1. **Problem statement:** (a) Find the value of $z_1$ such that the area under the standard normal curve from 0 to $z_1$ is 67%. 2. **Formula and rules:** The standard normal distribution has mean 0 and standard deviation 1. The total area under the curve is 1. The area from 0 to $z_1$ corresponds to $P(0 < Z < z_1) = 0.67$. 3. **Find $z_1$:** Since the standard normal curve is symmetric, the area from $-z_1$ to $z_1$ would be twice this, but here only from 0 to $z_1$ is 0.67. The cumulative area from $-\\infty$ to 0 is 0.5. So the cumulative area from $-\infty$ to $z_1$ is $0.5 + 0.67 = 1.17$, which is impossible. This means the problem likely means the shaded area from 0 to $z_1$ is 0.67, so the cumulative area from $-\infty$ to $z_1$ is $0.5 + 0.67 = 1.17$ which cannot be. Instead, the shaded area is 67% of the data, so the area from $-\infty$ to $z_1$ is 0.67. Therefore, $P(Z < z_1) = 0.67$. 4. **Use the standard normal table or inverse CDF:** $z_1 = \Phi^{-1}(0.67) \approx 0.44$ --- 5. **Problem (b)(i):** Mary's scores: Maths = 65, English = 68. Maths mean = 70, SD = 15. English mean = 72, SD = 10. Calculate z-scores: $$z_{Maths} = \frac{65 - 70}{15} = \frac{-5}{15} = -0.33$$ $$z_{English} = \frac{68 - 72}{10} = \frac{-4}{10} = -0.4$$ Mary's z-score in Maths is higher (less negative) than in English, so she did better relative to the class in Maths. --- 6. **Problem (b)(ii):** Top 15% in English get A grade. Find the least mark $x$ such that $P(X \geq x) = 0.15$. This means $P(X < x) = 0.85$. Find $z$ such that $P(Z < z) = 0.85$. From standard normal table, $z \approx 1.04$. Convert back to marks: $$x = \mu + z \sigma = 72 + 1.04 \times 10 = 72 + 10.4 = 82.4$$ Least whole number mark is 83. --- 7. **Problem (b)(iii):** Estimate percentage of students scoring between 52 and 82 in English. Mean = 72, SD = 10. Calculate z-scores: $$z_1 = \frac{52 - 72}{10} = -2$$ $$z_2 = \frac{82 - 72}{10} = 1$$ Using empirical rule: - About 95% of data lies within $\pm 2$ SD. - From $z=-2$ to $z=1$ covers from -2 SD to +1 SD. Area from $-2$ to $1$ is approximately: $P(Z < 1) - P(Z < -2) = 0.8413 - 0.0228 = 0.8185$ or 81.85%. **Final answers:** (a) $z_1 \approx 0.44$ (b)(i) Mary did better in Maths because her z-score is higher ($-0.33$ vs $-0.4$). (b)(ii) Least mark for A grade is 83. (b)(iii) Approximately 82% of students scored between 52 and 82 in English.