Step 1 :Given values are: sample mean (\(x_{bar}\)) = \$26369.38, z-score (z) for 95% confidence level = 1.96, standard deviation (\(\sigma\)) = \$8500, and sample size (n) = 20.
Step 2 :Calculate the margin of error using the formula: \(margin\_of\_error = z \times \frac{\sigma}{\sqrt{n}}\).
Step 3 :Substitute the given values into the formula: \(margin\_of\_error = 1.96 \times \frac{8500}{\sqrt{20}}\), which gives \(margin\_of\_error \approx \$3725.29\).
Step 4 :Calculate the confidence interval using the formula: \(lower\_limit = x_{bar} - margin\_of\_error\) and \(upper\_limit = x_{bar} + margin\_of\_error\).
Step 5 :Substitute the values into the formula: \(lower\_limit = 26369.38 - 3725.29 \approx \$22644.09\) and \(upper\_limit = 26369.38 + 3725.29 \approx \$30094.67\).
Step 6 :Final Answer: The 95% confidence interval for the mean cost, \(\mu\), of all recent weddings in this country is from \(\boxed{\$22644.09}\) to \(\boxed{\$30094.67}\).