1. **Problem statement:** We have data on the number of students and their average daily reading times in intervals. We need to answer three parts: (i) Find the maximum number of students who could have read for more than 100 minutes. (ii) Estimate the median reading time. (iii) Explain the limitation of the sampling method. --- 2. **Given data:** | Minutes spent reading per day | Number of students | |-------------------------------|--------------------| | 0 - 30 | 16 | | 30 - 60 | 35 | | 60 - 90 | 62 | | 90 - 120 | 59 | | 120 - 150 | 18 | | 150 - 180 | 10 | Total students = 200 --- ### Part (i) Maximum number of students reading more than 100 minutes 3. The intervals that include reading times more than 100 minutes are: - Part of 90 - 120 (from 100 to 120) - 120 - 150 - 150 - 180 4. To find the maximum number, assume all students in 120 - 150 and 150 - 180 read more than 100 minutes. 5. For 90 - 120 interval (59 students), only those reading more than 100 minutes count. The interval length is 30 minutes (90 to 120). 6. The portion from 100 to 120 is 20 minutes out of 30 minutes. 7. Assuming uniform distribution, number of students reading more than 100 minutes in 90 - 120 interval: $$59 \times \frac{20}{30} = 59 \times \frac{2}{3} = 39.33 \approx 39$$ 8. Total maximum number reading more than 100 minutes: $$39 + 18 + 10 = 67$$ --- ### Part (ii) Estimate the median reading time 9. Median is the middle value when data is ordered. For 200 students, median is the average of 100th and 101st values. 10. Calculate cumulative frequencies: | Interval | Frequency | Cumulative Frequency | |------------|-----------|----------------------| | 0 - 30 | 16 | 16 | | 30 - 60 | 35 | 51 | | 60 - 90 | 62 | 113 | | 90 - 120 | 59 | 172 | | 120 - 150 | 18 | 190 | | 150 - 180 | 10 | 200 | 11. The 100th and 101st students lie in the 60 - 90 interval (cumulative frequency 51 to 113). 12. Use linear interpolation to estimate median within 60 - 90 interval: - Lower boundary $L = 60$ - Frequency in interval $f = 62$ - Cumulative frequency before interval $CF = 51$ - Median position $M = 100.5$ (average of 100 and 101) - Interval width $w = 30$ 13. Median formula: $$\text{Median} = L + \left(\frac{M - CF}{f}\right) \times w$$ 14. Substitute values: $$\text{Median} = 60 + \left(\frac{100.5 - 51}{62}\right) \times 30 = 60 + \left(\frac{49.5}{62}\right) \times 30$$ 15. Calculate: $$\frac{49.5}{62} \approx 0.7984$$ $$0.7984 \times 30 = 23.95$$ $$\text{Median} \approx 60 + 23.95 = 83.95$$ 16. So, the estimated median reading time is approximately 84 minutes. --- ### Part (iii) Limitation of sampling method 17. The students were chosen as the first 200 who came into the school on Monday morning. 18. This is a convenience sample, not random. 19. It may not represent all students because: - Time of day may bias who is present. - Students arriving first may have different habits. - It excludes students absent or arriving later. 20. Therefore, conclusions may not generalize to all students in Ireland. --- **Final answers:** (i) Maximum number reading more than 100 minutes: **67** (ii) Estimated median reading time: **approximately 84 minutes** (iii) Sampling method limits validity because it is a convenience sample and may not represent the whole population fairly.