1. **State the problem:** We want to find the P-value for testing the hypothesis \(H_0: \mu = 6\) against \(H_A: \mu < 6\) given a sample mean critical region \(\overline{X} < 5.8\), sample size \(n=100\), population standard deviation \(\sigma=1.5\). 2. **Formula and explanation:** The test statistic for the sample mean when population standard deviation is known is $$Z = \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}}$$ where \(\mu_0 = 6\) is the hypothesized mean. 3. **Calculate the test statistic:** $$Z = \frac{5.8 - 6}{1.5 / \sqrt{100}} = \frac{-0.2}{1.5 / 10} = \frac{-0.2}{0.15} = -1.33$$ 4. **Find the P-value:** Since the alternative hypothesis is \(\mu < 6\), the P-value is the probability that \(Z\) is less than \(-1.33\). Using standard normal distribution tables or a calculator: $$P(Z < -1.33) \approx 0.0918$$ 5. **Final answer:** The P-value rounded to two decimal places is **0.09**. This means there is about a 9% chance of observing a sample mean less than 5.8 if the true mean is 6.