76. D **Educational Text Transcription and Explanation**---**Questions on Statistical Sampling and Proportions****75. According to Quantcast analytics of CrossFit.com, 59% of CrossFitters have children.** Suppose we take a random sample of 40 CrossFitters and find the sample proportion (\(\hat{p}\)) that have children. We repeat this process many times.- **(a) What do you expect the mean of the \(\hat{p}\)’s to be?**The expected mean of the sample proportions \(\hat{p}\) is equal to the population proportion. Hence, the expected mean is 0.59 or 59%.- **(b) What do you expect the standard deviation of the \(\hat{p}\)'s to be?**The standard deviation of the sample proportion \(\hat{p}\) can be calculated using the formula:\[\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}}\]where \( p = 0.59 \) and \( n = 40 \). Plugging in these values, we calculate \(\sigma_{\hat{p}}\).- **(c) What do you expect the shape of the \(\hat{p}\)'s to be?**The shape of the distribution of \(\hat{p}\) is expected to be approximately normal, especially since the sample size (40) is relatively large.**76. According to Quantcast analytics of CrossFit.com, 40% of CrossFitters have post-graduate degrees.** Suppose we take a random sample of 60 CrossFitters.- **(a) Describe the sampling distribution of the sample proportion of CrossFitters who have post-graduate degrees for samples of size 60.**The sampling distribution of the sample proportion \(\hat{p}\) will be approximately normal because the sample size is large. The mean of this distribution will be 0.40 or 40%. The standard deviation can be calculated using the formula:\[\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}}\]where \( p = 0.40 \) and \( n = 60 \). The distribution's standard deviation can be calculated by plugging these values into the formula.---Note: This explanation is meant to provide insight into understanding basic statistical sampling concepts Certainly! Here is the transcription of the text from the image, devised for an educational website:---### Statistical Analysis on Sample Proportions#### Study 76: CrossFitters and Post-Graduate Degrees- **(b)** Find the probability that at least half of the sample 60 CrossFitters has post-graduate degrees.- **(c)** Describe the sampling distribution of the sample proportion of CrossFitters who have post-graduate degrees for samples of size 100.- **(d)** Find the probability that between 25% and 50% of the sample 100 CrossFitters has post-graduate degrees.#### Study 77: Gender Distribution in University StudentsAt a university, 60% of the 7,400 students are female. The student newspaper reports results of a survey of a random sample of 200 students about various topics involving alcohol abuse, such as participation in binge drinking.- **(a)** Using the binomial distribution, find the probability that at most 104 students in a sample of 200 are female.- **(b)** Describe the sampling distribution of the sample proportion of females in samples of size 200 at this university.- **(c)** Suppose that out of a sample of 200 students, 52% are female (which is 104 out of 200). Find the probability that the sample proportion of females for a random sample of 200 students is at most 0.52.- **(d)** Find the probability that between 50 and 60 percent of students in a random sample of 200 are female.--- Each segment of these studies aims to explore probabilities related to academic and social trends through statistical sampling and probability distributions. Participants are encouraged to use theoretical distributions such as the binomial and normal approximations to calculate and interpret these probabilities.

Name: 76. D **Educational Text Transcription and Explanation**---**Questions
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Description: 76. D **Educational Text Transcription and Explanation**---**Questions on Statistical Sampling and Proportions****75. According to Quantcast analytics of CrossFit.com, 59% of CrossFitters have children.** Suppose we take a random sample of 40 CrossFi