1. **Problem statement:** We have a frequency distribution of students' long jump distances in intervals and need to answer several questions about the data. 2. **Given data:** - Intervals (cm): 0-80, 80-160, 160-240, 240-320, 320-400 - Number of students: 0, 2, 7, 12, 3 3. **(a) Number of students who jumped less than 240 cm:** This includes students in intervals 0-80, 80-160, and 160-240. Calculate: $$0 + 2 + 7 = 9$$ So, 9 students jumped less than 240 cm. 4. **(c) Maximum number of students who could have jumped more than 200 cm:** - The intervals 240-320 and 320-400 are fully above 200 cm, so all 12 + 3 = 15 students in these intervals jumped more than 200 cm. - The interval 160-240 partially overlaps with distances above 200 cm. The number of students in 160-240 is 7. To find the maximum number who jumped more than 200 cm, assume all 7 students in 160-240 jumped more than 200 cm. Therefore, maximum number: $$15 + 7 = 22$$ 5. **(d) Estimate the mean distance using mid-interval values:** - Mid-interval values: - 0-80: $\frac{0 + 80}{2} = 40$ - 80-160: $\frac{80 + 160}{2} = 120$ - 160-240: $\frac{160 + 240}{2} = 200$ - 240-320: $\frac{240 + 320}{2} = 280$ - 320-400: $\frac{320 + 400}{2} = 360$ - Multiply mid-interval by frequency: - $40 \times 0 = 0$ - $120 \times 2 = 240$ - $200 \times 7 = 1400$ - $280 \times 12 = 3360$ - $360 \times 3 = 1080$ - Sum of products: $$0 + 240 + 1400 + 3360 + 1080 = 6080$$ - Total number of students: $$0 + 2 + 7 + 12 + 3 = 24$$ - Mean distance: $$\frac{6080}{24} = 253.33$$ Rounded to nearest centimetre: $$253$$ --- **Final answers:** - (a) 9 students - (c) Maximum 22 students - (d) Mean distance approximately 253 cm