Histogram area represents frequency. With unequal intervals, height must show frequency divided by class width.
A histogram groups a continuous variable into intervals. Adjacent bars touch because the intervals are continuous. Define boundaries clearly: "5 to under 10 minutes" avoids counting a 10-minute journey twice.
Frequency density = frequency / class width. A group with eight people across a five-minute interval has density 1.6 people per minute. A group with 12 people across a ten-minute interval has density 1.2. Its bar is lower but wider, so it contains more people.
To construct the histogram, put the numerical variable on the horizontal axis and density on the vertical axis. Give each bar its actual interval width. Recover frequency by multiplying height by width; adding heights does not give the sample size.
Grouping loses exact values. The display shows a distribution of walking times but cannot identify every person's time. Different bin boundaries can change the apparent pattern, so retain the original data and explain the grouping.
Step by step
Calculate widths
Subtract each lower boundary from its upper boundary.
Calculate heights
Divide each frequency by its class width.
Check the areas
Height times width should reproduce every class frequency.
Worked example: Read one bar
The 20-to-under-40-minute class has height 0.4 people per minute and width 20 minutes. Its frequency is 0.4 x 20 = 8 people, the same as the much narrower 5-to-under-10-minute class.
Watch out for this
The tallest histogram bar always has the highest frequency.
For equal-width intervals, taller bars represent more observations. With unequal widths, compare bar areas.
Check your understanding
A 10-minute interval contains 15 journeys. What density should be plotted?
- 1.5 journeys per minute.
- 15 journeys per minute.
- 150 journeys per minute.