1. **Problem statement:** Given the frequency table of fish caught: Number of fish: 0, 1, 2, 3, 4, 5, 6 Frequency: 6, 20, 45, 70, 35, 10, 2 Calculate: i) Mean number of fish caught ii) Modal number of fish caught iii) Median number of fish caught iv) Range in the number of fish caught 2. **Formulas and rules:** - Mean: $$\text{mean} = \frac{\sum (x \times f)}{\sum f}$$ where $x$ is the number of fish and $f$ is the frequency. - Mode: The value of $x$ with the highest frequency. - Median: The middle value when data is ordered; find the cumulative frequency and locate the middle position. - Range: $$\text{range} = \text{max value} - \text{min value}$$ 3. **Calculations:** - Total frequency: $$6 + 20 + 45 + 70 + 35 + 10 + 2 = 188$$ - Calculate $\sum (x \times f)$: $$0 \times 6 = 0$$ $$1 \times 20 = 20$$ $$2 \times 45 = 90$$ $$3 \times 70 = 210$$ $$4 \times 35 = 140$$ $$5 \times 10 = 50$$ $$6 \times 2 = 12$$ Sum: $$0 + 20 + 90 + 210 + 140 + 50 + 12 = 522$$ - Mean: $$\frac{522}{188} \approx 2.7766$$ - Mode: Highest frequency is 70 at number of fish = 3, so mode = 3. - Median: Total frequency = 188, so median position = $$\frac{188 + 1}{2} = 94.5$$th value. Cumulative frequencies: 0 fish: 6 1 fish: 6 + 20 = 26 2 fish: 26 + 45 = 71 3 fish: 71 + 70 = 141 Since 94.5 lies between 71 and 141, median number of fish = 3. - Range: Max number of fish = 6, Min number = 0 Range = $$6 - 0 = 6$$ 4. **Assumption for part b:** The rate of fish caught per hour remains constant over the next four hours as it was in the first two hours. **Final answers:** i) Mean = 2.78 (rounded to 2 decimal places) ii) Mode = 3 iii) Median = 3 iv) Range = 6 Assumption: The catch rate is constant over time.