1. **State the problem:** We have a frequency distribution of garden areas with some missing frequencies. We know the total number of responses is 100, and the frequency in the interval $20 < x \leq 30$ is twice the frequency in $30 < x \leq 40$. We need to estimate the mean garden area to 1 decimal place. 2. **Set variables:** Let the frequency in $30 < x \leq 40$ be $f$. Then the frequency in $20 < x \leq 30$ is $2f$. 3. **Write the total frequency equation:** $$16 + 44 + 2f + f + 4 = 100$$ Simplify: $$64 + 3f = 100$$ $$3f = 36$$ $$f = 12$$ 4. **Find missing frequencies:** Frequency in $30 < x \leq 40$ is $12$. Frequency in $20 < x \leq 30$ is $2 \times 12 = 24$. 5. **Calculate midpoints of intervals:** - $0 < x \leq 10$: midpoint $= 5$ - $10 < x \leq 20$: midpoint $= 15$ - $20 < x \leq 30$: midpoint $= 25$ - $30 < x \leq 40$: midpoint $= 35$ - $40 < x \leq 50$: midpoint $= 45$ 6. **Calculate weighted sum of midpoints and frequencies:** $$\text{Sum} = 16 \times 5 + 44 \times 15 + 24 \times 25 + 12 \times 35 + 4 \times 45$$ Calculate each term: $$16 \times 5 = 80$$ $$44 \times 15 = 660$$ $$24 \times 25 = 600$$ $$12 \times 35 = 420$$ $$4 \times 45 = 180$$ Sum all: $$80 + 660 + 600 + 420 + 180 = 1940$$ 7. **Calculate mean:** $$\text{Mean} = \frac{\text{Sum of (midpoint} \times \text{frequency)}}{\text{Total frequency}} = \frac{1940}{100} = 19.4$$ **Final answer:** The estimated mean garden area is **19.4 m²**.