1. **State the problem:** We need to estimate the median height of 60 trees using the histogram data, which gives frequency densities over height intervals. 2. **Recall the formula:** Frequency for each bar = frequency density \( \times \) class width. 3. **Calculate frequencies for each class:** - Class 1 (0 to 10 m): width = 10, frequency density = 1, frequency = $1 \times 10 = 10$ - Class 2 (10 to 15 m): width = 5, frequency density = 5, frequency = $5 \times 5 = 25$ - Class 3 (15 to 20 m): width = 5, frequency density = 1.5, frequency = $1.5 \times 5 = 7.5$ - Class 4 (20 to 25 m): width = 5, frequency density = 0.5, frequency = $0.5 \times 5 = 2.5$ - Class 5 (25 to 50 m): width = 25, frequency density = 0.2, frequency = $0.2 \times 25 = 5$ 4. **Check total frequency:** $10 + 25 + 7.5 + 2.5 + 5 = 50$ which is less than 60, so we must adjust or assume the total frequency is 50 (likely a typo or rounding). We proceed with 50 as total frequency. 5. **Find median class:** Median position = $\frac{50}{2} = 25$th tree. Cumulative frequencies: - Up to class 1: 10 - Up to class 2: 10 + 25 = 35 Since 25th tree lies in class 2 (10 to 15 m). 6. **Apply median formula:** $$\text{Median} = L + \left(\frac{\frac{N}{2} - F}{f}\right) \times w$$ Where: - $L = 10$ (lower boundary of median class) - $N = 50$ (total frequency) - $F = 10$ (cumulative frequency before median class) - $f = 25$ (frequency of median class) - $w = 5$ (width of median class) 7. **Calculate:** $$\text{Median} = 10 + \left(\frac{25 - 10}{25}\right) \times 5 = 10 + \left(\frac{15}{25}\right) \times 5 = 10 + 3 = 13$$ 8. **Final answer:** The estimated median height is **13.0 m**. Note: If total frequency is actually 60, the method is the same but frequencies or density might need adjustment. Here, we used given data as is.