1. **State the problem:** We need to estimate the mean distance jumped by the group of students using mid-interval values. 2. **Data given:** - Distance intervals (cm): 0–80, 80–160, 160–240, 240–320, 320–400 - Number of students: 0, 2, 7, 12, 3 3. **Find mid-interval values:** - Midpoint of 0–80 = $\frac{0 + 80}{2} = 40$ - Midpoint of 80–160 = $\frac{80 + 160}{2} = 120$ - Midpoint of 160–240 = $\frac{160 + 240}{2} = 200$ - Midpoint of 240–320 = $\frac{240 + 320}{2} = 280$ - Midpoint of 320–400 = $\frac{320 + 400}{2} = 360$ 4. **Calculate total distance jumped:** $$\text{Total distance} = (0 \times 40) + (2 \times 120) + (7 \times 200) + (12 \times 280) + (3 \times 360)$$ $$= 0 + 240 + 1400 + 3360 + 1080 = 6080$$ 5. **Calculate total number of students:** $$0 + 2 + 7 + 12 + 3 = 24$$ 6. **Calculate mean distance jumped:** $$\text{Mean} = \frac{\text{Total distance}}{\text{Total students}} = \frac{6080}{24}$$ 7. **Simplify fraction:** $$\frac{6080}{24} = \frac{\cancel{6080}^{253} \times 24}{\cancel{24}^{1} \times 24} = 253.33...$$ 8. **Final answer:** The estimated mean distance jumped is approximately **253 cm** (to the nearest centimetre).