In a histogram with unequal class widths, the vertical axis shows frequency density, not frequency. Frequency density = frequency ÷ class width. This ensures that the area of each bar — not its height — represents the frequency for that class, making comparisons fair when classes vary in width.

Why is frequency density necessary for unequal class widths?

If you plotted raw frequency on the y-axis when class widths differ, a wide class with moderate frequency would appear as tall as a narrow class with the same frequency — even though the data is spread over a much larger interval. Frequency density corrects this by giving wider classes shorter bars.

Key formula: frequency density = frequency ÷ class width

Inverse (reading from a histogram): frequency = frequency density × class width

The area of each bar (height × width = frequency density × class width) equals the frequency. This is why histograms are said to use area to represent frequency.

How do you draw a histogram from a frequency table?

Worked example 1 — complete histogram

Times (minutes) taken by 100 students to complete an assignment:

Time (min) Frequency Class width Frequency density
0 ≤ t < 10 20 10 2.0
10 ≤ t < 20 35 10 3.5
20 ≤ t < 30 25 10 2.5
30 ≤ t < 50 16 20 0.8
50 ≤ t < 90 4 40 0.1

Steps:

  1. Calculate frequency density for each class: divide frequency by class width.
  2. Draw a horizontal axis from 0 to 90, labelled with the variable and units.
  3. Draw a vertical axis labelled Frequency density — this label earns a mark.
  4. Draw each bar: width from lower to upper class boundary; height equals frequency density.
  5. Bars must touch — no gaps. Continuous data means there is no gap between classes.

Check: multiply frequency density by class width for every bar to recover the original frequency. For the 30–50 bar: 0.8 × 20 = 16 ✓

How do you find frequency from a histogram?

Worked example 2 — reading back from a histogram

A histogram bar runs from 60 to 80 (class width 20) with frequency density 1.5.

Frequency = 1.5 × 20 = 30

Answer: 30 values

If only part of a bar is needed — "how many values lie between 65 and 80?" — use the area of that section only, assuming uniform distribution within the class:

Frequency ≈ 1.5 × (80 − 65) = 1.5 × 15 = 22.5 ≈ 23

How do you complete a histogram with one bar missing?

Worked example 3 — missing bar

The total frequency is 80. Four bars account for 68 values. The missing class is 40 ≤ t < 60 (class width 20).

Missing frequency = 80 − 68 = 12

Frequency density = 12 ÷ 20 = 0.6

Draw a bar from 40 to 60 with height 0.6 on the frequency density axis.

How do you estimate the median from a histogram?

Work through bars from left to right, accumulating area (frequency) until you reach n/2.

Worked example 4 — median from histogram

Using worked example 1 (n = 100, median at 50th value):

Cumulative frequencies: 20, 55, 80, 96, 100.

The 50th value falls in the class 10 ≤ t < 20 (cumulative goes from 20 to 55). Values needed within this bar: 50 − 20 = 30 out of 35.

Median ≈ 10 + (30/35) × 10 = 10 + 8.57 ≈ 18.6 minutes

Estimated median: 18.6 minutes

What are the most common mistakes?

Mistake How to fix it
Frequency on y-axis instead of frequency density Always label the y-axis "Frequency density"
Reading bar height as frequency Frequency = height × width (the area)
Gaps between bars Continuous data: bars always touch
Dividing wrong way fd = frequency ÷ class width
Omitting y-axis label Write "Frequency density" — it is a mark

Frequently asked questions

What is the difference between a histogram and a bar chart?

A bar chart represents discrete or categorical data; the height of each bar shows frequency and gaps between bars indicate distinct categories. A histogram represents continuous grouped data; bars must touch, and frequency is shown by area, not height. When all class widths are equal the visual difference disappears, but the conceptual distinction remains.

Why does frequency = frequency density × class width?

Because frequency density is defined as frequency per unit of class width. Rearranging gives frequency = frequency density × class width — the same as multiplying a rate by the size of the interval. Geometrically, this is the area of the rectangular bar (base × height = class width × frequency density).

Can frequency density be a decimal or greater than 1?

Yes on both counts. Frequency density has no upper bound and no requirement to be an integer. A narrow class with many values can have a very large frequency density; a wide class with few values can have a very small decimal frequency density such as 0.1. Both are perfectly valid outcomes.

What if all class widths are equal?

With equal class widths, frequency and frequency density are proportional, so the histogram looks the same whether you use frequency or frequency density on the y-axis. It is always safe to use frequency density — but you must still label the axis correctly. Many GCSE questions with equal widths still ask for frequency density to test conceptual understanding.


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