1. The problem involves interpreting a histogram with frequency density on the vertical axis and height intervals on the horizontal axis. 2. The histogram bars represent frequency density over height intervals: - Bar 1: Interval 0 to 10 meters, height 1 - Bar 2: Interval 10 to 15 meters, height 5 - Bar 3: Interval 15 to 25 meters, height 1.5 - Bar 4: Interval 25 to 40 meters, height 0.5 - Bar 5: Interval 40 to 50 meters, height 0 3. Frequency density is defined as $$\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}$$. 4. To find the frequency for each bar, multiply the frequency density by the class width: - Bar 1 frequency: $1 \times (10 - 0) = 10$ - Bar 2 frequency: $5 \times (15 - 10) = 25$ - Bar 3 frequency: $1.5 \times (25 - 15) = 15$ - Bar 4 frequency: $0.5 \times (40 - 25) = 7.5$ - Bar 5 frequency: $0 \times (50 - 40) = 0$ 5. The total frequency is the sum of all frequencies: $$10 + 25 + 15 + 7.5 + 0 = 57.5$$ 6. This histogram shows how frequency density varies with height intervals, and the area of each bar (height \times width) represents the frequency for that interval. 7. Understanding this helps interpret histograms where class widths differ, ensuring accurate frequency calculations.