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Imagine two income groups: 10 people earning between 0and0 and 10k, and 10 people earning between 10kand10k and 50k. If you draw both bars with the same height, the second bar is four times wider — making the high-income group look massive even though both groups have the exact same number of people.
Our eyes naturally judge quantity by visual area, not just height. When class intervals have unequal widths, standard bar charts deceive the eye.
How do statisticians fix this visual distortion so that bar areas tell the true story?

The Area Rule and Frequency Density

In 1891, statistician Karl Pearson introduced the term histogram to describe diagrams where total geometric area represents total frequency. To keep the visual area proportional to the count, the height of each bar cannot be the raw frequency; it must be the frequency density.
Since the area of a rectangle is Area=Width×Height\text{Area} = \text{Width} \times \text{Height}, setting Area=Frequency\text{Area} = \text{Frequency} means Frequency=Class Width×Frequency Density\text{Frequency} = \text{Class Width} \times \text{Frequency Density}. Rearranging this gives our key formula:
What does this look like in practice when working with real grouped data?

Step-by-Step Construction

Suppose we record the waiting times (in minutes) for 80 patients at a medical clinic. Because the wait times spread out widely, the intervals are grouped into unequal widths: [0,10)[0, 10), [10,20)[10, 20), [20,40)[20, 40), and [40,70)[40, 70).
To plot the histogram, place waiting time along the horizontal xx-axis and frequency density on the vertical yy-axis. Each bar spans from its lower class boundary to its upper class boundary with height equal to its frequency density.