1. **State the problem:** Calculate the median from the given frequency distribution and find the number of students getting marks between the first and third quartile. 2. **Given data:** | Marks | 0-20 | 20-40 | 40-50 | 50-60 | 60-80 | 80-100 | |----------------|-------|-------|-------|-------|-------|--------| | No. of students| 50 | 100 | 150 | 90 | 60 | 50 | 3. **Calculate total number of students:** $$N = 50 + 100 + 150 + 90 + 60 + 50 = 500$$ 4. **Find cumulative frequencies:** | Marks | 0-20 | 20-40 | 40-50 | 50-60 | 60-80 | 80-100 | |----------------|-------|-------|-------|-------|-------|--------| | Frequency (f) | 50 | 100 | 150 | 90 | 60 | 50 | | Cumulative (CF)| 50 | 150 | 300 | 390 | 450 | 500 | 5. **Median class:** The median position is at $$\frac{N}{2} = \frac{500}{2} = 250$$. The cumulative frequency just greater than or equal to 250 is 300, so median class is 40-50. 6. **Median formula:** $$\text{Median} = L + \left(\frac{\frac{N}{2} - F}{f_m}\right) \times h$$ Where: - $L$ = lower boundary of median class = 40 - $N$ = total frequency = 500 - $F$ = cumulative frequency before median class = 150 - $f_m$ = frequency of median class = 150 - $h$ = class width = 50 - 40 = 10 7. **Calculate median:** $$\text{Median} = 40 + \left(\frac{250 - 150}{150}\right) \times 10 = 40 + \left(\frac{100}{150}\right) \times 10$$ $$= 40 + \frac{2}{3} \times 10 = 40 + \frac{20}{3} = 40 + 6.67 = 46.67$$ 8. **Find first quartile (Q1) and third quartile (Q3) positions:** $$Q1 = \frac{N}{4} = \frac{500}{4} = 125$$ $$Q3 = \frac{3N}{4} = \frac{3 \times 500}{4} = 375$$ 9. **Q1 class:** cumulative frequency just greater than or equal to 125 is 150, so Q1 class is 20-40. 10. **Q1 calculation:** $$L = 20, F = 50, f_m = 100, h = 40 - 20 = 20$$ $$Q1 = 20 + \left(\frac{125 - 50}{100}\right) \times 20 = 20 + \left(\frac{75}{100}\right) \times 20 = 20 + 15 = 35$$ 11. **Q3 class:** cumulative frequency just greater than or equal to 375 is 390, so Q3 class is 50-60. 12. **Q3 calculation:** $$L = 50, F = 300, f_m = 90, h = 60 - 50 = 10$$ $$Q3 = 50 + \left(\frac{375 - 300}{90}\right) \times 10 = 50 + \left(\frac{75}{90}\right) \times 10 = 50 + 8.33 = 58.33$$ 13. **Number of students between Q1 and Q3:** Students between marks 35 and 58.33 are those between first and third quartile. 14. **Calculate cumulative frequency at Q1 and Q3:** - At Q1 (125th student), cumulative frequency is 125. - At Q3 (375th student), cumulative frequency is 375. 15. **Number of students between Q1 and Q3:** $$375 - 125 = 250$$ **Final answers:** - Median = 46.67 - Number of students between first and third quartile = 250