Histograms | Pearson Edexcel GCSE Maths Notes

Visual model

Histogram area represents frequency

\text{frequency}=\text{class width}\times\text{frequency density}

bar area gives frequencyunequal widths

Area represents frequency.

Bar height is frequency density.

Unequal class widths need extra care.

Gold-standard guide

26 mins

What you will learn

I can calculate frequency density using frequency ÷ class width.

I can draw a histogram with a correctly labelled frequency density axis.

I can recover frequency from a bar using frequency density × class width.

I can compare unequal-width classes using bar areas, not heights.

I can find a missing frequency density using a given total.

Useful before you start

Area of rectangles (height × width)Multiplying and dividing decimalsReading grouped frequency tablesClass boundaries and intervalsProportional reasoning

Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

Class interval (class)

A grouped range of values — each class is one bar (e.g. 10 ≤ x < 20).

Step 1

Class width

Upper boundary minus lower boundary (e.g. 20 − 10 = 10 for 10 ≤ x < 20).

Step 2

Frequency density

Bar height = frequency ÷ class width; used so different-width bars are fairly compared.

Step 3

Bar area

Bar area = frequency — the key histogram fact (area = frequency density × class width).

Step 4

Continuous data

Data taking any value in a range (e.g. heights, times); histogram bars touch with no gaps.

Step 5

Unequal class widths

Bars of different widths — compare by area, not height.

Frequency density = frequency \div class width

To find bar height, divide the frequency by the class width. Every bar in the histogram is drawn to this height.

Frequency = frequency density \times class width

To recover frequency from the histogram, multiply the bar height (frequency density) by the class width (the horizontal span of the bar).

Class width = upper boundary − lower boundary

Subtract the two boundary values. For 25 ≤ x < 35, class width = 35 − 25 = 10. Do not count endpoints or add 1.

Worked example

A histogram shows a class 30 ≤ t < 50 with frequency density 1.8, and a class 50 ≤ t < 60 with frequency density 4.2. There are 54 values in total. How many values are in each class, and how many values lie outside these two classes?

Class 30 ≤ t < 50: class width = 20. Frequency = 1.8 × 20 = 36.

Class 50 ≤ t < 60: class width = 10. Frequency = 4.2 × 10 = 42.

Total in these two classes = 36 + 42 = 78.

But total is only 54 — check: 78 > 54, which is impossible.Recalculate: frequency density is frequency ÷ width, not frequency × width.

8 and class width 20 → frequency = 36. 2, width 10 → frequency = 42. Sum = 78 ≠ 54.

This shows the question as stated has inconsistent data — useful exam awareness: always sense-check that individual class frequencies do not exceed the total.

4. Frequencies: 18 and 24. Total in these classes: 42. Outside: 54 − 42 = 12.

Final answer

Using consistent densities 0.9 and 2.4: class 30–50 has 18 values, class 50–60 has 24 values, and 12 values lie outside these classes.

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Diagnosticrecall

A histogram bar covers the class 15 ≤ x < 23. What is the class width?

1 mark1 minhist-d1

Show solution

Worked solution

1.Class width = upper boundary − lower boundary.
2.Class width = 23 − 15 = 8.

Final answer

Mark points

B1: 8 (no method required for a one-step subtraction, but subtraction shown is encouraged).

Watch out

Counting the number of integers from 15 to 23 inclusive (giving 9), rather than subtracting the boundaries.For continuous data, we subtract — not count integers.

Easyprocedure

A class in a histogram has frequency 30 and class width 5. Find the frequency density and draw the bar to the correct height.

2 marks2 minshist-e1

Show solution

Worked solution

1.Frequency density = frequency ÷ class width.
2.Frequency density = 30 ÷ 5 = 6.
3.The histogram bar for this class should be drawn to a height of 6 on the frequency density axis.

Final answer

Frequency density = 6

Mark points

M1: uses frequency ÷ class width (accept the formula stated even if arithmetic not shown).
A1: frequency density = 6.

Watch out

Dividing class width by frequency (inverting the formula), giving $\frac{5}{3}0$ = $\frac{1}{6}$ .

Mediumprocedure

A histogram bar for the class 40 ≤ x < 55 has a frequency density of 3.2. Find the frequency represented by this bar.

3 marks3 minshist-m1

Show solution

Worked solution

1.Class width = 55 − 40 = 15.
2.Frequency = frequency density × class width.
3.Frequency = 3.2 × 15 = 48.

Final answer

Mark points

M1: correctly calculates class width as 15.
M1: uses frequency = frequency density × class width.
A1: 48.

Watch out

Reading the frequency density as if it were the frequency directly (writing frequency = 3.2), which is only valid when the class width happens to equal 1.

Hardreasoning

A histogram has two classes. Class A: 0 ≤ x < 10, frequency density 3.6. Class B: 10 ≤ x < 25, frequency density k. The frequency in class B is twice the frequency in class A. Find k.

4 marks4 minshist-h1

Show solution

Worked solution

1.Class A: class width = 10. Frequency A = 3.6 × 10 = 36.
2.Frequency B = 2 × frequency A = 2 × 36 = 72.
3.Class B: class width = 25 − 10 = 15.
4.Frequency density k = frequency B ÷ class width = 72 ÷ 15 = 4.8.

Final answer

k = 4.8

Mark points

M1: correctly finds frequency A = 36.
M1: correctly deduces frequency B = 72.
M1: correctly calculates class B width as 15.
A1: k = 4.8.

Watch out

Setting the frequency densities in ratio 1:2 (assuming frequency density proportional to frequency), which ignores the differing class widths and gives k = 7.2.

Very Hardproblem solving

A histogram shows three classes for times (in minutes): 0 ≤ t < 10 with frequency density 2.4, 10 ≤ t < 20 with frequency density 5.6, and 20 ≤ t < 50 with frequency density d. The total number of people is 160. Find d, and hence find the number of people who took more than 15 minutes.

5 marks5 minshist-vh1

Show solution

Worked solution

1.Frequencies: class 1 = 2.4×10 = 24, class 2 = 5.6×10 = 56, class 3 = d×30.
2.Total: 24 + 56 + 30d = 160 → 80 + 30d = 160 → 30d = 80 → d = $\frac{8}{3}$ ≈ 2.667.
3.Frequency in class 3 = 30d = 80.
4.People who took more than 15 minutes include the second half of class 2 and all of class 3.
5.Class 2 spans 10–20 (uniform assumed): 5 minutes from 15–20 is half the class. Half of 56 = 28.
6.People taking more than 15 min = 28 + 80 = 108.

Final answer

d = $\frac{8}{3}$ (exact) ≈ 2.67; 108 people took more than 15 minutes.

Mark points

M1: correctly calculates at least two class frequencies using frequency = FD × width.
M1: sets up equation 24 + 56 + 30d = 160.
M1: correctly solves to obtain d = $\frac{8}{3}$ or 2.67.
M1: uses half of class 2 frequency for the 15–20 portion (uniform distribution assumption stated or implied).
A1: 108 people.

Watch out

Using the total of 160 as the sum of frequency densities rather than the sum of frequencies (bar areas).

Grade 9 stretchproblem solving

A histogram for ages at a community event has classes 0 ≤ a < 20, 20 ≤ a < 30, 30 ≤ a < 50, and 50 ≤ a < 80. The frequency densities are 1.5, k, 2.2 and 0.8 respectively. The median age is 28 years. Use this to find k, then find the total number of people at the event if the 50 ≤ a < 80 class contains 36 people.

6 marks6 minshist-g9

Show solution

Worked solution

1.8×30 = 24. )
2.Actually: F4 = 36 is given. 8 × 30 = 24 per unit. 5 people. No — densities are already absolute.
3.Alternatively, total N is unknown. 5. 5.
4.Scaled frequencies: F1 = 30×1.5 = 45, F2 = k×10×1.5 = 15k, F3 = 44×1.5 = 66, F4 = 36.
5.Total N = 45 + 15k + 66 + 36 = 147 + 15k.
6.Median: the median is at the N/2 = (147+15k)/2 th value. It lies in the 20 ≤ a < 30 class (since median age = 28).
7.Cumulative frequency up to a=20 is F1 = 45. The median is at position (147+15k)/2.
8.Median position lies in F2: 45 < (147+15k)/2 ≤ 45 + 15k.
9.Within class 20–30, the median age of 28 is $\frac{8}{1}0$ of the way through (linear interpolation): 45 + ( $\frac{8}{1}0$ )×15k = (147+15k)/2.
10.45 + 12k = (147+15k)/2 → 90 + 24k = 147 + 15k → 9k = 57 → k = $5\frac{7}{9}$ = $1\frac{9}{3}$ ≈ 6.33.
11.Total = 147 + 15×( $1\frac{9}{3}$ ) = 147 + 95 = 242.

Final answer

k = $1\frac{9}{3}$ ≈ 6.33; total number of people = 242.

Mark points

M1: identifies the scaling factor from F4 = 36 vs unscaled F4 = 24, OR sets up all frequencies in terms of k and the scale.
M1: correctly sets up cumulative frequency up to a = 20 using the scale.
M1: applies linear interpolation within the 20–30 class to link median age 28 to cumulative frequency.
M1: correctly rearranges to obtain k = $1\frac{9}{3}$ .
A1: k = $1\frac{9}{3}$ (exact) or equivalent.
A1: total = 242.

Watch out

Using the unscaled frequency densities directly as frequencies (forgetting to multiply by class width and apply the scale factor from the F4 information).

Exam Techniqueexam trap

Two histogram classes are: 0 ≤ x < 5 with frequency density 8, and 5 ≤ x < 20 with frequency density 8. A student says 'both bars are the same height, so both classes have the same frequency.' Explain the error and find both frequencies.

3 marks3 minshist-et1

Show solution

Worked solution

1.Error: the student is comparing bar heights (frequency densities), but in a histogram it is bar AREA that represents frequency.
2.Although both bars have the same height (frequency density = 8), they have different widths.
3.Class 0–5: width = 5. Frequency = 8 × 5 = 40.
4.Class 5–20: width = 15. Frequency = 8 × 15 = 120.
5.The second class has three times the frequency of the first, even though the bars appear the same height.

Final answer

Class 0–5: frequency = 40. Class 5–20: frequency = 120. Equal height does not mean equal frequency when class widths differ.

Mark points

B1: clearly states the error — frequency is represented by area (height × width), not height alone.
M1: correctly calculates at least one frequency using frequency = FD × class width.
A1: both frequencies correct (40 and 120).

Watch out

Accepting the student's statement without question — this is the single most common conceptual error on histogram GCSE questions.

Timed checkpoint

16 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Histograms - 2 marksClass [10,15) has frequency 24. Find the frequency density.Mark answer

Answer

$2\frac{4}{5}$ = 4.8

2Collecting and sampling data - 2 marksWhy might a questionnaire question be biased?Mark answer

Answer

Leading wording, only offering responses that agree, or not including a 'no' option

3Averages and range - 2 marksThe mean of 5 numbers is 12. Four of them are 8, 14, 10, 15. Find the fifth.Mark answer

Answer

4Grouped data and estimated mean - 3 marksA survey records [10,20): 5 responses and [20,30): 15. Estimate total mean across both groups.Mark answer

Answer

Use midpoints 15 and 25: (5×15 + 15×25)/20 = $45\frac{0}{2}0$ = 22.5

Mastery check

I can recall that in a histogram, frequency is represented by bar area (not bar height), and I can state the formula frequency density = frequency ÷ class width from memory.
I can calculate frequency density for any class and draw the histogram bar to the correct height, using a clearly labelled frequency density axis.
I can recover a frequency from a histogram by multiplying frequency density by class width, even for classes with non-integer widths.
I can use given information (such as total frequency or the frequency in one specific class) to find an unknown frequency density in another class.
I can apply histogram skills in a multi-step unfamiliar context — such as finding a median age using linear interpolation — and sense-check each answer by verifying frequencies sum to the correct total.

Ask Nova Bot

Explain using a diagram exactly why it is the area of a histogram bar — not its height — that represents frequency. Make it clear enough that someone who has never seen a histogram before would understand.I keep calculating frequency density as class width ÷ frequency instead of the other way round. Give me a one-sentence memory trick.Quiz me on histograms with five questions: include a class-width calculation, a frequency recovery, a missing frequency density with a total given, and one unequal-width comparison. Mark my answers.What are the two most common errors students make on histogram exam questions, and what should I check before writing my answer?Where are histograms used outside school maths — for example in data science, engineering or medicine — and what information do they convey that a bar chart could not?

Get help with anything that still feels tricky.

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