1. **State the problem:** We have ages of eldest children summarized by a box-and-whisker plot with minimum 2, lower quartile 6, median 7, upper quartile 10, and maximum 18. (a) Find the largest value of $c$ (age of eldest child) that is not an outlier. (b) Given regression lines $a = \frac{7}{4}c + 20$ and $c = \frac{1}{2}a - 9$, (i) estimate the child's age if adult is 42, (ii) find mean age of adults. 2. **Formula and rules for outliers:** Outliers are values outside the range: $$\text{Lower bound} = Q_1 - 1.5 \times IQR$$ $$\text{Upper bound} = Q_3 + 1.5 \times IQR$$ where $IQR = Q_3 - Q_1$. 3. **Calculate IQR:** $$IQR = 10 - 6 = 4$$ 4. **Calculate upper bound for outliers:** $$\text{Upper bound} = 10 + 1.5 \times 4 = 10 + 6 = 16$$ 5. **Interpretation:** Largest $c$ not an outlier is 16 years old. 6. **Part (b)(i) Estimate child's age for adult 42:** Use regression $c = \frac{1}{2}a - 9$: $$c = \frac{1}{2} \times 42 - 9 = 21 - 9 = 12$$ 7. **Part (b)(ii) Find mean age of adults:** Use regression $a = \frac{7}{4}c + 20$ and mean $c$. 8. **Find mean $c$ from box plot:** Approximate mean $c$ as average of min, Q1, median, Q3, max: $$\frac{2 + 6 + 7 + 10 + 18}{5} = \frac{43}{5} = 8.6$$ 9. **Calculate mean $a$:** $$a = \frac{7}{4} \times 8.6 + 20 = 15.05 + 20 = 35.05$$ **Final answers:** - Largest $c$ not an outlier: 16 - Estimated child's age for adult 42: 12 - Mean adult age: 35.05