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Frequency density = frequency ÷ class width, and it goes on the vertical axis of a histogram with unequal class widths. Free GCSE Maths notes and 13 FAQs.
Histograms and Cumulative Frequency Curves is part of Charts, Diagrams and Distributions in GCSE AQA Higher Maths.
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GCSE AQA Higher
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Frequency density = frequency ÷ class width, and it goes on the vertical axis of a histogram with unequal class widths. Free GCSE Maths notes and 13 FAQs.
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Use this lesson to stay inside Charts, Diagrams and Distributions while connecting the detail back to the wider Statistics topic.
Read the core explanations from Histograms and Cumulative Frequency Curves before testing yourself from memory.
Raw data rarely arrives in a neat list. In the real world — think hospital waiting times, exam scores, or rainfall measurements — data comes grouped into classes. Two diagrams handle this beautifully: histograms and cumulative frequency graphs. Both appear regularly on Higher-tier papers, and both reward students who understand what the axes actually represent.
The key insight that unlocks both diagrams: with grouped data, you can never know exact values, only which class they fall into. These diagrams let you estimate medians, spreads, and proportions from that grouped information — and that estimation skill is exactly what examiners test.
Start with the simpler case: all class intervals are the same width. Here, the height of each bar is proportional to frequency, so you can read frequencies directly from the vertical axis. The bars touch each other — no gaps — because the data is continuous.
Equal-width histograms look similar to bar charts, but the no-gaps rule signals continuous data. The moment class widths differ, you must switch to frequency density on the vertical axis — which is where most marks are won or lost.
Here is where histograms become genuinely different from anything else on the GCSE course. When class widths vary, a wider bar would look bigger even if it contained fewer data values — so height alone becomes misleading. The fix is to use frequency density on the vertical axis instead.
The golden rule: area = frequency. Every mark-winning move in a histogram question comes back to this. To find the frequency from a histogram bar, multiply its height (frequency density) by its width (class width). To draw a bar, divide frequency by class width to get the height.
Worked example: Drawing a Histogram with Unequal Class Widths
The table shows the times, in minutes, taken by 60 students to complete a puzzle. Draw a histogram to represent this data. | Time (minutes) | Frequency | |---|---| | 0 < t ≤ 5 | 6 | | 5 < t ≤ 10 | 15 | | 10 < t ≤ 20 | 24 | | 20 < t ≤ 30 | 10 | | 30 < t ≤ 50 | 5 |
Answer: Histogram with bars at heights 1.2, 3.0, 2.4, 1.0, and 0.25 for the five classes respectively. Vertical axis labelled 'Frequency density'.
Examiners often give you a completed histogram and ask you to find a frequency, an estimate of the mean, or the number of values above a threshold. The method is always the same: use area = frequency to work backwards from the bar.
Worked example: Finding a Frequency from a Histogram
A histogram shows a bar for the class $15 < t \leq 25$ with a frequency density of 3.6. Calculate the frequency for this class.
Answer: The frequency for the class $15 < t \leq 25$ is 36.
Now for the second major diagram. A cumulative frequency graph — sometimes called an ogive — lets you estimate the median and quartiles from grouped data. Think of it as a running total: after each class, you add up all the frequencies so far.
The shape of a cumulative frequency curve is always an S-shape (a stretched S, sometimes called a sigmoid). If your curve bends the wrong way or looks like a straight line, check your cumulative totals — you have likely added incorrectly.
Worked example: Constructing a Cumulative Frequency Graph
The table shows the heights, in cm, of 80 plants. Construct a cumulative frequency graph and use it to estimate the median and interquartile range. | Height (cm) | Frequency | |---|---| | 0 < h ≤ 10 | 4 | | 10 < h ≤ 20 | 12 | | 20 < h ≤ 30 | 26 | | 30 < h ≤ 40 | 22 | | 40 < h ≤ 50 | 16 |
Answer: Median ≈ 29 cm, Lower Quartile ≈ 22 cm, Upper Quartile ≈ 38 cm, IQR ≈ 16 cm.
Both diagrams handle grouped continuous data, but they answer different questions. A histogram shows you the shape of a distribution — where data is concentrated and how spread out it is across classes. A cumulative frequency graph lets you estimate specific statistics such as the median, quartiles, and percentiles.
| Feature | Histogram | Cumulative Frequency Graph |
|---|---|---|
| Vertical axis | Frequency density (unequal widths) or frequency (equal widths) | Cumulative frequency |
| Points plotted at | Bars drawn between class boundaries | Upper class boundaries |
| Shape | Bars (no gaps) | Smooth S-shaped curve |
| Use to find | Frequency in a class, shape of distribution | Median, quartiles, IQR, percentiles |
| Key rule | Area of bar = frequency | Read across then down from the curve |
The methods only stick when you apply them to fresh data. Work through these problems using the techniques from earlier sections. For each histogram question, start by checking whether class widths are equal — that determines which method to use.
These two diagram types account for several marks on Higher-tier papers, and the same errors appear year after year. Knowing where students drop marks is as valuable as knowing the methods themselves.
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Move to the related lessons or back to the topic page when you need the wider sequence before exam-style practice.
Frequency density = frequency ÷ class width, and it goes on the vertical axis of a histogram with unequal class widths. For a class from 10 to 20 minutes with a frequency of 24, the frequency density is 24 ÷ 10 = 2.4.
Work out the running totals, plot each one at the upper class boundary, start at (0, 0), then join the points with a smooth S-shaped curve. For frequencies of 4, 12, 26, 22 and 16, the cumulative frequencies are 4, 16, 42, 64 and 80.
Read across from n/2 on the cumulative frequency axis to the curve, then down. For 80 plants, the median is at 40, giving about 29 cm; the quartiles are at 20 and 60, giving 22 cm and 38 cm, so the IQR is 38 - 22 = 16 cm.
An ogive is another name for a cumulative frequency graph, which plots running totals of frequency against upper class boundaries. It is used to estimate the median, quartiles and interquartile range from grouped data.
Histogram bars touch, with no gaps, because the data is continuous. Bar charts have gaps because each bar is a different group.
Use area = frequency: multiply the frequency density by the class width. For the class 15 to 25 with a frequency density of 3.6, the frequency is 3.6 × 10 = 36.
A histogram shows the shape of a distribution and the frequency in each class, with area = frequency. A cumulative frequency graph is a smooth S-shaped curve used to estimate the median, quartiles, IQR and percentiles.
The area of a histogram bar equals the frequency for that class: frequency = frequency density × class width. For a bar from 30 to 50 with height 0.25, the frequency is 0.25 × 20 = 5.
Points on a cumulative frequency graph are plotted at the upper class boundary, not the midpoint, with a starting point at (0, 0) when appropriate. For heights up to 10, 20 and 30 cm, the points are (10, 4), (20, 16) and (30, 42).
A cumulative frequency graph is a smooth, stretched S-shape. If the curve bends the wrong way or looks like a straight line, the cumulative totals have likely been added wrongly.
The lower quartile is found by reading across from n/4 on the cumulative frequency axis to the curve, then down. For 80 plants, read across from 20, giving a lower quartile of about 22 cm.
The upper quartile is found by reading across from 3n/4 on the cumulative frequency axis to the curve, then down. For 80 plants, 3 × 80 ÷ 4 = 60, giving an upper quartile of about 38 cm.
A histogram is checked by multiplying each bar's height by its width to get back the frequencies, then adding them. The total of all areas should equal the total frequency, such as 6 + 15 + 24 + 10 + 5 = 60.