Probability and statistics · GCSE Maths
Histograms
GCSE Maths histograms: frequency density equals frequency divided by class width, so unequal widths still have area equal to frequency, with a full reading and drawing method.
Frequency density = frequency ÷ class width. The area of each bar is the frequency. Unequal widths are why you cannot plot frequency on the vertical axis.
The important bits
What you need to know
- 1
A histogram uses frequency density on the vertical axis when class widths are unequal. Frequency density = frequency / class width.
- 2
The area of a bar equals the frequency (or is proportional to it if the scale is scaled). A bar from 10 to 20 of height 3 has width 10 and frequency 30.
- 3
Class width is the difference of the class boundaries you plot, for example 10 ≤ t < 20 has width 10. Watch 0–10, 10–15, 15–30 as widths 10, 5, 15.
- 4
Equal-width bars can use frequency on the vertical axis and still look like a histogram, but exam questions with mixed widths will punish a frequency axis.
- 5
To draw: for each class, compute width, then density = frequency / width, then draw a bar of that height across the class interval. Label frequency density, not frequency.
- 6
To read a frequency from a drawn histogram: frequency = density × width. Count the squares in the bar if a key says “1 square = 4 people”, which is an area key.
- 7
The modal class is the class with the greatest frequency density (tallest bar), not automatically the widest bar, and not the greatest frequency if a wide class is short.
- 8
Histograms show continuous grouped data. A bar chart of favourite colours is not a histogram: gaps and equal categories are a different diagram.
Quotations worth analysing
Short evidence. Real method.
“frequency density = frequency / class width”
This formula is the method mark for both drawing and reading. A vertical axis labelled “frequency” on unequal widths is a wrong diagram.
“The area of the bar represents the frequency”
Height is not frequency when widths differ. A short wide bar can hold more people than a tall thin bar. Compare areas, not heights, for totals.
“Unequal class widths”
A 20-minute class will naturally collect more people than a 5-minute class at the same density. Density makes the comparison fair.
Go deeper
Why density, not frequency, goes up the side
Suppose 8 people took 0–10 minutes, 15 took 10–15 minutes, and 12 took 15–35 minutes. Widths 10, 5, 20. If you plot frequencies 8, 15, 12 as bar heights, the middle class looks busiest, but it is only five minutes wide; the rate is 15/5 = 3 people per minute, while the first class is 8/10 = 0.8 and the last is 12/20 = 0.6. The tallest density bar is 10–15, which is the modal class in the histogram sense. Drawing frequency bars would make the first class look comparable to the last, which it is not per unit time. Always compute a density column in the table before you draw. Label the vertical axis “frequency density”. If the question gives a key such as 1 cm² = 5 students, you are being told the area scale; count squares × 5, do not read a height as a head count.
Go deeper
Reading a frequency from a finished histogram
A bar runs from 20 to 30 with height 4 on a frequency-density scale. Width 10, so frequency = 4 × 10 = 40. A bar from 30 to 50 of height 1.5 has width 20, frequency 30. To estimate how many are between 25 and 40, split across bars: half of the 20–30 bar if you assume even spread (20 people), plus half of the 30–50 bar (15 people), total 35. That even-spread assumption is the same modelling choice as a grouped mean using midpoints; say “estimate” if the question does. When the paper prints a grid and a key “2 small squares = 1 person”, ignore the density formula and count squares, because they have given you the area scale directly. Still check that a full bar’s square-count matches density × width if both are readable — a mismatch means you misread the scale.
Go deeper
Modal class, mean estimates, and not a bar chart
Modal class on a histogram is the class of greatest frequency density, the tallest bar. A very wide class can have the largest frequency and still not be modal if it is short. Estimated mean still uses midpoints and frequencies, not densities: find each frequency from area first, then Σfx / Σf. Do not average the densities. Gaps between bars belong to bar charts of discrete or categorical data. Histogram bars for continuous data typically touch, using the class boundaries. If a class is 10 < x ≤ 20, the bar still spans 10 to 20 on the axis. A frequency polygon would plot the frequencies (or densities, depending on the course) at midpoints and join with straight lines; do not confuse that sketch with the histogram bars the question asked you to draw.
See the idea in action
Times to complete a puzzle, in minutes: 0 ≤ t < 10, frequency 8; 10 ≤ t < 15, frequency 15; 15 ≤ t < 35, frequency 12. Draw the histogram values and find the modal class. Step 1: Class widths: 10, 5, 20. Step 2: Frequency density = frequency ÷ width: 8/10 = 0.8, 15/5 = 3, 12/20 = 0.6. Step 3: Bars: 0–10 height 0.8; 10–15 height 3; 15–35 height 0.6. Vertical axis labelled frequency density. Step 4: Modal class is 10 ≤ t < 15, the tallest bar. Check areas: 0.8 × 10 = 8, 3 × 5 = 15, 0.6 × 20 = 12, which recover the frequencies.
Exam technique
Turn knowledge into marks
Add a frequency-density column to the table before you draw. When reading, write frequency = density × width on the bar you are using. Never label the vertical axis “frequency” if the widths are unequal.
Common mistakes
Do not give these marks away
- 01
Plotting frequency on the vertical axis when class widths are unequal, so wide classes look unfairly tall.
- 02
Using the tallest bar as the greatest frequency without multiplying by width, or calling a wide short bar the modal class.
- 03
Forgetting to divide by class width, or using the midpoint as if it were the width.
A class 10 ≤ t < 20 contains 24 people. The frequency density is
A2.4
B24
C10
D0.42
Show the answer
2.4. Width = 10, density = 24 / 10 = 2.4. 24 is the frequency, not the density. 10 is the width. 0.42 would be 10/24, the reciprocal.
Quick questions
If this is the bit you searched
Why do histograms use frequency density?
So that the area of each bar equals the frequency. When class widths are unequal, plotting frequency as height would make wide classes look bigger than they are per unit.
How do you calculate frequency density?
Frequency density = frequency ÷ class width. Rearrange: frequency = frequency density × class width, which is how you read a bar.
What is the modal class on a histogram?
The class with the greatest frequency density, that is the tallest bar, not necessarily the class with the largest frequency.
Is a bar chart the same as a histogram?
No. Bar charts compare categories and usually have gaps. Histograms show continuous grouped data; bars typically touch, and area represents frequency.