1. The problem asks to find the P-value for a right-tailed test with a test statistic $z = 1.36$. 2. In a right-tailed test, the P-value is the probability that the test statistic is greater than the observed value, i.e., $P(Z > 1.36)$ where $Z$ is a standard normal variable. 3. We use the standard normal distribution table or a calculator to find $P(Z \leq 1.36)$, the cumulative probability up to 1.36. 4. From the standard normal table, $P(Z \leq 1.36) = 0.9131$. 5. Since the test is right-tailed, the P-value is: $$P\text{-value} = 1 - P(Z \leq 1.36) = 1 - 0.9131 = 0.0869$$ 6. Therefore, the correct P-value is $0.0869$. This matches the second option given.