1. **Problem Statement:** What percentage of students spend less than 2.5 hours per day on social media? 2. **Given Information:** Mean $\mu = 3.5$ hours, Standard Deviation $\sigma = 1$ hour. 3. **Formula for z-score:** $$z = \frac{X - \mu}{\sigma}$$ where $X$ is the value of interest. 4. **Calculate the z-score for $X = 2.5$ hours:** $$z = \frac{2.5 - 3.5}{1} = \frac{-1}{1} = -1$$ 5. **Interpretation:** A z-score of $-1$ means 2.5 hours is 1 standard deviation below the mean. 6. **Use the z-table:** The z-table gives the cumulative probability to the left of $z = -1$. From the z-table, $P(Z < -1) \approx 0.1587$ or 15.87%. 7. **Conclusion:** Approximately 15.87% of students spend less than 2.5 hours per day on social media.