \(\mu_{\hat{p}}\), where \(\hat{p}\) is the sample proportion of coffee drinkers.

The mean of the sampling distribution of the sample proportion \(\hat{p}\) is given by:

\[
\mu_{\hat{p}} = p
\]

where \(p\) is the population proportion of adults who drink coffee daily. In this case, \(p = 0.63\).

So, the mean \(\mu_{\hat{p}} = 0.63\).

Would you like a further explanation or do you have any questions about this concept?

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Here are five related questions for further practice:
1. How would you calculate the standard deviation of the sampling distribution of \(\hat{p}\)?
2. What is the probability that in a sample of 275 adults, fewer than 60% drink coffee?
3. If the sample size was increased to 500 adults, how would that affect the sampling distribution of \(\hat{p}\)?
4. What conditions must be met for the sampling distribution of \(\hat{p}\) to be approximately normal?
5. How would you interpret a sample proportion of 0.65 in the context of this problem?

**Tip:** In sampling distribution problems, always check if the sample size is large enough to apply the Central Limit Theorem for proportions!

The National Coffee Association reported that 63% of U.S. adults drink coffee daily. A random sample of 275 U.S. adults is selected. Find the mean \(\mu_{\hat{p}}\).

Learn how to find the mean of the sampling distribution of proportions for a scenario where 63% of U.S. adults drink coffee daily. This problem involves key statistical concepts such as sampling distributions and the Central Limit Theorem for Proportions. Suitable for students in Grades 10-12.

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