Frequency Polygons and Histograms with Equal Class Widths — AQA GCSE Maths Revision Notes

What you'll learn

This topic covers two ways of displaying grouped continuous data: the histogram, a bar diagram whose bars touch, and the frequency polygon, a line graph joining one point per class. By the end of this guide you should be able to draw each of them, find the midpoint of a class, and read information back off a completed diagram.

You should also be able to say how a histogram differs from a bar chart, estimate the mean of grouped data using midpoints, identify the modal class, and compare two distributions drawn on the same axes.

The organising idea is the midpoint stands for the whole class. Once data has been grouped, the individual values are gone: a class recorded as 10 ≤ x < 20 with a frequency of 7 tells you that seven values fell somewhere in that range, but not where. The midpoint is the single best representative of them, and it is what you plot for a frequency polygon and what you multiply by for an estimated mean. It also explains why every answer from grouped data is an estimate — the real values were lost when the grouping happened.

Key terms and definitions

Grouped data — data sorted into class intervals rather than listed individually.

Class interval — a range of values, written like 10 ≤ x < 20.

Class width — the upper boundary minus the lower boundary. For 10 ≤ x < 20 the width is 10.

Midpoint — the middle of a class, found by adding the two boundaries and halving.

Frequency polygon — a line graph plotting frequency against midpoint, with the points joined by straight lines.

Histogram — a diagram for continuous data whose bars touch, with no gaps.

Modal class — the class with the highest frequency. With grouped data there is no single mode.

Continuous data — data that can take any value in a range, such as height, mass or time.

Core concepts

Why the bars touch

A histogram displays continuous data, where one class ends exactly where the next begins. There is no gap between 19.999 and 20, so there is no gap between the bars.

A bar chart shows separate categories — favourite colours, modes of transport — which have nothing between them, so its bars are drawn apart. The gap, or its absence, is the visible difference between the two diagrams, and questions do ask for it.

Histograms with equal class widths

When all the classes are the same width, the height of each bar is simply the frequency. Nothing has to be calculated.

That convenience is exactly what fails when the widths are unequal, which is why frequency density exists as a separate topic. For equal widths, height and frequency are the same thing.

The horizontal axis is a continuous number line, so it is labelled with the class boundaries rather than with class names.

Finding a midpoint

Add the lower and upper boundaries and divide by two.

For 0 ≤ x < 10, the midpoint is 5. For 10 ≤ x < 20 it is 15. For 30 ≤ x < 40 it is 35.

The midpoint is not the class width, and the two are easy to confuse. For 20 ≤ x < 30, the width is 10 while the midpoint is 25.

Drawing a frequency polygon

Plot the frequency against the midpoint of each class, then join consecutive points with straight lines — never a curve, and never joined to the axis unless the question supplies classes with zero frequency at each end.

Plotting at boundaries instead of midpoints shifts the whole polygon sideways, which is the error that most often loses the marks here.

A frequency polygon can be drawn straight from the table, without drawing the histogram first. Where a histogram has already been drawn, the polygon joins the midpoints of the tops of the bars.

Why polygons are useful

Two frequency polygons can be drawn on the same axes and compared at a glance, which is awkward with two sets of bars.

The shape also carries meaning. A polygon with its peak on the left describes data bunched at the low end; a symmetrical peak in the middle describes data spread evenly about a central value.

The modal class

With grouped data the individual values are unknown, so there is no single mode. The modal class is the class with the highest frequency, and the answer must be given as a class interval — "20 ≤ x < 30", not "25".

Estimating the mean

Because the raw values are gone, the mean can only be estimated, and midpoints do the work.

Multiply each midpoint by its frequency, add those products, then divide by the total frequency.

For classes with midpoints 5, 15 and 25 and frequencies 4, 10 and 6: the products are 20, 150 and 150, totalling 320, and the total frequency is 20, so the estimated mean is 16.

Dividing by the number of classes rather than by the total frequency is the standard error. There were 20 values, not 3.

Worked examples

Example 1: Midpoints and class widths

For the classes 0 ≤ x < 20, 20 ≤ x < 40 and 40 ≤ x < 60, state the class width and the midpoint of each.

Each class width is the upper boundary minus the lower, so every width is 20.

The midpoints are the averages of the boundaries: 10, 30 and 50.

Notice that the widths are all equal while the midpoints all differ, which is the distinction worth keeping clear.

Example 2: Drawing a frequency polygon

Times in seconds are grouped as 0 ≤ t < 10 with frequency 3, 10 ≤ t < 20 with frequency 8, 20 ≤ t < 30 with frequency 11 and 30 ≤ t < 40 with frequency 4. Describe the points to plot.

The midpoints are 5, 15, 25 and 35.

So the points are (5, 3), (15, 8), (25, 11) and (35, 4), joined by straight lines.

The peak is at the third point, so the modal class is 20 ≤ t < 30, and the answer is given as that interval rather than as the value 25.

Example 3: Estimating the mean

Using the same data, estimate the mean time.

Multiply each midpoint by its frequency: 5 × 3 = 15, 15 × 8 = 120, 25 × 11 = 275, 35 × 4 = 140.

Add the products: 15 + 120 + 275 + 140 = 550.

The total frequency is 3 + 8 + 11 + 4 = 26.

The estimated mean is 550 ÷ 26 = 21.2 seconds to one decimal place.

It is an estimate because the midpoints stood in for values that could have been anywhere in their classes.

Common mistakes and how to avoid them

Plotting at boundaries instead of midpoints. A frequency polygon uses midpoints, which shifts every point to the centre of its class.

Confusing midpoint with class width. For 20 ≤ x < 30 the midpoint is 25 and the width is 10.

Leaving gaps between histogram bars. Continuous data means the bars touch. Gaps belong to bar charts.

Joining frequency polygon points with a curve. Use straight lines.

Giving the modal class as a single number. It is a class interval.

Dividing by the number of classes when estimating the mean. Divide by the total frequency.

Calling an estimated mean exact. Grouping destroyed the original values, so the answer is an estimate and should be described as one.

Exam technique for "Frequency Polygons and Histograms"

Add a midpoint column to the table before plotting anything. It takes a few seconds, prevents the boundary error, and is often worth a mark in its own right.

For an estimated mean, add a second column for midpoint × frequency and total both columns. Method marks are awarded for the products and for dividing by the correct total.

Use a ruler for a frequency polygon and plot each point clearly, since accuracy marks depend on the positions.

Say "estimate" when the question involves grouped data, and give the modal class as an interval.

When comparing two polygons, describe the data rather than the lines: say which set has the higher peak and where each is bunched, in the context of what was measured.

Check that the bars in a histogram touch before moving on.

Quick revision summary

Once data is grouped the original values are lost, so the midpoint stands for its whole class — which is why answers from grouped data are always estimates.

The midpoint is the average of the two class boundaries: 10 ≤ x < 20 gives 15. The class width is the difference between them, which is 10 — the two are not the same.

A histogram shows continuous data, so its bars touch. A bar chart shows separate categories, so its bars are apart. With equal class widths, the bar height is simply the frequency.

A frequency polygon plots frequency against midpoint and joins the points with straight lines. Two polygons on the same axes compare easily.

The modal class is the class with the highest frequency, and the answer is an interval.

To estimate the mean, multiply each midpoint by its frequency, add the products, and divide by the total frequency — not by the number of classes.