Histograms and Frequency Density — AQA GCSE Maths Revision Notes
What you'll learn
This topic covers histograms drawn for grouped data whose classes are not all the same width. By the end of this guide you should be able to calculate frequency density, draw such a histogram, and work backwards from a completed one to recover the frequency of any class.
You should also be able to explain why frequency density is necessary, find the frequency contained in part of a bar, and estimate the median and other values from a histogram.
The organising idea is the one sentence that governs every question here: area represents frequency. On an ordinary histogram with equal classes the height happens to give the frequency too, and that coincidence is what makes unequal widths so confusing — the habit of reading heights carries over and quietly breaks. Once you fix on area rather than height, every calculation in the topic is the same triangle of relationships: frequency is height times width, height is frequency divided by width, and width is frequency divided by height.
Key terms and definitions
Histogram — a diagram for continuous grouped data in which the area of each bar represents the frequency.
Frequency density — the height of each bar, calculated as frequency divided by class width.
Class width — the upper boundary of a class minus the lower boundary.
Frequency — the number of data values in a class, given by the area of its bar.
Continuous data — data taking any value in a range, such as height, mass or time.
Modal class — for a histogram with unequal widths, the class with the greatest frequency density, not the tallest total frequency.
Core concepts
Why frequency density is needed
Suppose one class covers 0 to 10 with 20 values and another covers 10 to 50 with 40 values. Drawing bars whose heights are 20 and 40 makes the second look twice as important. But it is four times as wide, so the values inside it are spread much more thinly — it is actually the less crowded region.
Plotting frequency as height therefore gives wider classes an unfair visual weight, and the diagram misleads.
Frequency density fixes this by dividing out the width. The first class has density 20 ÷ 10 = 2 and the second has 40 ÷ 40 = 1, so the first bar is drawn twice as tall — which correctly shows that the data is more densely packed there.
The formula, and its rearrangements
frequency density = frequency ÷ class width
This is the formula to memorise, but most exam questions need it the other way round:
frequency = frequency density × class width
which is exactly the area of the bar. A third rearrangement, class width = frequency ÷ frequency density, appears occasionally.
Questions usually supply two of the three quantities and ask for the third, so knowing which rearrangement you need is most of the work.
Drawing the histogram
Work out the class width for each class, divide the frequency by it to get the frequency density, and plot that as the bar height.
Label the vertical axis frequency density, not frequency. An unlabelled or wrongly labelled axis loses a mark even when every bar is correct.
The bars touch, because the data is continuous, and the horizontal axis is a continuous scale.
Reading a frequency back off a histogram
This is the commonest question type, and it runs the formula backwards.
Read the bar's height from the vertical axis, work out its width from the horizontal axis, and multiply. A bar of density 3 spanning a class of width 10 contains 3 × 10 = 30 values.
Reading the height as the frequency gives 3 rather than 30, and that single mistake accounts for most of the marks lost in this topic.
Comparing two classes
Because frequency is an area, a taller bar does not necessarily contain more data.
Take class A with width 5 and density 6, and class B with width 10 and density 4. Class A is the taller bar, but its frequency is 5 × 6 = 30 while class B's is 10 × 4 = 40. Class B contains more data despite being shorter.
The same reasoning fixes the modal class. The class with the greatest density is the most crowded, and that is what a histogram question means by modal, but it need not be the class holding the most values.
Part of a bar
Some questions ask how many values lie in a range that cuts across a bar — for instance, how many are below 25 when a class runs from 20 to 30.
Assume the values are spread evenly through the class, and take the matching fraction of its frequency. Half the class means half the frequency.
If the class 20 ≤ x < 30 has frequency 45, then the values below 25 number 45 × (5 ÷ 10) = 22.5, which would be rounded to a sensible whole number, typically 22 or 23, with the assumption stated.
The phrase examiners look for is that the data is assumed to be evenly distributed within the class.
Estimating the median
The median lies at half the total frequency, so first find each class's frequency by area and total them.
Add the frequencies in order until you pass half the total, which identifies the class containing the median, then work out how far into that class you need to go and take that fraction of its width.
The answer is an estimate, because the individual values inside each class are unknown.
Worked examples
Example 1: Calculating frequency density
A class covers 50 ≤ x < 80 and contains 60 values. Find its frequency density.
The class width is 80 − 50 = 30.
Frequency density = 60 ÷ 30 = 2.
Note that the answer is much smaller than the frequency, which is normal. Frequency density is a rate, not a count.
Example 2: Recovering a frequency
A bar has a frequency density of 1.5 and spans a class of width 8. How many data values does it contain?
Frequency is the area of the bar, so multiply the height by the width.
Frequency = 1.5 × 8 = 12 values.
Reading the height of 1.5 as the answer is the error to avoid. The height is a density; only the area is a count.
Example 3: Comparing two classes
Class A has width 5 and frequency density 6. Class B has width 10 and frequency density 4. Which class contains more data?
Class A's frequency is 5 × 6 = 30.
Class B's frequency is 10 × 4 = 40.
Class B contains more data, even though class A is drawn as the taller bar. Class A is the more crowded, which is a different statement from containing more values.
Common mistakes and how to avoid them
Reading the height as the frequency. The height is the frequency density. Multiply by the class width to get the frequency.
Multiplying instead of dividing when finding density. Density is frequency divided by width, so it is usually a smaller number than the frequency.
Labelling the vertical axis "frequency". It must say frequency density.
Assuming the tallest bar holds the most data. A narrow tall bar can hold fewer values than a wide short one.
Leaving gaps between the bars. The data is continuous, so the bars touch.
Forgetting to state the even-distribution assumption. Part-of-a-bar answers need it.
Treating the answer as exact. Everything read from a histogram is an estimate, because the original values were lost in the grouping.
Exam technique for "Histograms and Frequency Density"
Add two columns to the table — class width and frequency density — before drawing anything. The working is where the method marks are, and the columns make the arithmetic visible.
Write the formula you are using at the start of each calculation, in the direction you need it: "frequency = density × width" for reading off, "density = frequency ÷ width" for drawing.
Label the vertical axis frequency density every time.
For part-of-a-bar questions, write the fraction of the class you are taking and state that the data is assumed evenly distributed.
Check whether a "which has more" question is asking about frequency or about density — the two can point to different classes, and the question decides which is wanted.
Round frequencies to sensible whole numbers, since they count data values.
Quick revision summary
On a histogram, area represents frequency. The height is the frequency density, and only the area is a count.
Frequency density = frequency ÷ class width, and rearranged, frequency = frequency density × class width.
Frequency density exists because plotting frequency as height would give wide classes unfair visual weight. Dividing by the width shows how crowded each class is.
To read a frequency off a histogram, multiply the bar's height by its width. A density of 3 over a width of 10 means 30 values.
A taller bar does not always contain more data: width 5 with density 6 gives 30 values, while width 10 with density 4 gives 40.
For part of a class, assume the values are evenly distributed and take that fraction of the frequency.
Label the vertical axis frequency density, keep the bars touching, and treat every result as an estimate.