1. **Stating the problem:** We are given two grouped frequency distributions and asked to solve question 4 (Q4). Since the user did not specify the exact question, we assume Q4 relates to calculating the mean from the grouped frequency table. 2. **Formula for mean of grouped data:** The mean $\bar{x}$ is given by: $$\bar{x} = \frac{\sum f x}{\sum f}$$ where $f$ is the frequency and $x$ is the midpoint of each class interval. 3. **Calculate midpoints for the exam mark groups:** - 20-29: $\frac{20+29}{2} = 24.5$ - 30-39: $\frac{30+39}{2} = 34.5$ - 40-49: $\frac{40+49}{2} = 44.5$ - 50-59: $\frac{50+59}{2} = 54.5$ - 60-69: $\frac{60+69}{2} = 64.5$ - 70-79: $\frac{70+79}{2} = 74.5$ - 80-89: $\frac{80+89}{2} = 84.5$ 4. **Use Frequency group A to calculate $\sum f x$:** $$\sum f x = 1\times24.5 + 3\times34.5 + 6\times44.5 + 6\times54.5 + 11\times64.5 + 10\times74.5 + 8\times84.5$$ Calculate each term: - $1\times24.5 = 24.5$ - $3\times34.5 = 103.5$ - $6\times44.5 = 267$ - $6\times54.5 = 327$ - $11\times64.5 = 709.5$ - $10\times74.5 = 745$ - $8\times84.5 = 676$ Sum these: $$24.5 + 103.5 + 267 + 327 + 709.5 + 745 + 676 = 2852.5$$ 5. **Sum of frequencies for group A:** $$1 + 3 + 6 + 6 + 11 + 10 + 8 = 45$$ 6. **Calculate mean:** $$\bar{x} = \frac{2852.5}{45}$$ Show cancellation: $$\bar{x} = \frac{\cancel{2852.5}}{\cancel{45}}$$ Calculate: $$\bar{x} \approx 63.39$$ 7. **Interpretation:** The mean exam mark for frequency group A is approximately 63.39. **Final answer:** $$\boxed{63.39}$$