Histograms and Frequency Density
A histogram represents a grouped numerical distribution using touching rectangles. The essential idea is that rectangle area represents frequency, so unequal class widths require careful choice of height.
Higher. This includes both equal-width and unequal-width histograms in the current specifications. Use class intervals and grouped-data summaries. You will divide frequencies by class widths.
Why area must represent frequency
In a histogram, the horizontal axis is a numerical scale. Each rectangle spans its class boundaries; adjacent classes touch. A wider class covers a larger interval, so comparing raw heights as frequencies would distort the display.
Frequency density = frequency ÷ class width.
Frequency = class width × frequency density. The product is the rectangle area measured in the axis units, not its area in square centimetres on your screen.
Worked example — equal widths first
Three classes of width 5 have frequencies 10, 15 and 20. Their densities are 2, 3 and 4. Because all widths are equal, heights are proportional to frequencies. A frequency-labelled axis is sometimes used for equal-width displays; read its label. For unequal widths, height alone cannot represent frequency.
Construct an unequal-width histogram
Find every class width, calculate every density, then choose a uniform horizontal scale and a labelled vertical density scale. Draw each rectangle across its true interval.
| Time | f | w | f/w |
|---|---|---|---|
| 0–10 | 4 | 10 | 0.4 |
| 10–20 | 12 | 10 | 1.2 |
| 20–40 | 16 | 20 | 0.8 |
| 40–60 | 8 | 20 | 0.4 |
| All | 40 | — | — |
Worked example — calculate and draw one rectangle
For 20–40 minutes, width = 40 − 20 = 20 and density = 16/20 = 0.8 journeys per minute. Draw it twice as wide as 10–20, with height 0.8. Its area is 20 × 0.8 = 16 journeys.
The 10–20 class has the tallest rectangle but contains 12 journeys; the wider 20–40 class contains 16. The largest class frequency comes from area, not height. A frequency table’s modal class is the class with greatest frequency; the tallest histogram bar identifies greatest density. State which you mean.
Recover frequencies and estimate parts of a class
Worked example — read a whole class
A rectangle covers 40–60 minutes and has density 0.4. Frequency = 20 × 0.4 = 8. Adding all four areas gives 4 + 12 + 16 + 8 = 40, matching the sample size.
Worked example — estimate a subinterval
To estimate the count from 20 up to 30 minutes, take width 10 inside the 20–40 class: 10 × 0.8 = 8 journeys. This assumes the observations are spread uniformly through that class. The exact count cannot be recovered from the histogram.
The estimated proportion from 20 up to 30 minutes is 8/40 = 20%. Do not treat the density 0.8 as either a count or a probability.
Use all relevant rectangle parts when an interval crosses a class boundary. For 15 up to 30 minutes, the estimate is 5 × 1.2 + 10 × 0.8 = 14 journeys. Once frequencies have been recovered, use class midpoints to estimate a mean: here (5 × 4 + 15 × 12 + 30 × 16 + 50 × 8)/40 = 27 min. For a median estimate, form running totals and use the method in the next cumulative-frequency lesson.
Find an unknown vertical scale using area ratios
Some questions give a histogram without a numerical density scale, together with one class frequency. Areas are still proportional to frequencies. Use the known rectangle to find how much frequency one area unit represents.
Worked example — calibrate the grid
The first rectangle has width 4 and height 3, so its grid area is 12. Its frequency is 24, so one grid-area unit represents 2 observations.
Second area = (10 − 4) × 2 = 12, giving frequency 24. Third area = (20 − 10) × 1.5 = 15, giving frequency 30. Total = 24 + 24 + 30 = 78.
The displayed heights 3, 2 and 1.5 are not the actual densities. Multiplying by 2 gives densities 6, 4 and 3. Use a given frequency to calibrate the scale before calling height “frequency density”.
Use the correct class boundaries
Read the measurement convention. For a continuous measurement rounded to the nearest whole unit, class boundaries usually lie halfway between neighbouring recorded values. This accounts for the full range of measurements that round into each class.
Worked example — rounded measurements away from zero
Heights recorded to the nearest centimetre in classes 150–154, 155–159 and 160–169 have continuous boundaries 149.5–154.5, 154.5–159.5 and 159.5–169.5. Widths are 5, 5 and 10, not 4, 4 and 9.
For frequencies 10, 15 and 20, densities are 2, 3 and 2. The wider last class contains more observations than the first even though their densities match.
When boundaries are already written as inequalities, use them directly. Do not automatically add or subtract 0.5 from every question.
Think Like a Mathematician
If a class is twice as wide, the same frequency must be spread across twice the width, so its density must halve. This keeps area—and therefore the represented count—unchanged.
Common Mistakes
Height as count: use width × density for frequency.
Ignoring unequal widths: calculate each width separately.
Wrong horizontal spacing: a class of width 20 must be twice as wide as one of width 10.
Exact claims within classes: a partial-class count is an estimate based on an assumption.
Practice
1. A class 10–18 has frequency 24. Find its width and density.
Width = 18 − 10 = 8; density = 24/8 = 3.
2. A rectangle spans 20–35 and has density 1.6. Find the frequency.
15 × 1.6 = 24.
3. Two classes have widths 5 and 10 and both have frequency 20. Find their densities.
20/5 = 4 and 20/10 = 2. The wider rectangle is half as tall.
4. In the journey histogram, which class contains the most journeys and which has greatest density?
Greatest frequency: 20–40 min, 16 journeys. Greatest density: 10–20 min, density 1.2.
5. Estimate the number from 30 up to 50 minutes in that histogram. State the assumption.
10 × 0.8 + 10 × 0.4 = 12 journeys, assuming uniform spread within each relevant class.
6. In the grid-scale histogram, what frequency does grid area 7.5 represent?
Each area unit represents 2 observations, so 7.5 × 2 = 15.
7. Measurements rounded to the nearest cm are grouped 25–29. Give the continuous boundaries and width.
Boundaries 24.5 and 29.5 cm; width 5 cm.
8. A class of width 12 has frequency 36. It is split into two equal-width parts. Can you know each new frequency exactly?
No. Assuming uniform spread gives an estimate of 18 in each, with density 3, but the original grouped data do not reveal the exact split.
Frequently Asked Questions
Why do histogram bars touch?
They cover adjacent numerical intervals on a continuous scale. A gap would suggest an interval with no observations or a break in the data.
Can the tallest bar have fewer observations?
Yes. With unequal widths, a narrow bar can have high density but smaller area and frequency than a wider bar.