Question

8.16 guest ages at a ski mountain typically have a right - skewed distribution. assume the standard deviation (σ) of age is 14.5 years. (a) even though the population distribution of age is right - skewed, what will be the shape of the distribution of (overline{x}), the average age, in a random sample of 40 guests? (b) from a random sample of 40 guests, the sample mean is 36.4 years. calculate a 99 percent confidence interval for (mu), the true mean age of ski mountain guests.

Explanation:

Step1: Determine the shape of the sampling distribution

According to the Central Limit Theorem, when the sample size \(n\geq30\), the sampling distribution of the sample mean \(\bar{X}\) is approximately normal. Here, \(n = 40\geq30\).

Step2: Calculate the standard error

The formula for the standard error \(SE=\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 14.5\) and \(n = 40\), then \(SE=\frac{14.5}{\sqrt{40}}\approx\frac{14.5}{6.3246}\approx2.29\)

Step3: Find the z - score for a 99% confidence interval

For a 99% confidence interval, the significance level \(\alpha=1 - 0.99=0.01\), and \(\alpha/2=0.005\). Looking up in the standard normal table, \(z_{\alpha/2}=z_{0.005} = 2.576\)

Step4: Calculate the margin of error

The margin of error \(E = z_{\alpha/2}\times SE\). Substituting the values, \(E=2.576\times2.29\approx5.9\)

Step5: Calculate the confidence interval

The confidence interval is \(\bar{X}\pm E\). Given \(\bar{X} = 36.4\), the lower limit is \(36.4-5.9 = 30.5\) and the upper limit is \(36.4 + 5.9=42.3\)

Answer:

(a) The distribution of \(\bar{X}\) is approximately normal.
(b) The 99% confidence interval for \(\mu\) is \((30.5,42.3)\)