Histograms & Frequency Polygons (Edexcel GCSE Statistics)

Revision Note

Frequency Polygons

What is a frequency polygon?

  • Frequency polygons are a very simple way of showing frequencies for continuous, grouped data

    • They give a quick guide to how frequencies change from one class interval to the next

What are the key features of a frequency polygon?

  • Apart from plotting and joining up points with straight lines there are 3 rules for frequency polygons:

    • Plot points at the midpoint of class intervals

    • Unless one of the frequencies is 0 do not join the frequency polygon to the x-axis

    • Do not join the first point to the last one

  • The result is not actually a polygon

    • It's more of an 'open' polygon that ‘floats’ in mid-air!

  • You may be asked to draw a frequency polygon and/or use it to make comments and compare data

How do I draw a frequency polygon?

  • This is most easily shown by an example

    • e.g.  The lengths of 59 songs, in seconds, are recorded in the table below

Song length
t seconds

Frequency

120 ≤ t < 150

4

150 ≤ t < 180

10

180 ≤ t < 210

24

210 ≤ t < 240

18

240 ≤ t < 270

3

  • Frequencies are plotted at the midpoints of the class intervals, so in this case we would plot the points (135, 4), (165, 10), (195, 24), (225, 18) and (255, 3).
    Join these up with straight lines (but do not join the last to the first!)

Song Length FP, IGCSE & GCSE Maths revision notes
  • If you have a histogram with equal class widths

    • then a frequency polygon can be drawn

      • by marking the points at the centres of the tops of all the histogram bars

      • and joining those points together to make the 'polygon'

How do I use and interpret a frequency polygon?

  • Think about what you could you say about the data above, particularly by looking at the diagram only

    • The two things to look for are averages and spread

      • The modal class is 180 ≤ t < 210

        • Because the graph reaches its highest point there

        • It would be acceptable to say that 195 seconds is (an estimate of) the modal song length

      • The diagram (rather than the table) shows (an estimate of) the range of song lengths is 255 – 135 = 120 seconds

        • i.e. midpoint of highest class interval minus midpoint of lowest class interval

    • If 2 frequency polygons are drawn on the same graph comparisons between the 2 sets of data can be made

Examiner Tips and Tricks

  • Jot down the midpoints next to the frequencies so you are not trying to work them out in your head while also concentrating on actually plotting the points

Worked Example

A local council ran a campaign to encourage households to waste less food.

To compare the impact of the campaign the council recorded the weight of food waste produced by 30 households in a week both before and after the campaign.

The results are shown in the table below.

Food waste
w kg

Frequency
(before campaign)

Frequency
(after campaign)

1 ≤ w < 1.4

3

5

1.4 ≤ w < 1.8

4

8

1.8 ≤ w < 2.2

8

14

2.2 ≤ w < 2.6

10

3

2.6 ≤ w < 3

5

1

(a) On the same diagram, draw two frequency polygons, one for before the council’s campaign and one for after.

Remember to include a key to show which frequency polygon is which.

Food-Waste-FP, downloadable IGCSE & GCSE Maths revision notes

(b) Comment on whether you think the council’s campaign has been successful or not and give a reason why.

Remember to look for average(s) and/or spread
The mode (average) is appropriate in this case.

The council campaign has been successful as the modal amount of waste has decreased from 2.4 kg of food waste per week to 2 kg

Frequency Density

What is frequency density?

  • Frequency density is given by the formula

    • frequency space density equals fraction numerator frequency over denominator class space width end fraction

      • This formula is not on the exam formula sheet, so you need to remember it

  • Frequency density is used with grouped data (i.e. data grouped by class intervals)

    • It is useful when the class intervals are of unequal width

    • It provides a measure of how dense data is within its class interval

      • relative to the width of the interval

    • For example,

      • 10 data values spread over a class interval of width 20 would have a frequency density of 10 over 20 equals 1 half

      • 20 data values spread over a class interval of width 100 would have a frequency density of 20 over 100 equals 1 fifth

      • As 1 half greater than 1 fifth 

        • the data in the first interval is more dense (closer together) than in the second interval

        • even though the second interval has twice as many data values

How do I calculate frequency density?

  • In questions it is usual to be presented with grouped data in a table

  • So add two extra columns to the table

    • one to work out and write down the class width of each interval

    • the second to then work out the frequency density for each group (row)

Worked Example

The table below shows information regarding the average speeds travelled by trains in a region of the UK.

The data is to be plotted on a histogram.

Average speed
s m/s

Frequency

20 less or equal than s less than 40

5

40 less or equal than s less than 50

15

50 less or equal than s less than 55

28

55 less or equal than s less than 60

38

60 less or equal than s less than 70

14

Work out the frequency density for each class interval.

Add two columns to the table - one for class width, one for frequency density
Use  frequency space density equals fraction numerator frequency over denominator class space width end fraction
Writing the calculation in each box helps to maintain accuracy

Average speed
s m/s

Frequency

Class width

Frequency density

20 less or equal than s less than 40

5

40 - 20 = 20

5 ÷ 20 = 0.25

40 less or equal than s less than 50

15

50 - 40 = 10

15 ÷ 10 = 1.5

50 less or equal than s less than 55

28

55 - 50 = 5

28 ÷ 5 = 5.6

55 less or equal than s less than 60

38

60 - 55 = 5

38 ÷ 5 = 7.6

60 less or equal than s less than 70

14

70 - 60 = 10

14 ÷ 10 = 1.4

Histograms

What is a histogram?

  • A histogram looks similar to a bar chart, but there are important differences

  • Bar charts are used for discrete (and possibly non-numerical) data

    • In a bar chart, the height (or length) of a bar determines the frequency

    • There are usually gaps between the bars

  • Histograms are used with continuous data, grouped into class intervals (usually of unequal width)

    • In a histogram, the area of a bar determines the frequency

      • This means it is difficult to tell anything simply from looking at a histogram

      • Some basic calculations will be needed for conclusions and comparisons to be made

    • There are no gaps between the bars

How do I draw a histogram?

  • Drawing a histogram first requires the calculation of the frequency densities for each class interval (group)

    • Use  table row cell frequency space density end cell equals cell fraction numerator frequency over denominator class space width end fraction end cell end table

    • Exam questions often ask you to finish an incomplete histogram, rather than start with a blank graph

  • Once the frequency densities are known, the bars (rectangles) for each class interval can be drawn

    • with widths being measured on the horizontal (x) axis

    • and the height of each bar (the frequency density) being measured on the vertical (y) axis

    • As the data is continuous, the bars will be touching

How do I interpret a histogram?

  • It is important to remember that the frequency density (y-) axis does not tell us frequency

    • The area of the bar is equal to the frequency

      • table row frequency equals cell frequency space density cross times class space width end cell end table

  • Note that a very simple histogram may have equal class widths

    • In this case the y-axis may be labelled 'frequency' instead of 'frequency density'

    • If 'frequency' is on the y-axis, then you can use the heights of the bars to directly determine the frequencies

  • You may be asked to estimate the frequency of part of a bar/class interval within a histogram

    • Find the area of the bar for the part of the interval required

    • Once area is known, frequency can be found as above

  • You can use histograms for two data sets to compare the data distributions

    • but only if they have the same class intervals and the same frequency density scales

Examiner Tips and Tricks

  • Always work out and write down the frequency densities

    • It is easy to make errors and lose marks by going straight to the graph

    • Method marks may depend on showing you know to use frequency density rather than frequency

  • The frequency density axis will not always be labelled

    • Look carefully at the scale, it is unlikely to be 1 unit to 1 square

Worked Example

A histogram is shown below representing the distances achieved by some athletes throwing a javelin.

Histogram Question Bars 1, IGCSE & GCSE Maths revision notes

There are two classes missing from the histogram.  These are:

Distance, x m

Frequency

60 less or equal than x less than 70

8

80 less or equal than x less than 100

2

 Add these to the histogram.

Before completing the histogram, remember to show clearly you've worked out the missing frequency densities

Distance, x m

Frequency

Class width

Frequency density

60 less or equal than x less than 70

8

70 - 60 = 10

8 ÷ 10 = 0.8

80 less or equal than x less than 100

2

100 - 80 = 20

2 ÷ 20 = 0.1

Now the bars can be drawn on the histogram
They should stretch along the x-axis from the start to the end of the class interval
The heights will be equal to the frequency densities

Histogram Question Bars 1, IGCSE & GCSE Maths revision notes

Worked Example

The table below and the corresponding histogram show the weight, in kg, of some newborn bottlenose dolphins.

Weight
w kg

Frequency

4 ≤ w < 8

4

8 ≤ w < 10

16

10 ≤ w < 12

19

12 ≤ w < 15

12

15 ≤ w < 30

A partially complete histogram for the data in the question

(a) Use the histogram to complete the table.

The frequency for the 15 ≤ w < 30 class interval is missing

The bar for that class interval on the histogram has a

  • height (frequency density) of 0.6

  • width of 30-15 = 15

Rearrange  table row cell frequency space density end cell equals cell fraction numerator frequency over denominator class space width end fraction end cell end table  to get

table row frequency equals cell frequency space density cross times class space width end cell end table

table row frequency equals cell 0.6 cross times 15 equals 9 end cell end table

Weight
w kg

Frequency

4 ≤ w < 8

4

8 ≤ w < 10

16

10 ≤ w < 12

19

12 ≤ w < 15

12

15 ≤ w < 30

9

(b) Use the table to complete the histogram.

The bar for the 4 ≤ w < 8 class interval is missing

That class interval has a

  • frequency of 16

  • width of 10-8 = 2

Use  table row cell frequency space density end cell equals cell fraction numerator frequency over denominator class space width end fraction end cell end table  to find the frequency density

table row cell frequency space density end cell equals cell 16 over 2 equals 8 end cell end table

Draw a bar with that height on the histogram, between 8 and 10 on the horizontal axis

A completed histogram for the data in the question

(c) Estimate the number of dolphins whose weight is greater than 13 kg.

We know from part (a) that there are 9 dolphins in the 15 ≤ w < 30 class interval
So we need to estimate the number of dolphins that are in the interval 13 ≤ w < 15

For 13 ≤ w < 15, the histogram shows that

  • the frequency density is 4

  • the width is 15-13 = 2

The histogram from the question with the bar between 13 and 15 highlighted

Now use  table row frequency equals cell frequency space density cross times class space width end cell end table  to estimate the number of dolphins in the 13 ≤ w < 15 interval
(Note that using the histogram in this way is actually a form of linear interpolation)

frequency equals 4 cross times 2 equals 8

This is only an estimate because we don't actually know that dolphins are evenly distributed across the entire 12 ≤ w < 15 class interval

Now the total number of dolphins with a weight greater than 13 kg can be estimated

8 plus 9 equals 17

There are approximately 17 dolphins with a weight greater than 13 kg

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Roger B

Author: Roger B

Expertise: Maths

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.