1. **State the problem:** We want to find the proportion of the class that scored above 74% given the test average (mean) is 85%, median is 89%, and standard deviation (SD) is 8%. 2. **Assumption:** We assume the test scores are approximately normally distributed since mean, median, and SD are given. 3. **Formula:** To find the proportion above a certain score $x$, we use the standard normal distribution. First, convert $x=74$ to a z-score: $$z = \frac{x - \mu}{\sigma}$$ where $\mu=85$ and $\sigma=8$. 4. **Calculate z-score:** $$z = \frac{74 - 85}{8} = \frac{-11}{8} = -1.375$$ 5. **Interpret z-score:** A z-score of -1.375 means 74 is 1.375 standard deviations below the mean. 6. **Find proportion above 74:** Using standard normal distribution tables or a calculator, find $P(Z > -1.375)$. Since the normal distribution is symmetric, $$P(Z > -1.375) = 1 - P(Z < -1.375) = 1 - \Phi(-1.375)$$ But $\Phi(-1.375)$ is the cumulative probability up to -1.375. From tables or calculator, $$\Phi(-1.375) \approx 0.084$$ Therefore, $$P(Z > -1.375) = 1 - 0.084 = 0.916$$ 7. **Final answer:** Approximately 91.6% of the class scored above 74%.