1. **Problem statement:** Given a normal distribution with mean $\mu=25$ and standard deviation $\sigma=7.5$, find the percentage of scores: a) Greater than 40 b) Lower than 40 c) Between 28 and 42 2. **Formula and rules:** We use the standard normal distribution (Z) formula to convert scores to Z-scores: $$Z = \frac{X - \mu}{\sigma}$$ where $X$ is the score, $\mu$ is the mean, and $\sigma$ is the standard deviation. Then, use standard normal distribution tables or a calculator to find probabilities. 3. **Step-by-step calculations:** a) Greater than 40: Calculate $Z$ for $X=40$: $$Z = \frac{40 - 25}{7.5} = \frac{15}{7.5} = 2$$ The probability that $Z$ is greater than 2 is $P(Z > 2) = 1 - P(Z \leq 2)$. From standard normal tables, $P(Z \leq 2) \approx 0.9772$. So, $P(Z > 2) = 1 - 0.9772 = 0.0228$ or 2.28%. b) Lower than 40: This is $P(X < 40) = P(Z < 2) = 0.9772$ or 97.72%. c) Between 28 and 42: Calculate $Z$ for $X=28$: $$Z = \frac{28 - 25}{7.5} = \frac{3}{7.5} = 0.4$$ Calculate $Z$ for $X=42$: $$Z = \frac{42 - 25}{7.5} = \frac{17}{7.5} \approx 2.27$$ From tables, $P(Z < 0.4) \approx 0.6554$ and $P(Z < 2.27) \approx 0.9884$. So, $P(28 < X < 42) = P(0.4 < Z < 2.27) = 0.9884 - 0.6554 = 0.333$ or 33.3%. 4. **Final answers:** - Greater than 40: 2.28% - Lower than 40: 97.72% - Between 28 and 42: 33.3%