Histograms and frequency density

A histogram displays grouped continuous data. Its defining rule is

$$\boxed{\text{frequency is proportional to bar area}.}$$

This matters whenever class widths differ. A class covering twice as much of the horizontal axis has twice the area at the same height, so bar height cannot simply be frequency. Instead, the vertical axis usually shows frequency density:

$$\boxed{\text{frequency density}=\frac{\text{frequency}}{\text{class width}}.}$$

Equivalently,

$$\boxed{\text{frequency}=\text{class width}\times\text{frequency density}.}$$

Prerequisites

You should be able to:

• interpret grouped continuous data and class boundaries;
• calculate the width of an interval;
• find the area of a rectangle;
• use inequalities such as $20\leq x<30$.

Review presenting and interpreting data if class intervals or frequency tables are unfamiliar. Histograms are then used to estimate quantities in averages and measures of spread.

Histogram or bar chart?

These diagrams are not interchangeable.

Feature	Bar chart	Histogram
Typical data	qualitative or discrete	grouped continuous
Horizontal axis	categories	continuous numerical scale
Gaps between bars	usually present	absent between adjacent classes
Frequency represented by	height	area
Vertical axis	frequency, relative frequency or percentage	usually frequency density

Touching bars alone do not make a histogram. The horizontal scale must encode numerical class widths, and each bar’s area must represent its frequency.

The tallest bar need not contain the most observations

Height shows frequency per unit of class width. A shorter but much wider bar can have the greater area and therefore the greater frequency.

Class width and frequency density

For a class $a\leq x<b$,

$$\text{class width}=b-a.$$

The units help explain frequency density. If frequency is a count and time is measured in seconds, then

$$\frac{\text{observations}}{\text{seconds}}$$

is a density of observations along the time axis. It is not a physical rate of events happening each second. It is only the scale needed to make area equal frequency.

Worked example 1: calculate densities

The times $t$, in minutes, taken by $60$ students to complete a task are grouped below.

Time, $t$	Frequency, $f$	Class width	Frequency density, $f/w$
$0\leq t<10$	$8$	$10$	$0.8$
$10\leq t<15$	$9$	$5$	$1.8$
$15\leq t<25$	$24$	$10$	$2.4$
$25\leq t<40$	$15$	$15$	$1.0$
$40\leq t<60$	$4$	$20$	$0.2$

For the class $10\leq t<15$,

$$\text{frequency density}=\frac95=1.8.$$

For $25\leq t<40$,

$$\text{frequency density}=\frac{15}{15}=1.$$

Although the second of these classes contains more students, its bar is lower because those students are spread across a wider interval. Its area is still

$$15\times1=15,$$

equal to its frequency. The first area is $5\times1.8=9$.

Self-check 1

A class $32\leq x<44$ has frequency $30$. Find its class width and frequency density.

Answer

The width is

$$44-32=12.$$

Therefore

$$\text{frequency density}=\frac{30}{12}=2.5.$$

Constructing a histogram

Use this reliable procedure.

1. Write every class using its continuous boundaries.
2. Calculate each class width.
3. Calculate $f/w$ for every class.
4. Draw the horizontal axis to scale and mark the class boundaries.
5. Label the vertical axis frequency density and choose a suitable scale.
6. Draw touching rectangles across the exact intervals at the calculated heights.

For Worked example 1, the first rectangle runs from $0$ to $10$ and has height $0.8$. The second runs from $10$ to $15$ and has height $1.8$. Do not give every class the same drawn width: a width of $5$ minutes must occupy half the horizontal distance of a width of $10$ minutes.

Equal class widths

When every class has the same width, frequency and frequency density differ by the same constant factor. A frequency vertical axis then produces proportional areas, so some questions use frequency as the height. Frequency density remains the general method and is essential for unequal widths.

Rounded measurements and class boundaries

If lengths are measured to the nearest centimetre, the labelled group $10$ to $14$ has continuous boundaries

$$9.5\leq x<14.5,$$

so its width is $5$, not $4$. Adjacent groups $10$ to $14$ and $15$ to $19$ meet at $14.5$ and leave no gap.

If the table already states $10\leq x<15$, its boundaries are explicit. Do not adjust them again.

Worked example 2: boundaries from rounded data

Masses are recorded to the nearest kilogram.

Recorded mass, kg	Frequency
$50$ to $54$	$6$
$55$ to $64$	$20$
$65$ to $79$	$18$

The continuous intervals and densities are:

Continuous class	Width	Frequency density
$49.5\leq m<54.5$	$5$	$6/5=1.2$
$54.5\leq m<64.5$	$10$	$20/10=2.0$
$64.5\leq m<79.5$	$15$	$18/15=1.2$

The first and third bars have equal height but different areas. Their frequencies are $6$ and $18$ because the third bar is three times as wide.

Reading frequencies from a histogram

If frequency density is labelled, recover a class frequency directly:

$$f=w\times d.$$

Worked example 3: recover a frequency

A histogram bar covers $12\leq x<20$ and has frequency density $3.5$. Its width is

$$20-12=8,$$

so its frequency is

$$f=8\times3.5=28.$$

If a second bar covers $20\leq x<35$ at density $1.6$, its frequency is

$$(35-20)\times1.6=24.$$

The first bar is both taller and larger in area, so it has the greater frequency.

When the vertical scale is missing

Sometimes a histogram gives one known frequency but no numerical density scale. Areas are still proportional to frequencies. A known bar calibrates the diagram.

If a known class has frequency $f_0$, drawn width $w_0$ and drawn height $h_0$, then its drawn area $w_0h_0$ represents $f_0$. For another bar,

$$\frac{f}{f_0}=\frac{wh}{w_0h_0}.$$

You may instead use the known class to find the scale factor between a measured diagram height and frequency density. Be consistent with the actual horizontal scale.

Worked example 4: use area ratios

On a histogram with no labelled vertical scale:

• class $0\leq x<10$ has frequency $18$ and drawn height $3$ cm;
• class $10\leq x<25$ has drawn height $2$ cm;
• horizontal lengths are drawn in proportion to the true class widths.

The drawn areas are proportional to

$$A_1=10\times3=30$$

and

$$A_2=15\times2=30.$$

The areas are equal, so the frequencies are equal. The second frequency is therefore

$$\boxed{18}.$$

Notice that the second bar is lower but wider.

Self-check 2

A histogram has no vertical scale. A class of width $4$ has drawn height $6$ cm and frequency $12$. Another class has width $10$ and drawn height $3$ cm. Find its frequency.

Answer

The area ratio is

$$\frac{10\times3}{4\times6}=\frac{30}{24}=\frac54.$$

Hence the second frequency is

$$12\times\frac54=15.$$

Estimating frequencies within a class

A histogram preserves the class totals, not the exact positions of observations inside each class. To estimate a frequency in part of a bar, assume observations are distributed evenly through that class. The estimate is proportional to horizontal width:

$$\text{estimated frequency in part} =\frac{\text{width of part}}{\text{class width}}\times\text{class frequency}.$$

This is the same as finding the relevant part of the bar’s area.

Worked example 5: estimate from part of a bar

The class $20\leq t<35$ has frequency density $2.4$. Estimate how many observations satisfy $20\leq t<28$.

The required interval has width

$$28-20=8.$$

Its area is

$$8\times2.4=19.2.$$

So the histogram estimate is about $19$ observations.

The non-integer intermediate result is acceptable because this is an estimate based on an even spread. The actual count must be a whole number, but the grouped data do not reveal it.

Alternatively, the whole class has estimated frequency

$$15\times2.4=36,$$

and the required fraction is $8/15$, giving

$$\frac8{15}\times36=19.2.$$

Worked example 6: combine complete and partial classes

A histogram represents $80$ journey times. The relevant bars are:

Time, $t$	Frequency density
$10\leq t<20$	$1.5$
$20\leq t<30$	$3.2$
$30\leq t<50$	$1.1$

Estimate the number of journeys with $15\leq t<38$.

Split the interval at class boundaries:

$$[15,20),\qquad [20,30),\qquad [30,38).$$

The corresponding areas are

$$\begin{aligned} (20-15)(1.5)&=7.5,\\ (30-20)(3.2)&=32,\\ (38-30)(1.1)&=8.8. \end{aligned}$$

Therefore the estimate is

$$7.5+32+8.8=48.3,$$

or about $48$ journeys.

An estimate is not hidden exact data

Dividing part of a bar assumes uniform density within that class. Two raw data sets can produce the same histogram but have different numbers below an internal value such as $28$.

Shape and the modal class

The distribution’s shape can be described as symmetric, positively skewed, negatively skewed, unimodal or bimodal. Interpret this cautiously because grouping can hide detail.

For unequal class widths, the modal class is the class with the greatest frequency density, not necessarily the greatest frequency. It is the interval with the greatest concentration of observations per unit width.

In Worked example 1, $15\leq t<25$ is the modal class because its density $2.4$ is the largest. It also happens to have the greatest frequency, but that need not occur.

Self-check 3

Class A has width $5$ and frequency $18$. Class B has width $12$ and frequency $30$.

1. Which class has the greater frequency?
2. Which class produces the taller histogram bar?

Answer

1. Class B has the greater frequency because $30>18$.
2. The densities are

$$d_A=\frac{18}{5}=3.6, \qquad d_B=\frac{30}{12}=2.5.$$

Therefore Class A produces the taller bar and is the modal class of these two.

Common misconceptions

Misconception	Correction
Height always equals frequency	Area represents frequency; height is usually frequency density.
Class width is the number of integer labels	Width is upper boundary minus lower boundary.
A class $10$ to $19$ has width $9$	For whole-number rounded data its boundaries are $9.5$ and $19.5$, so its width is $10$.
The tallest bar has the greatest frequency	It has the greatest density. Compare areas for frequencies.
Every estimate from a histogram is exact	Estimates inside a class assume an even spread.
Bars may be equally wide for convenience	Their horizontal widths must follow the numerical scale.

Exam method

Before accepting a histogram calculation, ask:

1. Have I used continuous class boundaries?
2. Did I calculate width as upper boundary minus lower boundary?
3. Am I comparing heights or areas for the quantity asked?
4. If I split a class, have I stated or recognised the uniformity assumption?
5. Are all axes and units labelled?

The identity

$$\boxed{f=wd}$$

is the central check. If any two of frequency $f$, class width $w$ and frequency density $d$ are known, the third follows.

Mixed self-check

The lifetimes $x$, in hours, of $100$ components are summarised by a histogram.

Lifetime, $x$	Frequency density
$0\leq x<20$	$0.6$
$20\leq x<50$	$1.4$
$50\leq x<70$	$1.8$
$70\leq x<90$	unknown

1. Find the frequencies in the first three classes.
2. Find the frequency density for $70\leq x<90$.
3. State the modal class.
4. Estimate how many components lasted between $35$ and $60$ hours.

Answer

1. Using $f=wd$:

$$\begin{aligned} f_1&=20(0.6)=12,\\ f_2&=30(1.4)=42,\\ f_3&=20(1.8)=36. \end{aligned}$$

2. The first three frequencies total

$$12+42+36=90.$$

The final frequency is $100-90=10$. Its width is $20$, so

$$d=\frac{10}{20}=0.5.$$

3. The greatest density is $1.8$, so the modal class is

$$50\leq x<70.$$

4. From $35$ to $50$, the width is $15$ and density is $1.4$. From $50$ to $60$, the width is $10$ and density is $1.8$. Hence

$$15(1.4)+10(1.8)=21+18=39.$$

The estimate is $39$ components.

Next steps

• Use class midpoints to estimate grouped means and standard deviations in averages and measures of spread.
• Compare distribution shape, centre and spread in presenting and interpreting data.
• Learn how unusual values affect summaries in outliers and cleaning data.
• Apply statistical summaries to paired variables in correlation and regression.