1. **Problem Statement:** We have a normally distributed variable $X$ representing exam marks with mean $\mu = 45$ and variance $\sigma^2 = 13$. (a) Find the proportion of candidates scoring less than 42. (b) With 50 candidates taking the exam, find how many score over 50 and thus get a prize. --- 2. **Step 1: Calculate the standard deviation:** Since variance $\sigma^2 = 13$, standard deviation $\sigma = \sqrt{13} \approx 3.605$. 3. **Step 2: Standardize the scores** for part (a) and (b). For any score $x$, the standardized $Z$-score is: $$Z = \frac{x - \mu}{\sigma}$$ ---- ### Part (a): Proportion scoring less than 42 4. Calculate $Z$ for $x=42$: $$Z = \frac{42 - 45}{3.605} = \frac{-3}{3.605} \approx -0.832$$ 5. Find $P(X < 42) = P(Z < -0.832)$. Using standard normal distribution tables or a calculator: $$P(Z < -0.832) \approx 0.203$$ Thus, about 20.3% score less than 42. --- ### Part (b): Number of candidates scoring over 50 6. Calculate $Z$ for $x=50$: $$Z = \frac{50 - 45}{3.605} = \frac{5}{3.605} \approx 1.387$$ 7. Find $P(X > 50) = P(Z > 1.387) = 1 - P(Z \leq 1.387)$. From standard normal tables: $$P(Z \leq 1.387) \approx 0.917$$ Hence, $$P(X > 50) = 1 - 0.917 = 0.083$$ 8. Expected number of prize winners among 50 candidates: $$50 \times 0.083 = 4.15$$ Rounded to nearest whole number: $$4$$ candidates receive a prize. --- **Final Answers:** (a) Proportion scoring less than 42 is approximately **0.203**. (b) Number of candidates scoring over 50 and getting a prize is approximately **4**.