1. **Problem statement:** We have a histogram representing the lengths of 575 items in a candle maker's workshop. We need to find: a) The proportion of items less than 25 cm long. b) The estimated number of items between 12.4 cm and 36.8 cm long. c) The length of the shortest item that will not be recycled if the shortest 20% are recycled. 2. **Understanding the histogram:** The histogram shows frequency density on the y-axis and length intervals on the x-axis. Frequency density \(= \frac{\text{frequency}}{\text{class width}}\). Frequency for each class = frequency density \(\times\) class width. 3. **Calculate frequencies for each class:** - Class 0 to 10 cm: height = 5, width = 10 Frequency = 5 \(\times\) 10 = 50 - Class 10 to 20 cm: height = 25, width = 10 Frequency = 25 \(\times\) 10 = 250 - Class 20 to 30 cm: height = 10, width = 10 Frequency = 10 \(\times\) 10 = 100 - Class 30 to 50 cm: height = 5, width = 20 Frequency = 5 \(\times\) 20 = 100 4. **Check total frequency:** Total frequency = 50 + 250 + 100 + 100 = 500 Given total items = 575, so there might be some rounding or missing data, but we proceed with these frequencies for estimation. --- ### a) Proportion of items less than 25 cm long 5. Items less than 25 cm include: - All items in 0-10 cm: 50 - All items in 10-20 cm: 250 - Part of items in 20-30 cm between 20 and 25 cm. 6. Calculate frequency for 20-25 cm: Class width = 10 cm, frequency = 100 Frequency density = 10 Width from 20 to 25 = 5 cm Frequency = 10 \(\times\) 5 = 50 7. Total items less than 25 cm = 50 + 250 + 50 = 350 8. Proportion = \(\frac{350}{575} \approx 0.6087\) --- ### b) Estimate number of items between 12.4 cm and 36.8 cm 9. Break the interval into parts: - 12.4 to 20 cm (within 10-20 cm class) - 20 to 30 cm class - 30 to 36.8 cm (within 30-50 cm class) 10. Calculate frequencies: - 10-20 cm class frequency density = 25 Width from 12.4 to 20 = 7.6 cm Frequency = 25 \(\times\) 7.6 = 190 - 20-30 cm class frequency = 100 (full class) - 30-50 cm class frequency density = 5 Width from 30 to 36.8 = 6.8 cm Frequency = 5 \(\times\) 6.8 = 34 11. Total frequency = 190 + 100 + 34 = 324 --- ### c) Length of shortest item not recycled (shortest 20% recycled) 12. Total items = 575 20% of 575 = 0.20 \(\times\) 575 = 115 items to be recycled 13. Find length \(L\) such that cumulative frequency up to \(L\) is 115. 14. Cumulative frequencies: - 0-10 cm: 50 - 10-20 cm: 50 + 250 = 300 15. Since 115 < 300, \(L\) is in 10-20 cm class. 16. Frequency density in 10-20 cm = 25 17. Number of items needed beyond 10 cm to reach 115: 115 - 50 = 65 18. Length beyond 10 cm: \(\text{length} = \frac{\text{frequency}}{\text{frequency density}} = \frac{65}{25} = 2.6\) cm 19. So, \(L = 10 + 2.6 = 12.6\) cm --- **Final answers:** - a) Proportion less than 25 cm = **0.609** (rounded to 3 decimal places) - b) Estimated number between 12.4 cm and 36.8 cm = **324** items - c) Length of shortest item not recycled = **12.6 cm**