LearnYMA01WST01 · Statistics 1Representation and summary of dataHistograms and cumulative frequency

Representation and summary of data · Subtopic 2 of 4

Histograms and Cumulative Frequency

In a histogram, area (not height) shows frequency, which is what makes unequal class widths fair. A cumulative frequency graph turns the same table into a curve from which the median and quartiles can be read.

Cambridge 9709 Paper 5Edexcel YMA01 WST01Edexcel 9MA0 Paper 3AQA 7357 Paper 3≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Find class boundaries and class widths for grouped data.
Calculate frequency density and draw histograms with unequal widths.
Read frequencies back from a histogram, including a part of a class.
Draw cumulative frequency graphs and estimate the median and quartiles.

Pause and think

A table groups ages as $0$ – $9$ ( $20$ people) and $10$ – $49$ ( $40$ people). On a bar chart with bar height $=$ frequency, the second bar is twice as tall. Is it fair to say "twice as many people are in each year of age" there?

Section 2 of 14

Core idea

Start hereWhen bars lie

$12$ people took 20–30 minutes; $10$ took 10–15 minutes. Draw bars of heights $12$ and $10$ and the 20–30 bar looks huge, partly just because it's twice as wide. Histograms fix this by making area, not height, show the count.

The big idea

Histogram bar height = frequency density = frequency ÷ class width, so bar area = frequency. Cumulative frequency graphs (plotted at upper boundaries) estimate medians and quartiles by interpolation.

1. Frequency density

\text{frequency density}=\frac{\text{frequency}}{\text{class width}}\quad\Longleftrightarrow\quad\text{frequency}=\text{height}\times\text{width}

Time taken (minutes): a histogram with unequal classes

Heights:

\tfrac{8}{10}=0.8

\tfrac{10}{5}=2

\tfrac{14}{5}=2.8

\tfrac{12}{10}=1.2

\tfrac{6}{20}=0.3

. Each bar's area (width × height) gives back the frequency.

Worked exampleBuilding and reading the histogram

Classes 0–10: 8; 10–15: 10; 15–20: 14; 20–30: 12; 30–50: 6. Find the densities, and estimate how many took 12–18 minutes.

Goal: Heights; an estimate.
Strategy: Divide by widths; then area over the interval.

$0.8,\ 2,\ 2.8,\ 1.2,\ 0.3$
$\frac{8}{10},\frac{10}{5},\frac{14}5,\frac{12}{10},\frac6{20}$ .
$12\text{–}15:\ 3\times2=6;\quad15\text{–}18:\ 3\times2.8=8.4$
Area = width × height.

AnswerAbout

14.4

, i.e.

14

people.

✓ CheckTotal area

8+10+14+12+6=50

= total frequency ✓.

Common misconceptionBoundaries For whole-number data grouped as 10–14, 15–19, the true boundaries are

9.5

14.5

19.5

(width

5

). Ages "10–14" run from

10

up to

15

2. Cumulative frequency

Cumulative frequency for the same 50 people

Go across from

25

to the graph, then down: the median time is about

17.5

minutes. For quartiles go across from

12.5

and

37.5

Worked exampleMedian by interpolation

Estimate the median time.

Goal: Median.
Strategy: Find the class containing the $25$ th person; assume values spread evenly within it.

$\text{cumulative: }8,18,32,44,50$
Running totals.
$25\text{th}\in15\text{–}20\ \ (18\text{ before},\ 14\text{ in it})$
Locate.
$15+\frac{25-18}{14}\times5=17.5$
Interpolate.

AnswerAbout

17.5

min.

✓ CheckMatches reading the graph at

25

✓.

Worked exampleEstimated mean using midpoints

Estimate the mean.

Goal: Mean.
Strategy: $\frac{\sum fx}{\sum f}$ with class midpoints.

$\sum fx=5(8)+12.5(10)+17.5(14)+25(12)+40(6)$
Midpoints $5,12.5,17.5,25,40$ .
$=40+125+245+300+240=950$
Add.
$950\div50=19$
Divide.

Answer

19

min (an estimate: exact values unknown).

✓ CheckMean > median (

17.5

): consistent with the long right tail ✓.

Pause and think

A class 40–60 has frequency 15. Frequency density?

Summary

Area = frequency.
Cumulative frequency at upper boundaries.
Interpolation and midpoints give estimates.

Frequency density: Frequency per unit width.

HistoryPearson names the histogram. The word "histogram" was introduced by the statistician Karl Pearson in a lecture in 1891. He probably built it from the Greek histos, a mast or upright, for the standing bars, though historians are not certain.

Fun factThe diagram that changed hospitals. In 1858 Florence Nightingale published a "rose diagram", a circular cousin of the histogram, showing that far more soldiers in the Crimean War died of preventable disease than of wounds. Its area-based design persuaded the government to reform army hospitals. She became the first woman elected to the Royal Statistical Society.

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