Question 3 6 pts 37 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 10.3 pounds and a population standard deviation of 2.4 pounds. What is the $95 \%$ confidence interval for the true mean weight, $\mu$, of all packages received by the parcel service? Round your answer to the nearest ten-thousandth. (e.g. 0.35889 would enter as 0.3589 ) between and - Previous Next
Solution
Step 1 ：Given values are: sample mean \(X = 10.3\), Z-score for 95% confidence level \(Z = 1.96\), population standard deviation \(\sigma = 2.4\), and sample size \(n = 37\).
Step 2 ：Calculate the margin of error using the formula: \(Z \times \frac{\sigma}{\sqrt{n}}\). Substituting the given values, we get the margin of error as \(1.96 \times \frac{2.4}{\sqrt{37}} = 0.7733328362844006\).
Step 3 ：Calculate the confidence interval using the formula: \(X \pm\) margin of error. Substituting the given values, we get the confidence interval as \(10.3 \pm 0.7733328362844006\).
Step 4 ：So, the lower limit of the confidence interval is \(10.3 - 0.7733328362844006 = 9.5266671637156\), and the upper limit of the confidence interval is \(10.3 + 0.7733328362844006 = 11.0733328362844\).
Step 5 ：Final Answer: The 95% confidence interval for the true mean weight of all packages received by the parcel service is between \(\boxed{9.5267}\) pounds and \(\boxed{11.0733}\) pounds.
