1. **State the problem:** We need to find the maximum score a student can have to be in the bottom 5% of the scores, given that the scores are normally distributed with mean $\mu = 65$ and standard deviation $\sigma = 20$. 2. **Understand the distribution:** The scores follow a normal distribution $X \sim N(65, 20^2)$. 3. **Find the z-score corresponding to the bottom 5%:** The bottom 5% means the 5th percentile. From standard normal distribution tables or using a calculator, the z-score for the 5th percentile is approximately $z = -1.645$. 4. **Use the z-score formula:** $$ z = \frac{X - \mu}{\sigma} $$ Rearranged to find $X$: $$ X = z \times \sigma + \mu $$ 5. **Calculate the maximum score:** $$ X = (-1.645) \times 20 + 65 = -32.9 + 65 = 32.1 $$ 6. **Interpretation:** The maximum score to be in the bottom 5% is approximately 32.1 marks. **Final answer:** The maximum score for a student to be eligible for the programme is **32.1 marks**.