1. **State the problem:** Find the median of the given grouped data representing marks and number of students. 2. **Given data:** Marks (class intervals) | Number of students (frequency, f) 0 and above: 40 10 and above: 28 20 and above: 16 30 and above: 10 40 and above: 8 50 and above: 0 3. **Interpretation:** The frequencies are cumulative frequencies (CF). To find the median, we first find the total number of students $N$ and then find the median class. 4. **Calculate total number of students:** $$N = 40$$ (since 0 and above includes all students) 5. **Find median position:** $$\text{Median position} = \frac{N}{2} = \frac{40}{2} = 20$$ 6. **Find the median class:** The median class is the class where the cumulative frequency just exceeds or equals 20. From the data: - 0 and above: 40 students - 10 and above: 28 students - 20 and above: 16 students Since 20 lies between 16 and 28, the median class is the class starting at 10 marks. 7. **Find class boundaries and frequencies:** - Median class: 10 to 20 - Lower boundary $l = 10$ - Frequency of median class $f = 28 - 16 = 12$ - Cumulative frequency before median class $CF = 16$ - Class width $h = 10$ 8. **Median formula:** $$\text{Median} = l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$$ 9. **Substitute values:** $$\text{Median} = 10 + \left(\frac{20 - 16}{12}\right) \times 10 = 10 + \left(\frac{4}{12}\right) \times 10$$ 10. **Simplify:** $$= 10 + \frac{1}{3} \times 10 = 10 + \frac{10}{3} = 10 + 3.33 = 13.33$$ **Final answer:** The median marks is approximately $13.33$.