1. **Problem statement:** We have 500 students with mean weight $\mu = 75.5$ kg and standard deviation $\sigma = 7.5$ kg. We assume weights are normally distributed. We want to find: - Number of students weighing less than 64 kg. - Number of students weighing more than 92.5 kg. - Number of students weighing between 60 and 77.5 kg. 2. **Formula and rules:** For a normal distribution, the z-score is calculated as: $$ z = \frac{X - \mu}{\sigma} $$ where $X$ is the value of interest. We use standard normal distribution tables or a calculator to find probabilities corresponding to z-scores. 3. **Step 1: Less than 64 kg** Calculate z-score: $$ z = \frac{64 - 75.5}{7.5} = \frac{-11.5}{7.5} = -1.5333 $$ From standard normal tables, $P(Z < -1.5333) \approx 0.0625$. Number of students: $$ 0.0625 \times 500 = 31.25 \approx 31 $$ students. 4. **Step 2: More than 92.5 kg** Calculate z-score: $$ z = \frac{92.5 - 75.5}{7.5} = \frac{17}{7.5} = 2.2667 $$ From tables, $P(Z < 2.2667) \approx 0.9883$. So, $$ P(Z > 2.2667) = 1 - 0.9883 = 0.0117 $$ Number of students: $$ 0.0117 \times 500 = 5.85 \approx 6 $$ students. 5. **Step 3: Between 60 and 77.5 kg** Calculate z-scores: $$ z_1 = \frac{60 - 75.5}{7.5} = \frac{-15.5}{7.5} = -2.0667 $$ $$ z_2 = \frac{77.5 - 75.5}{7.5} = \frac{2}{7.5} = 0.2667 $$ From tables: $$ P(Z < -2.0667) \approx 0.0194 $$ $$ P(Z < 0.2667) \approx 0.6054 $$ Probability between: $$ 0.6054 - 0.0194 = 0.586 $$ Number of students: $$ 0.586 \times 500 = 293 $$ students. **Final answers:** - Less than 64 kg: 31 students - More than 92.5 kg: 6 students - Between 60 and 77.5 kg: 293 students