Question: Pete collects the heights of 50 plants and records the data in the table below.
Complete the histogram for the data in the table.
Height (x cm) Frequency
0 < x \leq 15 15
15 < x \leq 20 10
20 < x \leq 35 15
35 < x \leq 40 5
40 < x \leq 50 5
Density
2
1.5
1
0.5
0
Height (x cm)
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1. **State the problem:** We have a frequency distribution of plant heights in intervals and need to complete the histogram by calculating the density for each interval.
2. **Recall the formula for histogram density:**
$$\text{Density} = \frac{\text{Frequency}}{\text{Class Width}}$$
This formula ensures the area of each bar (density \times width) equals the frequency.
3. **Calculate class widths:**
- For $0 < x \leq 15$, width = $15 - 0 = 15$
- For $15 < x \leq 20$, width = $20 - 15 = 5$
- For $20 < x \leq 35$, width = $35 - 20 = 15$
- For $35 < x \leq 40$, width = $40 - 35 = 5$
- For $40 < x \leq 50$, width = $50 - 40 = 10$
4. **Calculate densities:**
- Interval $0 < x \leq 15$:
$$\text{Density} = \frac{15}{15} = 1$$
- Interval $15 < x \leq 20$:
$$\text{Density} = \frac{10}{5} = 2$$
- Interval $20 < x \leq 35$:
$$\text{Density} = \frac{15}{15} = 1$$
- Interval $35 < x \leq 40$:
$$\text{Density} = \frac{5}{5} = 1$$
- Interval $40 < x \leq 50$:
$$\text{Density} = \frac{5}{10} = 0.5$$
5. **Compare with given densities:** The problem's density column shows values 2, 1.5, 1, 0.5, 0 which do not match the calculated densities exactly, so the histogram's polygon line is drawn approximately connecting points at heights 0, 1, 1.5, 1, 0.5, 0 over the intervals.
6. **Final answer:** The histogram bars have heights equal to the densities calculated above, and the polygon line connects points roughly at (15,0), (20,1), (35,0.67), (40,0) as described.
This completes the histogram for the given data.