1. **Problem statement:** We have 500 candidates with marks normally distributed with mean $\mu=40$ and standard deviation $\sigma=10$. We need to find: - Number of candidates passing if pass mark is 50. - Pass mark if 350 candidates are to pass. 2. **Recall the normal distribution properties:** The standardized variable is $Z=\frac{X-\mu}{\sigma}$ where $X$ is the mark. 3. **Number passing if pass mark is 50:** Calculate $Z$ for pass mark 50: $$Z=\frac{50-40}{10}=1$$ Find $P(X>50)=P(Z>1)$. From standard normal tables, $P(Z>1)=1-P(Z\leq1)=1-0.8413=0.1587$. Number passing $=0.1587 \times 500=79.35 \approx 79$ candidates. 4. **Pass mark if 350 candidates pass:** We want $P(X>k)=\frac{350}{500}=0.7$. So $P(X\leq k)=0.3$. Find $Z$ such that $P(Z\leq z)=0.3$. From tables, $z\approx -0.524$. Calculate pass mark $k$: $$k=\mu + z\sigma=40 + (-0.524) \times 10=40 - 5.24=34.76$$ **Final answers:** - Number passing if pass mark is 50: **79** candidates. - Pass mark if 350 candidates pass: **34.76** marks.