1. **State the problem:** Estimate the median value from the given frequency distribution: | Intervals | 0-6 | 6-12 | 12-18 | 18-24 | |-----------|-----|------|-------|-------| | Frequency | 6 | 8 | 12 | 10 | 4 | 2. **Formula for median in grouped data:** $$\text{Median} = L + \left(\frac{\frac{N}{2} - F}{f}\right) \times h$$ where: - $L$ = lower boundary of median class - $N$ = total frequency - $F$ = cumulative frequency before median class - $f$ = frequency of median class - $h$ = class width 3. **Calculate total frequency $N$:** $$N = 6 + 8 + 12 + 10 + 4 = 40$$ 4. **Find $\frac{N}{2}$:** $$\frac{N}{2} = \frac{40}{2} = 20$$ 5. **Find cumulative frequencies:** - Up to 0-6: 6 - Up to 6-12: 6 + 8 = 14 - Up to 12-18: 14 + 12 = 26 6. **Identify median class:** The median class is the class where cumulative frequency just exceeds $20$, which is the 12-18 interval. 7. **Assign values:** - $L = 12$ (lower boundary of median class) - $F = 14$ (cumulative frequency before median class) - $f = 12$ (frequency of median class) - $h = 6$ (class width, since intervals are 6 units wide) 8. **Calculate median:** $$\text{Median} = 12 + \left(\frac{20 - 14}{12}\right) \times 6 = 12 + \left(\frac{6}{12}\right) \times 6 = 12 + 0.5 \times 6 = 12 + 3 = 15$$ **Final answer:** The estimated median value is **15**.