A fitness center is interested in finding a $90 \%$ 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 234 members were looked at and their mean number of visits per week was 3.4 and the standard deviation was 2.7. Round answers to 3 decimal places where possible. a. To compute the confidence interval use a $?$ distribution. b. With $90 \%$ confidence the population mean number of visits per week is between and visits. c. If many groups of 234 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.

Step 1 ：We are given that the sample mean (\(\bar{x}\)) is 3.4, the standard deviation (s) is 2.7, and the sample size (n) is 234. The confidence level is 90%, so the Z-score is approximately 1.645.

Step 2 ：We can use these values to calculate the margin of error for the confidence interval. The formula for the margin of error is \(Z \frac{s}{\sqrt{n}}\).

Step 3 ：Substituting the given values into the formula, we get \(1.645 \frac{2.7}{\sqrt{234}}\).

Step 4 ：Calculating the above expression, we get the margin of error to be approximately 0.290.

Step 5 ：We can now calculate the confidence interval. The formula for the confidence interval is \(\bar{x} \pm \) margin of error.

Step 6 ：Substituting the given values into the formula, we get \(3.4 \pm 0.290\).

Step 7 ：Calculating the above expression, we get the confidence interval to be between 3.110 and 3.690.

Step 8 ：\(\boxed{\text{Final Answer: With 90% confidence, the population mean number of visits per week is between 3.110 and 3.690 visits.}}\)