Histograms
SPEC S3(H) • HIGHER TIER ONLY
Frequency density is the height, area is the frequency. The chart for grouped continuous data with unequal class widths.
Why Histograms? Higher Tier Only
Bar charts work for equal-width categories. But when class widths are unequal — say, 0–10, 10–20, 20–50 — plotting frequency directly makes the wider class look like it has more data than it does. Histograms fix that by putting frequency density on the vertical axis instead of frequency.
When to use a histogram
Data is continuous (height, time, mass) — not discrete.
Data is grouped into class intervals.
Class widths are unequal — this is when a histogram becomes essential.
MCE approach
Method — find class widths, compute frequency density (freq ÷ width) for each class, then plot density as height. Calculate — use the defining property: area of each bar = frequency of that class. Evaluate — check bar heights look sensible and the total area equals n.
Bar Chart vs Histogram Higher Tier Only
The two charts look similar but carry different information. Knowing when to use each is a key Higher skill.
FeatureBar chartHistogram
Data typeCategorical or discreteContinuous, grouped
BarsGaps between barsNo gaps — bars touch
X-axisLabels (categories)Continuous scale
Y-axisFrequencyFrequency density
Height meaningThe frequency itselfFreq ÷ class width
What equals the frequencyThe heightThe area of the bar
Why bar charts fail for unequal widths
Take classes 0–10 (freq 20) and 10–50 (freq 20). Same frequency, different widths. A bar chart shows two bars of equal height — misleading, because the data in the wider class is spread over four times the range. Frequency density rescales correctly: 20÷10 = 2 vs 20÷40 = 0.5. The shorter bar tells the truth.
Histogram Conventions Higher Tier Only
The four rules
No gaps between bars — the x-axis is a continuous scale.
Bars may have different widths (this is what distinguishes histograms from bar charts).
Y-axis must be labelled "Frequency density", not "Frequency".
Label the x-axis with the variable and its units.
When class widths are all equal
If every class has the same width, frequency density is just frequency ÷ (constant), so the histogram has the same shape as a bar chart with frequency on the y-axis. In practice, AQA histogram questions almost always use unequal widths — so frequency density is essentially always required. When widths are unequal, density is non-negotiable.
Frequency Density Formula Higher Tier Only
Frequency density
frequency density = frequency ÷ class width
Three-step routine
Find the width of each class: width = (upper bound) − (lower bound).
Divide the frequency by the width for each class.
Plot bars of the correct widths with heights equal to the density.
Units
If the variable is measured in cm, then frequency density has units "per cm". If time in minutes, "per minute". This is rarely written on exam answers but helps with sense-making.
Defining Property: Area = Frequency Higher Tier Only
The key identity
frequency = frequency density × class widthi.e. the area of each bar equals the frequency of its class.
This is the defining property of a histogram. Frequency density tells you the height, the class width tells you the width, and the product (area) gives the frequency. It's a rearrangement of the formula on the previous slide.
Consequences
Total frequency (n) equals the total area of all bars summed.
The frequency of any class is the area of that bar, not its height.
Reading a histogram always starts with multiplying height × width.
Don't read the height as the frequency
This is the single most common histogram error. If a bar has height 2.5 and width 10, the frequency of that class is 2.5 × 10 = 25, not 2.5. Students who skip the multiplication lose nearly every mark on the question.
Computing Densities — Step by Step Higher Tier Only
Always add two columns to a grouped frequency table: class width and frequency density. The widths don't all have to be the same — that's the point.
ClassFrequencyClass widthFrequency density
0 ≤ x < 101210 − 0 = 1012 ÷ 10 = 1.2
10 ≤ x < 202020 − 10 = 1020 ÷ 10 = 2.0
20 ≤ x < 401440 − 20 = 2014 ÷ 20 = 0.7
40 ≤ x < 801280 − 40 = 4012 ÷ 40 = 0.3
Sense-check your densities
The highest density bar is where the data is most "packed". The lowest density bar is where the data is most spread out — and that's often a wider class with fewer items per unit width.
Worked Example — Draw a Histogram
Higher Tier Only
The table shows the heights of 36 plants. Draw a histogram to represent the data. [3 marks]
Height h (cm)Frequency
100 ≤ h < 1206
120 ≤ h < 1308
130 ≤ h < 14012
140 ≤ h < 16010
MMethod. Add class width and frequency density columns. Plot bars with the correct widths and heights = density. No gaps.
Calculate.
The histogram
EEvaluate. The 130 ≤ h < 140 class has the highest density (1.2) — the data is most packed there. The 100 ≤ h < 120 class is wide but sparse (density 0.3). The areas (6 + 8 + 12 + 10 = 36) match the total frequency ✓.
Reading Frequencies from a Histogram Higher Tier Only
The inverse skill to drawing: given a histogram, find the frequency of a class.
Frequency from a histogram
frequency = height (density) × class width
Three-step routine
Read the height of the bar off the vertical axis — this is the frequency density.
Find the width of that bar from the horizontal axis.
Multiply — the area of the bar is the frequency of the class.
Misreading the vertical scale
AQA examiner reports flag this repeatedly: students read heights off the wrong scale. If the y-axis goes 0, 1, 2, 3, 4, 5 with each small square = 0.1, then the top of a bar at the 15th small square is 1.5, not 15 and not 0.15. Always check the scale before multiplying.
Worked Example — Reading a Histogram
Higher Tier Only
The histogram shows the masses of items produced by a factory sample. (a) Find the frequency of the class 10 ≤ m < 20. (b) How many items in total are in the sample? [4 marks]
MMethod. Frequency of a class = density × width. Total = sum of all class frequencies (the total area).
Calculate.
(a) Read bar 10 ≤ m < 20: height = 2.5, width = 10. Frequency = 2.5 × 10 = 25 items.
(b) Frequencies of all four bars:
•
0 ≤ m < 10: 2.0 × 10 = 20
•
10 ≤ m < 20: 2.5 × 10 = 25
•
20 ≤ m < 40: 0.5 × 20 = 10
•
40 ≤ m < 60: 0.25 × 20 = 5
Total = 20 + 25 + 10 + 5 = 60 items.
EEvaluate. The highest-density class (10–20, density 2.5) also has the highest frequency (25), but note that the wide low-density class 20–40 still has 10 items because it's twice as wide as the narrower classes. Area always wins.
Partial Classes — "Less Than X" Higher Tier Only
A common exam question asks for the number of items less than some value that cuts through the middle of a class. The trick is to split that class proportionally.
Partial-class method
Count all whole classes entirely below the target value.
For the class that contains the target, take (target − lower bound) × density to find the partial frequency.
Add them together.
Example with the WE2 histogram
"How many items are less than 14 g?" — class 0–10 gives 20 whole items. The target 14 is 4 units into the 10–20 class (density 2.5). Partial frequency = 4 × 2.5 = 10. Total: 20 + 10 = 30 items.
The equal-spread assumption
Splitting a class proportionally assumes the data within that class is uniformly distributed. This is the same assumption behind the "estimated mean" — it's why the answer is an estimate, not exact.
Scaling to a Larger Population Higher Tier Only
AQA examiners love this variant: a histogram represents a sample, and you're asked how many items in a much larger population share some property. The sample size clue is often easy to miss.
Scaling routine
Find the sample size n (total area of the histogram, or check the question).
Find the sub-frequency from the histogram (the area with the required property).
Express as a proportion: sub-frequency ÷ n.
Multiply the population size by that proportion.
The "sample of 80" trap
Examiner reports flag this recurring error: students compute the sub-frequency, then forget to scale, and write their sample frequency as the final answer. If the question says "the company produces 25,000 per day" or "the factory makes 600 every day", you must scale up from your sample total.
Worked Example — Partial Class & Scaling
Higher Tier Only
The sample of 60 items from the previous histogram came from a factory that makes 600 items each day. Estimate the number of items each day with a mass of less than 14 grams. [4 marks]
MMethod. (1) Find items less than 14 in the sample using the partial-class method. (2) Express as a proportion of 60. (3) Scale up to 600.
Calculate.
Items less than 14 in the sample:
•
0 ≤ m < 10: density 2.0 × width 10 = 20 items (whole class).
•
10 ≤ m < 14: density 2.5 × width 4 = 10 items (partial class).
Sample total < 14 g: 20 + 10 = 30 items out of 60.
Proportion = 30 ÷ 60 = 0.5 (or one half).
Scaled to daily production of 600: 0.5 × 600 = 300 items per day.
EEvaluate. 300 is less than 600 (can't exceed the daily total) and is a sensible proportion (half) given that almost half of the sample sits below 14 g. The answer is an estimate — it assumes the factory sample was representative and the 10–20 g class is evenly distributed.
Estimating the Median From a Histogram Higher Tier Only
Less common than drawing or reading, but it does appear. The approach is basically the partial-class method applied to finding n/2.
Five-step routine
Compute the frequency of each class (density × width).
Find the total n — the sum of all frequencies.
Target n ÷ 2.
Use running totals to find the class containing the n/2-th value.
Apply linear interpolation: median ≈ lower bound + (items still needed ÷ class density).
Quick example using WE2 data (n = 60)
n/2 = 30. Running totals: 20 (at m = 10), 45 (at m = 20). So the median is in 10 ≤ m < 20. Need 10 more items past m = 10. Density there is 2.5, so 10 / 2.5 = 4 units along. Estimated median ≈ 14 g.
Examiner Alerts Higher Tier Only
Alert 1 — Reading height as frequency
The height of a bar is frequency density, not frequency. To get the frequency, always multiply by the class width. This is the single most-penalised histogram error.
Alert 2 — Misreading the vertical scale
Examiner reports consistently flag scale misreads: reading 0.3 as 3, or 1.5 as 15. Always verify what each small square represents before computing frequencies.
Alert 3 — Using raw frequency on the y-axis
If the question gives a frequency table with unequal class widths and asks for a histogram, you must compute frequency densities. Plotting raw frequency gives bars of the wrong relative heights and typically scores zero.
Alert 4 — Forgetting to scale to the population
When the question gives a sample size and a daily/annual population total, scale up: (sub-frequency ÷ sample size) × population. Don't leave the sample frequency as your final answer.
Alert 5 — Gaps between bars
Histograms have no gaps. Data is continuous, so each class's upper bound is the next class's lower bound — the bars must touch. Gaps turn your histogram into a (wrong) bar chart.
Practice Question Higher Tier Only
The table shows the number of hours 40 students spent revising for a test.
Time t (hours)Frequency
0 ≤ t < 46
4 ≤ t < 814
8 ≤ t < 1212
12 ≤ t < 208
(a) Work out the frequency density for each class and state the height and width of each of the four bars. [3 marks]
(b) Estimate the number of students who spent less than 10 hours revising. [2 marks]
(c) The whole year group has 200 students. Estimate the number who spent less than 10 hours revising. [2 marks]
MMethod. (a) Density = freq ÷ width. (b) Partial-class method. (c) Proportion × 200.
Calculate.
(a) Densities: 6 ÷ 4 = 1.5; 14 ÷ 4 = 3.5; 12 ÷ 4 = 3.0; 8 ÷ 8 = 1.0. Bars: width 4 height 1.5; width 4 height 3.5; width 4 height 3.0; width 8 height 1.0. No gaps.
(b) Less than 10 hours:
•
0 ≤ t < 4 full: 6 students
•
4 ≤ t < 8 full: 14 students
•
8 ≤ t < 10 partial: density 3.0 × width 2 = 6 students
Total = 6 + 14 + 6 = 26 students.
(c) Proportion = 26 ÷ 40 = 0.65. Scaled to 200 students: 0.65 × 200 = 130 students.
EEvaluate. All frequencies recover to 40 (6 + 14 + 12 + 8) ✓. 26/40 is just under two-thirds — sensible given most students cluster in the 4–12 hour range. 130 is less than 200 ✓. Note the answer to (c) is an estimate — it assumes the 40-student sample is representative and the 8–12 class is uniformly distributed.
Summary — MCE Checkpoint Higher Tier Only
Key formulas
frequency density = frequency ÷ class width
frequency = frequency density × class width
Area of each bar = frequency of the class.
Histogram conventions
No gaps between bars — continuous data.
Bars may have different widths .
Y-axis = frequency density (not frequency).
Total area = n (the total frequency).
Common question patterns
Draw a histogram from a frequency table — compute densities then plot.
Read a frequency from a histogram — area (height × width) = frequency.
Partial class "less than X" — whole classes below + proportional chunk of the class containing X.
Scaling to population — (sub-frequency ÷ sample n) × population size.
Estimate median — find class with the n/2-th value, interpolate within.
Examiner traps
Reading height as frequency (must multiply by width).
Misreading the vertical scale (0.3 vs 3).
Using raw frequency on y-axis when widths are unequal.
Forgetting to scale up to the population.
Leaving gaps between bars.
✓ You should now be able to
Know when a histogram is the right chart (continuous grouped data, especially unequal class widths).
Compute frequency density from a frequency table.
Draw a histogram with no gaps, correct widths and correct heights.
Read the frequency of any class from a given histogram using area = frequency.
Apply the partial-class method for "less than X" questions.
Scale a sample result up to a larger population.
Estimate the median from a histogram using running totals and linear interpolation.
Next: Why Histograms? Higher Tier Only