1. **State the problem:** We want to find the percentage of data values greater than $x=22$ in a normal distribution with mean $\mu=24.8$ and standard deviation $\sigma=5.94$. 2. **Use the z-score formula:** $$z = \frac{x - \mu}{\sigma}$$ This formula converts the raw score $x$ into a standard normal variable $z$. 3. **Calculate the z-score for $x=22$:** $$z = \frac{22 - 24.8}{5.94} = \frac{\cancel{22 - 24.8}}{\cancel{5.94}} = -0.47$$ 4. **Interpret the z-score:** A $z$ of $-0.47$ means $22$ is $0.47$ standard deviations below the mean. 5. **Find the cumulative probability for $z=-0.47$:** Using standard normal tables or a calculator, the cumulative probability $P(Z < -0.47) = 0.3192$. 6. **Calculate the percentage greater than $22$:** $$P(X > 22) = 1 - P(Z < -0.47) = 1 - 0.3192 = 0.6808 = 68.08\%$$ **Final answer:** Approximately **68.08%** of the data is greater than $22$.