1. **Problem statement:** Find the probability that the mean weight of 12 randomly selected men is greater than 167 pounds, given the population mean $\mu = 172$ and population standard deviation $\sigma = 29$. The sample size is $n = 12$. 2. **Formula and explanation:** For the sampling distribution of the sample mean, the standard deviation (standard error) is $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$. The z-score formula for the sample mean is: $$ Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} $$ 3. **Calculate the standard error:** $$ \sigma_{\bar{x}} = \frac{29}{\sqrt{12}} \approx \frac{29}{3.464} \approx 8.37 $$ 4. **Calculate the z-score for $\bar{x} = 167$:** $$ Z = \frac{167 - 172}{8.37} = \frac{-5}{8.37} \approx -0.60 $$ 5. **Find the probability:** The probability that the sample mean is greater than 167 is: $$ P(\bar{x} > 167) = P(Z > -0.60) $$ From standard normal tables or a calculator, $P(Z < -0.60) \approx 0.2743$, so $$ P(Z > -0.60) = 1 - 0.2743 = 0.7257 $$ 6. **Interpretation:** There is approximately a 72.57% chance that the mean weight of 12 randomly selected men is greater than 167 pounds. 7. **Statement:** If we randomly select 12 men, the probability that their average weight exceeds 167 pounds is about 72.57%. This means it is quite likely for the sample mean to be above 167.