All subjects · Mathematics · WST01 (S1)

Histograms and frequency density

A histogram shows grouped continuous data. With unequal class widths, the area of each bar, not its height, represents the frequency.

Frequency density

For continuous data in classes, draw bars with no gaps between class boundaries. The area of each bar is proportional to the frequency. Usually the vertical axis is

frequency density = frequency ÷ class width.

So a wide class with the same frequency as a narrow one has a lower bar.

Time, t (minutes)Frequency
0 ≤ t < 1012
10 ≤ t < 1518
15 ≤ t < 2025
20 ≤ t < 3020
30 ≤ t < 5010

Class boundaries

Check whether data were rounded. Heights 'recorded to the nearest cm' as 150 to 154 have boundaries 149.5 and 154.5, so width 5. Ages in completed years 10 to 14 have boundaries 10 and 15. Always work out widths from the boundaries.

Scale problems

Exam questions often give the bar sizes in cm. Use one known bar to find the link between area and frequency.

Example: the class 10 ≤ t < 15 (frequency 18) is drawn 2 cm wide and 9 cm tall. Area = 18 cm2 represents 18 people, so 1 cm2 represents 1 person. The class 20 ≤ t < 30 is twice as wide, so 4 cm, and its area must be 20 cm2, so its height is 5 cm.

Estimating a frequency

To estimate how many values lie in part of a class, assume the values are spread evenly across the class. Between 25 and 40: half of the 20 ≤ t < 30 class (10) plus half of the 30 ≤ t < 50 class (5) = 15. Histograms are used because they show the shape of grouped continuous data and cope with unequal classes.

Worked example

Find the frequency density for the class 0 ≤ t < 10 and for 30 ≤ t < 50.

  1. 0 ≤ t < 10: 12 ÷ 10 = 1.2
  2. 30 ≤ t < 50: 10 ÷ 20 = 0.5

Answer: 1.2 and 0.5

Key idea: Frequency density = frequency ÷ class width; area is proportional to frequency; use one bar to fix the scale.

Check yourself

  1. Find the frequency density of the class 15 ≤ t < 20.
    Show answer

    5. 25 ÷ 5 = 5.

  2. Find the frequency density of the class 10 ≤ t < 15.
    Show answer

    3.6. 18 ÷ 5 = 3.6.

  3. Estimate the number of people with times between 25 and 40 minutes.
    Show answer

    15. Half of 20 (from 25 to 30) plus half of 10 (from 30 to 40): 10 + 5 = 15.

  4. On a histogram of these data the bar for 10 ≤ t < 15 is 2 cm wide and 9 cm tall. Find the height, in cm, of the bar for 20 ≤ t < 30.
    Show answer

    5 cm. Area 18 cm2 represents 18, so 1 cm2 = 1 person. The 20 to 30 class is 4 cm wide and needs area 20 cm2, so height = 20 ÷ 4 = 5 cm.

  5. Heights recorded to the nearest cm are grouped as 150 to 154. Find the class width.
    Show answer

    5 cm. Boundaries are 149.5 and 154.5, so width = 5 cm.

  6. On a histogram, a class of width 4 has frequency density 3.5. Find its frequency.
    Show answer

    14. Frequency = frequency density × width = 3.5 × 4 = 14.

  7. Why is a histogram suitable for these data?
    • The data are qualitative
    • Each value must be shown individually
    • The data are paired
    • The data are continuous and grouped into classes of unequal width
    Show answer

    The data are continuous and grouped into classes of unequal width. Histograms show the distribution of grouped continuous data, and the area rule handles unequal widths.

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Last updated 7 November 2026. Spotted a mistake? Tell us.