In a random sample of five mobile devices, the mean repair cost was $\$ 75.00$ and the standard deviation was $\$ 12.50$. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a $95 \%$ confidence interval for the population mean. Interpret the results. The $95 \%$ confidence interval for the population mean $\mu$ is (Round to two decimal places as needed.)

Step 1 ：Given that the sample mean (\(\bar{x}\)) is \$75.00, the standard deviation (s) is \$12.50, and the sample size (n) is 5.

Step 2 ：First, we need to calculate the degrees of freedom, which is n - 1 = 5 - 1 = 4.

Step 3 ：Next, we look up the t-score for a 95% confidence level and 4 degrees of freedom in a t-table or using a calculator. The t-score (t) is approximately 2.776.

Step 4 ：We then calculate the margin of error using the formula: Margin of Error = t * (s/√n) = 2.776 * (12.50/√5) = \$15.52.

Step 5 ：Finally, we construct the 95% confidence interval for the population mean by subtracting and adding the margin of error from/to the sample mean: (75 - 15.52, 75 + 15.52) = (\$59.48, \$90.52).

Step 6 ：\(\boxed{\text{Final Answer: The margin of error is approximately \$15.52. The 95% confidence interval for the population mean is approximately (\$59.48, \$90.52). This means that we are 95% confident that the true population mean repair cost lies within this interval.}}\)