A fitness center is interested in finding a $95 \%$ confidence interval for the mean number of days per week that Americans who are members of a fitness club go to their fitness center. Records of 214 members were looked at and their mean number of visits per week was 3.1 and the standard deviation was 1.6. a. To compute the confidence interval use a $t$ distribution. b. With $95 \%$ confidence the population mean number of visits per week is between and visits. c. If many groups of 214 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of visits per week and about percent will not contain the true population mean number of visits per week. Hint: Hints 준 Video 준 Textbook [?

Step 1 ：We are given that the sample mean (\(\bar{x}\)) is 3.1, the standard deviation (s) is 1.6, and the sample size (n) is 214. We want to find a 95% confidence interval for the mean number of visits per week.

Step 2 ：The formula for a confidence interval is \(\bar{x} \pm t \frac{s}{\sqrt{n}}\), where t is the t-score for the desired confidence level.

Step 3 ：We need to find the t-score for a 95% confidence level with n - 1 = 213 degrees of freedom. Using a t-distribution table or a statistical calculator, we find that the t-score is approximately 1.971.

Step 4 ：Substituting the given values into the formula, we get \(3.1 \pm 1.971 \frac{1.6}{\sqrt{214}}\).

Step 5 ：Calculating the above expression, we find that the 95% confidence interval for the mean number of visits per week is between 2.884 and 3.316.

Step 6 ：\(\boxed{\text{Final Answer: With 95% confidence, the population mean number of visits per week is between 2.884 and 3.316 visits.}}\)