IGCSE Maths · Area 9 · Statistics

Histograms

Histograms are Extended-only content, and almost every mark in them comes from one idea: on a histogram it is the area of a bar, not its height, that stands for the frequency. That is why the vertical axis is labelled 'Frequency density' and why the height is the frequency divided by the class width. This page covers the table, the diagram, reading a frequency from a bar, and the difference between a bar chart and a histogram.

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Syllabus coverage

  • Draw and interpret histograms (Extended only)
  • On a histogram the vertical axis is labelled 'Frequency density'
  • Frequency density is defined as frequency density = frequency \div class width
  • Calculate with frequency density, including finding a frequency from the area of a bar
  • Understand why frequency density, and not frequency, is plotted when the class widths are not all equal

Part of the IGCSE Maths syllabus 0580.

Histogram of the masses of 72 apples, with the frequency table it was drawn from0204060801001401800.00.20.40.60.81.0Mass (g)Frequency densityMass (g)FrequencyClass widthFrequencydensity0 to 204200.220 to 4010200.540 to 6018200.960 to 8016200.880 to 1008200.4100 to 14012400.3140 to 1804400.1Bar height is frequency density, so the two wide classes are short bars even though 12 and 4 apples fall in them.
A histogram of the masses of 72 apples beside the frequency table it was drawn from.

Why a histogram is not a bar chart

A bar chart shows how often each category occurs. The categories are separate - types of pet, days of the week, colours of car - so the bars are drawn with gaps between them, and the height of a bar is the frequency. Nothing about a bar chart depends on width, so every bar is the same width whatever the numbers.

A histogram shows continuous data that has been grouped into classes, such as masses, times or lengths. There are no gaps between the classes and no gaps between the bars: the bar for 20<m≤40 begins exactly where the bar for 0<m≤20 ends. That is why the bars of a histogram touch.

The consequence is that the width of a bar carries information. On a histogram it is the area of a bar, not its height, that stands for the frequency, and the height is a new quantity called frequency density. Once that idea is in place, the rest of the topic is arithmetic.

FeatureBar chartHistogram
DataDiscrete categories: types, days, namesContinuous grouped data: mass, time, length
Gaps between the barsYes, alwaysNo, the bars touch
What the height showsThe frequencyThe frequency density
What the area showsNothing in particularThe frequency
The widths of the barsAll the sameThe class width of each group
The favourite one-mark question in this topic is 'write down one difference between a bar chart and a histogram'. 'The bars touch in a histogram' and 'the area of a bar represents the frequency' are the two answers to have ready.

Frequency density: why the height alone will not do

If every class had the same width, the tallest bar would simply be the class with the most values and a histogram would be no different from a bar chart. The problem appears as soon as the widths differ. Suppose 12 apples weigh between 100 g and 140 g and 18 apples weigh between 40 g and 60 g. The second class holds more apples but covers only half as much mass, so plotting the frequencies as heights would make the second class look far more important than it is.

The fix is to divide by the width. Frequency density = frequency \div class width. It answers the question 'how many values per unit of the variable', so a wide class and a narrow class can be compared fairly. The class 40<m≤60 has density 1820=0.9 and the class 100<m≤140 has density 1240=0.3, so the first bar is three times as tall even though the two frequencies, 18 and 12, are close together.

The units of frequency density are 'values per gram', 'values per minute' and so on, which is why the numbers on the axis are usually small decimals. The syllabus's own definition is frequency density = frequency \div class width, and the vertical axis of a histogram is always labelled Frequency density, never just 'frequency'.

  • 40<m≤60: frequency 18, class width 20, frequency density 1820=0.9
  • 100<m≤140: frequency 12, class width 40, frequency density 1240=0.3
  • Rearranged: frequency = frequency density \times class width
  • Rearranged: class width = frequency \div frequency density
Frequency density is not a frequency, so it can be smaller than 1 and it is not a count of anything. If you work out a frequency density larger than the total frequency in the question, you have divided the wrong way round.

Drawing a histogram from a frequency table

Add two columns to the table before drawing anything: one for the class width and one for the frequency density. The class width is the difference between the upper and lower boundaries - for 40<m≤60 that is 60−40=20. The frequency density is the frequency divided by that width.

Draw each bar from one class boundary to the next, so consecutive bars share an edge, and label the vertical axis 'Frequency density' with a scale on it. Choose a scale that uses most of the grid: for densities up to 0.9, going up in steps of 0.1 to 1.0 works well. The horizontal axis carries the variable itself - mass in grams, time in minutes - with the class boundaries marked on it.

There is no rule that the bars must be the same width, and no rule that the tallest bar is the biggest class. Before you leave the diagram, check it against the table: the tallest bar must be the class with the largest frequency density, which is not always the class with the largest frequency.

Mass m (g)FrequencyClass widthFrequency density
0<m≤20420420=0.2
20<m≤4010201020=0.5
40<m≤6018201820=0.9
60<m≤8016201620=0.8
80<m≤100820820=0.4
100<m≤14012401240=0.3
140<m≤180440440=0.1
Total72
Look at the last two classes. They are twice as wide as the rest, so their bars are short: the class 100<m≤140 holds 12 apples but its bar is only 0.3 high, shorter than the bar for 80<m≤100, which holds only 8. That is frequency density doing its job.

Finding a frequency from the area of a bar

The area of a bar is its height multiplied by its width, and that area is the frequency of the class. So frequency = frequency density \times class width. If the bar for 80<m≤100 is 0.4 high, the class holds 0.4×20=8 apples.

This is how a histogram is read when the table has not been given. Read the height of the bar from the frequency density scale, multiply by the width of the class, and you have the frequency. The same rule works on part of a class: a bar 0.6 high covering only 10 g of a 20 g class stands for 0.6×10=6 values.

Work the other way round when the question gives a frequency and a bar height and asks for a width. A bar of height 0.5 standing for 15 values must cover 15÷0.5=30 units of the variable, so its class is 30 wide.

  • Frequency density 0.4, class width 20: frequency =0.4×20=8
  • Frequency density 0.6, class width 10: frequency =0.6×10=6
  • Frequency 15, frequency density 0.5: class width =15÷0.5=30
  • The total frequency is the total area of all the bars added together
The total area under a histogram is the total frequency. If the areas of your bars do not add up to the number of values in the question, at least one of your frequency densities is wrong.

Reading the frequency density scale

The vertical axis is a density scale, so its numbers are small: for the apple data the tallest bar reaches 0.9. Read the height against the gridlines rather than guessing, and take care when a bar top falls between two of them. On a scale going up in 0.1s, a bar top halfway between 0.3 and 0.4 is 0.35.

Questions often ask for the frequency of a class whose bar you have to measure. Read the height, multiply by the class width, and round only at the end. If the height reads 0.35 and the class is 40 g wide, the frequency is 0.35×40=14.

When two histograms are drawn for the same variable with different class widths, the bars cannot be compared by height at all. Merging two classes makes one wider bar whose height is the total frequency divided by the new, wider class width, so the shape of the diagram changes even though the data has not.

Read the height of a bar against the frequency density scale, never against the frequency of the class. A bar 0.35 high over a 40 g class is 14 apples, not 0.35 of anything.

Unequal class widths, and working backwards

Grouped data is usually tabulated with equal widths because that is easier to collect, but there is no requirement for it. When the last class is much wider than the others, or when two small classes have been merged into one, the widths differ and frequency density does the work.

Questions often give you enough to work backwards. If you know the total frequency, all the class widths and all but one of the frequencies, subtract to find the missing frequency and then divide by the width to get its density. If you know a bar height and a frequency, divide to get the width: a bar 1.5 high standing for 45 values covers 45÷1.5=30 units.

Never plot the frequency on the vertical axis when the widths are unequal. The bars would be the wrong shape and a wide class would look far too important. Even when all the widths happen to be equal, the axis is still labelled 'Frequency density', because that is what the numbers on it mean.

  • Equal widths: frequency density is proportional to frequency, so the histogram looks like a bar chart with no gaps
  • Unequal widths: plot frequency density, and check each bar's area against its frequency
  • Missing frequency: the total minus all the other frequencies
  • Missing class width: frequency \div frequency density
A histogram with unequal class widths and the frequency plotted on the vertical axis is a wrong answer however neat it looks. The axis label is part of the answer, so write it before you draw the bars.

Key terms you must be able to define

Histogram
A diagram for continuous grouped data in which the bars have no gaps and the area of each bar represents the frequency of its class.
Frequency density
Frequency divided by class width. It is the height of a bar on a histogram, and the syllabus defines it as frequency density = frequency \div class width.
Class width
The difference between the upper and lower boundaries of a class. For 40<m≤60 the class width is 60−40=20.
Class boundary
The end values of a class, such as 40 and 60 in 40<m≤60. The bars of a histogram are drawn between the class boundaries.
Continuous data
Data that can take any value in a range, such as mass, time or length, so it is grouped into classes rather than counted in categories.
Discrete data
Data that takes particular separate values, such as the number of pets in a household, usually shown on a bar chart.
Bar chart
A diagram for discrete categories in which the bars are the same width and separated by gaps, and the height of a bar is the frequency.
Area of a bar
Height multiplied by width on a histogram, which equals the frequency of the class.
Frequency
The number of values in a class. On a histogram it is found as frequency density \times class width.
Grouped frequency table
A table showing classes of continuous data and the frequency of each class.
Class interval
Another name for a class: the range of values it covers, written as 40<m≤60.
Unequal class widths
Classes of different widths in one table. They are the reason frequency density, and not frequency, is plotted on a histogram.
Vertical axis
The axis of a histogram, always labelled 'Frequency density'.
Modal class
The class with the greatest frequency. On a histogram with unequal class widths it is not necessarily the tallest bar.

In context

Cambridge sets 0580 in real situations - a recipe, a bank statement, a map, a garden, a graph of a car's journey. These show how the mathematics appears in a question, then you write your own working for the numbers you are given.

Working out the frequency densities

For the 72 apples, divide each frequency by its class width: 420=0.2, 1020=0.5, 1820=0.9, 1620=0.8, 820=0.4, 1240=0.3 and 440=0.1. The tallest bar is the class 40<m≤60 at 0.9.

Frequency from the area of a bar

A histogram of the 72 apples has a bar of height 0.4 over the class 80<m≤100. The class width is 20, so the class holds 0.4×20=8 apples. Checking the other way, 8÷20=0.4, which agrees.

The tallest bar is not the biggest class

The class 100<m≤140 holds 12 apples and the class 80<m≤100 holds 8, but the first bar is 0.3 high and the second is 0.4. The first class is 40 g wide and the second is only 20 g wide, and on a histogram it is the area that matters.

Merging two classes

Merging 0<m≤20, with 4 apples, and 20<m≤40, with 10 apples, gives the class 0<m≤40 with 14 apples and a width of 40. Its frequency density is 1440=0.35, between the original heights of 0.2 and 0.5, and the area 0.35×40=14 is unchanged.

Working backwards to a width

A bar on a histogram is 1.5 high and stands for 45 values. Its class width is 45÷1.5=30. Working backwards always means dividing the frequency by the frequency density, never multiplying.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What is plotted on the vertical axis of a histogram?
Frequency density. Never the frequency, unless the classes have all been stated to be the same width.
Q2How is frequency density calculated?
Frequency density = frequency / class width.
Q3How do you find a frequency from a histogram?
Find the area of the bar: frequency density x class width.
Q4How do you find a class width from a histogram?
Frequency / frequency density.
Q5Why are the bars of a histogram drawn touching?
The data is continuous and the classes meet, so there is no gap between one class boundary and the next.
Q6Give one difference between a bar chart and a histogram.
The bars of a bar chart have gaps and their height is the frequency; the bars of a histogram touch and their area is the frequency.
Q7What does the area of a bar on a histogram represent?
The frequency of that class.
Q8What does the total area of all the bars represent?
The total frequency.
Q9Why is frequency density used instead of frequency?
So that classes of different widths can be compared fairly: a wide class would look far too important if its frequency were plotted as a height.
Q10The class 40 < m <= 60 has frequency 18. What is its frequency density?
18 / 20 = 0.9. The class width is 60 - 40 = 20.
Q11The class 100 < m <= 140 has frequency 12. What is its frequency density?
12 / 40 = 0.3. The width is 40, not 100 and not 140.
Q12A histogram bar is 1.5 high and stands for 45 values. What is the class width?
45 / 1.5 = 30.
Q13Is the tallest bar on a histogram always the class with the greatest frequency?
No. It is the class with the greatest frequency density, and with unequal widths that class may hold fewer values than a wider one.
Q14Two classes, 0 < m <= 20 with frequency 4 and 20 < m <= 40 with frequency 10, are merged. What is the new frequency density?
Frequency 14 over a width of 40, so 14 / 40 = 0.35.
0 of 14 marked as known

Exam-style questions

7 questions, 17 marks. Write your answer, then reveal the mark scheme and tick each point you actually made. Answering honestly here is what makes the rest of the site useful.

Write downAO12 marks
The masses of 72 apples are recorded in the table: 0<m≤20: 4, 20<m≤40: 10, 40<m≤60: 18, 60<m≤80: 16, 80<m≤100: 8, 100<m≤140: 12 and 140<m≤180: 4, where m is in grams. Write down the class width of the class 100<m≤140 and the frequency density of the class 40<m≤60.
Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

  • B1class width 140−100=40not checked yet
  • B1frequency density 1820=0.9not checked yet
Examiner tip: The class width is the difference between the two boundaries, not the number of values in the class. The frequency density is the frequency divided by that width.
CalculateAO12 marks
A histogram is drawn for the 72 apples. The bar for the class 140<m≤180 is 0.1 high. Calculate the frequency that this bar represents.
Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

  • M1
  • A14 applesnot checked yet
Examiner tip: The frequency is the area of the bar, so multiply the height by the class width. The class width here is 40, not 140.
Work outAO23 marks
A second sample of apples is grouped as 0<m≤20 with frequency 5, 20<m≤60 with frequency 12 and 60<m≤100 with frequency 3. Work out the frequency density of each class.
Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

  • M1
  • A1
  • A1340=0.075not checked yet
Examiner tip: The second and third classes are 40 wide, so their densities come out far smaller than their frequencies suggest. Divide each frequency by its own width before comparing anything.
Work outAO23 marks
A histogram is drawn for the speeds of 200 cars. The class 0<s≤30 has frequency density 0.4 and the class 30<s≤50 has frequency density 1.2. Work out the total number of cars in these two classes.
Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

  • M1
  • M1
  • A112+24=36 carsnot checked yet
Examiner tip: Each frequency is the area of its own bar. The two classes are 30 and 20 wide, so you cannot compare the two heights directly or add them together.
ExplainAO12 marks
Explain why the vertical axis of a histogram is labelled 'Frequency density' and not 'Frequency'.

Mark scheme — tick each point you actually made

  • B1
  • B1
Examiner tip: The reason is about unequal widths. An answer that only says 'because the axis shows density' has restated the question and scores nothing.
StateAO11 mark
On a histogram, a bar of height 2.5 covers the class 10<x≤30. State the frequency of this class.
Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

  • B12.5×20=50not checked yet
Examiner tip: The class width is 30−10=20, and the frequency is the area of the bar.
Work outAO24 marks
A histogram shows the ages, in years, of 45 people. The bar for 0<a≤10 has frequency density 0.3 and the bar for 10<a≤20 has frequency density 0.6. Every other person is in the class 20<a≤40. Work out the frequency density of the bar for 20<a≤40.
Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

  • M1
  • M1
  • M1
  • A13620=1.8not checked yet
Examiner tip: Work out the two known frequencies from the areas first, subtract them from the total, then divide what is left by the class width of 20.

Exam technique for this topic

  • Write the class width column before the frequency density column, and write the division down. 'Frequency density = 18 \div 20 = 0.9' earns more than a bare 0.9, and it is what the mark scheme is looking for.
  • Label the vertical axis 'Frequency density' every time. An unlabelled axis on a histogram is a lost mark even when every bar is the right height.
  • Check your frequency densities against the frequencies. If the class with the biggest frequency does not have one of the biggest densities, look at the widths again: a wide class pulls its density down.
  • Decide what the question wants before you calculate. A frequency is the area of a bar, a frequency density is frequency \div class width, and a class width is frequency \div frequency density.
  • Read heights against the frequency density scale and use the gridlines. A bar top halfway between 0.6 and 0.7 is 0.65, not 6.5 and not 0.6.

Why students lose marks here

A student wroteThe vertical axis is frequency, so the bar for 40<m≤60 goes up to 18.
Why it loses marks18 is the frequency, but the axis of a histogram carries frequency density. The bar goes up to 1820=0.9, and it is the area 0.9×20=18 that stands for the frequency.
The answer that scoresPlot the frequency density 0.9 and label the vertical axis 'Frequency density'.
A student wroteThe bar for 100<m≤140 is three times the height of the bar for 0<m≤20, because 12 is three times 4.
Why it loses marksComparing frequencies when the class widths differ. The densities are 1240=0.3 and 420=0.2, so the first bar is taller, but only by a factor of 1.5.
The answer that scoresCompare frequency densities, not frequencies: 0.3 against 0.2.
A student wroteI left small gaps between the bars so that the classes are easier to see.
Why it loses marksThe bars of a histogram touch, because the data is continuous and the classes meet at the boundaries. Gaps turn the diagram into a bar chart, and the areas no longer represent the frequencies.
The answer that scoresDraw each bar from one class boundary to the next, so that consecutive bars share an edge.
A student wroteA bar is 0.5 high over a class 20 wide, so the class has frequency 0.5.
Why it loses marksThe height is the frequency density, not the frequency. The frequency is the area of the bar: 0.5×20=10.
The answer that scoresFrequency =0.5×20=10.
A student wroteThe total area of the bars is 1, because the frequencies add up to the whole sample.
Why it loses marksThe total area is the total frequency, which is 72 for the apples. Frequencies are counts, and they only add to 1 if the question has asked for proportions.
The answer that scoresThe total area of the bars is the total frequency: 4+10+18+16+8+12+4=72.