With unequal class widths, the histogram's bar heights must be the frequency density (frequency ÷ width), not the raw frequency, so that area still represents frequency. …

Since the class 20–40 has width 20 while the other classes have width 10, using the raw frequencies as bar heights would make the 20–40 bar look disproportionately tall for its true share of the data. The correct height is the frequency density: $\text{height} = \dfrac{\text{frequency}}{\text{class width}}$.

Verification (dual-check) — area must equal frequency for every bar:

• 0–10: area $= 10 \times 0.5 = 5$ ✓ matches frequency 5
• 10–20: area $= 10 \times 1.0 = 10$ ✓ matches frequency 10
• 20–40: area $= 20 \times 1.5 = 30$ ✓ matches frequency 30
• 40–50: area $= 10 \times 0.5 = 5$ ✓ matches frequency 5 …

The single most common error with unequal class widths is drawing all four bars using the raw frequencies (5, 10, 30, 5) as heights directly — this makes the 20–40 bar's area represent double its true frequency (area would …