1. **Problem Statement:** We want to find the 70th percentile $y^*$ of the fish length distribution, meaning 70% of fish lengths are less than $y^*$ and 30% are greater. 2. **Formula and Explanation:** The 70th percentile corresponds to a z-score $z_{0.30} = 0.52$ on the standard normal distribution, where 30% of the area lies to the right. 3. **Standardization Formula:** The relationship between $y^*$ and $z$ is given by: $$z = \frac{y^* - \mu}{\sigma}$$ where $\mu = 54$ (mean) and $\sigma = 4.5$ (standard deviation). 4. **Substitute Known Values:** Using $z = 0.52$, we have: $$0.52 = \frac{y^* - 54}{4.5}$$ 5. **Solve for $y^*$:** Multiply both sides by 4.5: $$y^* - 54 = 0.52 \times 4.5$$ $$y^* - 54 = 2.34$$ Add 54 to both sides: $$y^* = 54 + 2.34 = 56.34$$ 6. **Interpretation:** The 70th percentile of the fish length distribution is approximately $56.3$ mm, meaning 70% of fish are shorter than this length.