Histograms

AS · S1 · 13 min

A histogram shows the shape of a large set of grouped, continuous data. It looks like a bar chart, but it works on a different principle: the area of each bar, not its height, represents the frequency. That is what lets classes have different widths without distorting the picture. Paper 5 almost always includes a histogram question, and the marks go on three things: the correct class boundaries, the correct frequency densities, and an accurately drawn diagram with labelled axes.

Why area, not height

Imagine $30$ people aged $30$ to $49$ and $30$ people aged $20$ to $24$ . If both bars had height $30$ , the wider bar would look four times as important, yet both classes hold the same number of people. The fix is to make the bar heights show how densely packed the data are: $30$ people over $20$ years is $1.5$ people per year, and $30$ people over $5$ years is $6$ people per year. Then area $=$ height $\times$ width $=$ frequency, and the picture is honest.

Definition

The frequency density of a class is its frequency divided by its class width. In a histogram each bar is drawn between the class boundaries with height equal to the frequency density, so the area of each bar equals (or is proportional to) the class frequency.

Key result

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

Class boundaries

The bars of a histogram touch, so they must start and end at the class boundaries: the values where one class genuinely ends and the next begins. How you find them depends on how the data were recorded.

Key result

How the data are recorded	Class written as	Boundaries	Width
Continuous, given with inequalities	$10 \le x < 20$	$10$ and $20$	$10$
Continuous, rounded to the nearest unit	$10$ – $19$	$9.5$ and $19.5$	$10$
Discrete (whole numbers)	$10$ – $19$	$9.5$ and $19.5$	$10$
Age in completed years	$10$ – $19$	$10$ and $20$	$10$
Continuous, to $1$ d.p.	$2.5$ – $2.9$	$2.45$ and $2.95$	$0.5$

The rule behind the table: the lower boundary is the smallest value that would be recorded in the class, and the upper boundary is where the next class starts. A length of $19.7\ \text{cm}$ recorded to the nearest cm becomes $20$ , so it falls in the class $20$ – $29$ ; the boundary between " $10$ – $19$ " and " $20$ – $29$ " is therefore $19.5$ .

Age is the exception because people do not round their ages; they truncate. Someone who is $19$ years and $11$ months says they are $19$ , so " $10$ – $19$ " includes everyone up to (but not including) their $20$ th birthday.

The class width is upper boundary minus lower boundary, never the difference of the numbers written in the table: " $10$ – $19$ " has width $10$ , not $9$ .

The mid-point (class mark), used for estimating the mean, is halfway between the boundaries: for " $10$ – $19$ " rounded to the nearest unit it is $\tfrac{9.5 + 19.5}{2} = 14.5$ ; for ages " $10$ – $19$ " it is $15$ .

Drawing a histogram

Method

Find the class boundaries and class widths.
Calculate each frequency density, frequency $\div$ width. Add columns for these to the table.
Draw a horizontal axis with a continuous, uniform scale for the variable (label it with units) and a vertical axis labelled frequency density, with a uniform scale.
Draw each bar between its class boundaries with height equal to its frequency density. Bars touch; there are no gaps (unless a class has frequency $0$ ).
Check one bar: height $\times$ width should give back the frequency.

Here is a histogram for the journey times of $100$ people.

Time $t$ (minutes)	$0 \le t < 10$	$10 \le t < 15$	$15 \le t < 20$	$20 \le t < 30$	$30 \le t < 50$
Frequency	$8$	$14$	$23$	$30$	$25$
Class width	$10$	$5$	$5$	$10$	$20$
Frequency density	$0.8$	$2.8$	$4.6$	$3.0$	$1.25$

Bar heights are frequency densities; the area of each bar is the class frequency. The tallest bar marks the modal class, 15 to 20 minutes.

Notice that the class with the largest frequency ( $20 \le t < 30$ , $30$ people) is not the tallest bar. The modal class is the class with the highest frequency density, here $15 \le t < 20$ , because that is where the data are most concentrated.

Reading a histogram

Going backwards is just as common in exams: you are given the histogram (or the frequency densities) and asked for frequencies.

The frequency of a class is its frequency density times its width.
To estimate how many values lie in part of a class, assume the values are spread evenly through the class, and take the matching fraction of the bar's area: frequency density $\times$ the width of the part you want.

If the histogram is drawn with a vertical axis that is not frequency density (for example the question says "the bar has height $3\ \text{cm}$ "), then area is only proportional to frequency. Find the constant from a bar whose frequency you know: frequency $= k \times$ area.

Worked examples

Frequency densities with rounded data (routine)

The lengths of $80$ fish, measured to the nearest centimetre, are summarised.

Length (cm)	$10$ – $14$	$15$ – $19$	$20$ – $29$	$30$ – $49$
Frequency	$15$	$25$	$30$	$10$

Find the class boundaries and the frequency densities needed to draw a histogram.

Solution

Lengths are rounded to the nearest cm, so each class extends $0.5$ below and above the numbers written.

Length (cm)	Boundaries	Width	Frequency	Frequency density
$10$ – $14$	$9.5$ to $14.5$	$5$	$15$	$3$
$15$ – $19$	$14.5$ to $19.5$	$5$	$25$	$5$
$20$ – $29$	$19.5$ to $29.5$	$10$	$30$	$3$
$30$ – $49$	$29.5$ to $49.5$	$20$	$10$	$0.5$

The bars are drawn from $9.5$ to $49.5$ on the horizontal axis with these heights.

Discrete data in a histogram

The numbers of words in $70$ sentences are grouped as $1$ – $5$ , $6$ – $10$ , $11$ – $20$ and $21$ – $40$ , with frequencies $12$ , $18$ , $24$ and $16$ . Find the frequency densities, and state the modal class.

Solution

Word counts are whole numbers, so the gaps between classes are bridged at the half-way points: boundaries $0.5, 5.5, 10.5, 20.5, 40.5$ , widths $5, 5, 10, 20$ .

Frequency densities: $\tfrac{12}{5} = 2.4$ , $\tfrac{18}{5} = 3.6$ , $\tfrac{24}{10} = 2.4$ , $\tfrac{16}{20} = 0.8$ .

The modal class is $6$ – $10$ words (highest frequency density, $3.6$ ), even though $11$ – $20$ has the largest frequency.

Reading frequencies and part-classes

The masses, $m$ kg, of $120$ suitcases are shown in a histogram. The frequency densities are:

Mass $m$ (kg)	$0 \le m < 10$	$10 \le m < 15$	$15 \le m < 20$	$20 \le m < 30$	$30 \le m < 40$
Frequency density	$1.5$	$6$	$8$	$2.5$	$d$

(a) Find $d$ . (b) Estimate the number of suitcases with mass more than $18\ \text{kg}$ . (c) Estimate the number with mass between $12\ \text{kg}$ and $24\ \text{kg}$ .

Solution

(a) Frequencies are density $\times$ width: $1.5 \times 10 = 15$ , $6 \times 5 = 30$ , $8 \times 5 = 40$ , $2.5 \times 10 = 25$ . These total $110$ , so the last class has $120 - 110 = 10$ suitcases, and $d = \tfrac{10}{10} = 1$ .

(b) From $18$ to $20$ : $8 \times 2 = 16$ . Then all of the last two classes: $25 + 10 = 35$ . Estimate: $16 + 35 = 51$ suitcases.

(c) From $12$ to $15$ : $6 \times 3 = 18$ . From $15$ to $20$ : $40$ . From $20$ to $24$ : $2.5 \times 4 = 10$ . Estimate: $18 + 40 + 10 = 68$ suitcases.

These are estimates because we have assumed the masses are spread evenly within each class.

Bar dimensions on paper (exam-hard)

The masses $m$ grams of some objects are grouped as follows.

Mass $m$ (g)	$0 \le m < 4$	$4 \le m < 6$	$6 \le m < 10$	$10 \le m < 20$
Frequency	$20$	$24$	$16$	$10$

In a histogram drawn on paper, the bar for $0 \le m < 4$ is $2\ \text{cm}$ wide and $5\ \text{cm}$ high. Find the width and height of each of the other bars.

Solution

Widths. A class of width $4\ \text{g}$ is drawn $2\ \text{cm}$ wide, so the scale is $0.5\ \text{cm}$ per gram. The other widths are $2 \times 0.5 = 1\ \text{cm}$ , $4 \times 0.5 = 2\ \text{cm}$ and $10 \times 0.5 = 5\ \text{cm}$ .

Areas. The first bar has area $2 \times 5 = 10\ \text{cm}^2$ for a frequency of $20$ , so $1\ \text{cm}^2$ represents $2$ objects, i.e. area $= \tfrac{1}{2} \times$ frequency.

Class	Frequency	Area ( $\text{cm}^2$ )	Width (cm)	Height (cm)
$4 \le m < 6$	$24$	$12$	$1$	$12$
$6 \le m < 10$	$16$	$8$	$2$	$4$
$10 \le m < 20$	$10$	$5$	$5$	$1$

Check with frequency densities: $5, 12, 4, 1$ per gram. The heights $5, 12, 4, 1\ \text{cm}$ are exactly these, so the vertical scale is $1\ \text{cm}$ per unit of frequency density.

Unknown frequencies from a density condition (exam-hard)

A histogram is drawn for $200$ values grouped as $0 \le x < 10$ (frequency $a$ ), $10 \le x < 20$ (frequency $50$ ), $20 \le x < 40$ (frequency $b$ ) and $40 \le x < 80$ (frequency $30$ ). The bars for the first and third classes have the same height. Find $a$ and $b$ and state the modal class.

Solution

Equal heights means equal frequency densities: $\tfrac{a}{10} = \tfrac{b}{20}$ , so $b = 2a$ .

Total: $a + 50 + 2a + 30 = 200$ , so $3a = 120$ , $a = 40$ and $b = 80$ .

Frequency densities: $4$ , $5$ , $4$ , $0.75$ . The modal class is $10 \le x < 20$ , though the class $20 \le x < 40$ has the largest frequency.

Watch out

Using frequency as bar height with unequal classes. This is the most heavily penalised histogram error; the whole diagram scores almost nothing. Always calculate frequency density, even if the classes look equal (check them; often one is not).

Watch out

Wrong boundaries. Drawing " $10$ – $19$ " from $10$ to $19$ leaves gaps between bars and gives width $9$ . For rounded or discrete data, use $9.5$ to $19.5$ . For ages, use $10$ to $20$ .

Watch out

Labelling the vertical axis "frequency". It must say "frequency density". An unlabelled or mislabelled axis loses a mark even when the bars are right.

Exam tip

Show a table with boundaries, widths and frequency densities. This earns method marks even if the drawing is slightly off.
Choose scales that use most of the graph paper and are easy to plot (e.g. $2\ \text{cm}$ for $10$ units). Awkward scales cause plotting errors.
Bars must be accurate to within half a small square; use a sharp pencil and a ruler.
A histogram's horizontal axis starts at the lowest boundary; do not draw an extra bar or a gap from $0$ unless the data start there.
For any estimate from a histogram (part of a class, the median, the mean), say it is an estimate and why: values are assumed evenly spread within each class.

Summary

Area represents frequency; height is frequency density $=$ frequency $\div$ class width.
Bars run between class boundaries and touch.
Rounded continuous or discrete " $10$ – $19$ " gives boundaries $9.5$ and $19.5$ ; ages " $10$ – $19$ " give $10$ and $20$ ; " $10 \le x < 20$ " gives $10$ and $20$ .
Class width is the difference of the boundaries.
Frequency $=$ density $\times$ width; for part of a class, assume an even spread.
The modal class has the highest frequency density, not necessarily the highest frequency.
When the axis is not labelled in frequency density, frequency $= k \times$ area; find $k$ from a known bar.

Practice questions

Question

Write down the class boundaries and class width for: (a) masses $2.0$ – $2.4\ \text{kg}$ , measured to the nearest $0.1\ \text{kg}$ ; (b) the number of goals, $5$ – $9$ ; (c) ages $30$ – $39$ years; (d) $45 \le t < 60$ .
The heights of $80$ plants, measured to the nearest cm, are grouped as $5$ – $9$ ( $8$ plants), $10$ – $14$ ( $20$ ), $15$ – $24$ ( $32$ ) and $25$ – $44$ ( $20$ ). Calculate the frequency densities.
A histogram has bars for $0 \le x < 5$ (height $2.4$ ), $5 \le x < 10$ (height $4.8$ ), $10 \le x < 20$ (height $3.1$ ) and $20 \le x < 40$ (height $0.65$ ), where the heights are frequency densities. Find the total frequency.
For the histogram in question 3, estimate how many values lie between $8$ and $15$ .
Explain why, in question 3, the modal class is not the class with the largest frequency.
The times of $160$ runners are grouped as $20 \le t < 25$ (frequency $16$ ), $25 \le t < 30$ ( $48$ ), $30 \le t < 40$ ( $64$ ), $40 \le t < 60$ ( $32$ ). In a histogram the bar for $25 \le t < 30$ is $1.5\ \text{cm}$ wide and $6\ \text{cm}$ high. Find the width and height of the bar for $40 \le t < 60$ .
In a histogram of $150$ values, the classes are $0 \le x < 2$ , $2 \le x < 5$ , $5 \le x < 10$ and $10 \le x < 20$ . The first class has frequency $15$ . The second and third classes have the same frequency density, and the fourth class has frequency density $1.5$ . Find the frequencies of the second and third classes.
The numbers of pages in $60$ books are grouped as $100$ – $199$ ( $9$ books), $200$ – $249$ ( $18$ ), $250$ – $299$ ( $21$ ) and $300$ – $499$ ( $12$ ). Calculate the frequency densities, and estimate the number of books with more than $275$ pages.

Answers

(a) $1.95$ and $2.45$ , width $0.5\ \text{kg}$ . (b) Discrete: $4.5$ and $9.5$ , width $5$ . (c) Ages: $30$ and $40$ , width $10$ years. (d) $45$ and $60$ , width $15$ .
Boundaries $4.5, 9.5, 14.5, 24.5, 44.5$ ; widths $5, 5, 10, 20$ . Frequency densities $\tfrac{8}{5} = 1.6$ , $\tfrac{20}{5} = 4$ , $\tfrac{32}{10} = 3.2$ , $\tfrac{20}{20} = 1$ .
Frequencies: $2.4 \times 5 = 12$ , $4.8 \times 5 = 24$ , $3.1 \times 10 = 31$ , $0.65 \times 20 = 13$ . Total $= 80$ .
From $8$ to $10$ : $4.8 \times 2 = 9.6$ . From $10$ to $15$ : $3.1 \times 5 = 15.5$ . Estimate $9.6 + 15.5 = 25.1$ , so about $25$ values.
The class $10 \le x < 20$ has the largest frequency ( $31$ ) but it is twice as wide as $5 \le x < 10$ ; its values are spread over a wider interval. The modal class is the one where values are most concentrated, i.e. the highest frequency density: $5 \le x < 10$ (density $4.8$ ).
Width: $5$ minutes is drawn as $1.5\ \text{cm}$ , so $0.3\ \text{cm}$ per minute; the class $40 \le t < 60$ has width $20 \times 0.3 = 6\ \text{cm}$ . Area: $1.5 \times 6 = 9\ \text{cm}^2$ represents $48$ runners, so $1\ \text{cm}^2$ represents $\tfrac{48}{9} = \tfrac{16}{3}$ runners. $32$ runners need $32 \times \tfrac{3}{16} = 6\ \text{cm}^2$ , so height $= \tfrac{6}{6} = 1\ \text{cm}$ .
Fourth class: $1.5 \times 10 = 15$ . Let the common frequency density of classes 2 and 3 be $d$ ; their widths are $3$ and $5$ , so their frequencies are $3d$ and $5d$ . Then $15 + 3d + 5d + 15 = 150$ , so $8d = 120$ and $d = 15$ . The frequencies are $3 \times 15 = 45$ and $5 \times 15 = 75$ . (Check: $15 + 45 + 75 + 15 = 150$ .)
Pages are discrete: boundaries $99.5, 199.5, 249.5, 299.5, 499.5$ ; widths $100, 50, 50, 200$ . Frequency densities $0.09, 0.36, 0.42, 0.06$ . "More than $275$ pages" means $276$ or more, i.e. from $275.5$ . From $275.5$ to $299.5$ : $0.42 \times 24 = 10.08$ . Plus all $12$ in the last class. Estimate $10.08 + 12 = 22.08$ , about $22$ books.

Why area, not height

Class boundaries

Drawing a histogram

Reading a histogram

Worked examples

Practice questions

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