The mean and standard deviation of a random sample of $n$ measurements are equal to 34.9 and 3.5 , respectively. b. Find a $90 \%$ confidence interval for $\mu$ if $n=576$. c. Find the widths of the confidence intervals found in parts a and $\mathbf{b}$. What is the effect on the width of a confidence interval of quadrupling the sample size while holding the confidence coefficient fixed? b. The $90 \%$ confidence interval $\mu$ if $n=576$ is approximately (Round to three decimal places as needed.)

Step 1 ：Given that the sample mean (\(\bar{x}\)) is 34.9, the standard deviation (\(\sigma\)) is 3.5, and the sample size (\(n\)) is 576.

Step 2 ：The Z-score for a 90% confidence interval is approximately 1.645.

Step 3 ：The formula for the confidence interval for a population mean is \(\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\).

Step 4 ：Substitute the given values into the formula to find the margin of error: \(1.645 \times \frac{3.5}{\sqrt{576}}\), which is approximately 0.23989583333333336.

Step 5 ：Subtract the margin of error from the sample mean to find the lower bound of the confidence interval: \(34.9 - 0.23989583333333336\), which is approximately 34.66010416666666.

Step 6 ：Add the margin of error to the sample mean to find the upper bound of the confidence interval: \(34.9 + 0.23989583333333336\), which is approximately 35.139895833333334.

Step 7 ：Final Answer: The $90 \%$ confidence interval for \(\mu\) when \(n=576\) is approximately \(\boxed{(34.660, 35.140)}\).