1. The problem asks to find the p-value for a right-tailed test with 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: $$p\text{-value} = P(Z > 1.36)$$ where $Z$ is a standard normal variable. 3. Using standard normal distribution tables or a calculator, find the cumulative probability up to $z=1.36$: $$P(Z \leq 1.36) = 0.9131$$ 4. Since the p-value is the area to the right, calculate: $$p\text{-value} = 1 - P(Z \leq 1.36) = 1 - 0.9131 = 0.0869$$ 5. Therefore, the p-value for the test statistic $z=1.36$ in a right-tailed test is $0.0869$. Final answer: **p-value = 0.0869**