What you need to know

When displaying grouped data, especially continuous data, a histogram is often the best way to do it – specifically in cases where not all the groups/classes are the same width, and by “width” we mean the upper bound of the class takeaway the lower bound. Histograms are like bar charts with 2 key differences:

  • There are no gaps between the bars, and
  • It’s the area (as opposed to the height) of each bar that tells you the frequency of that class.

In order to make this work, when drawing a histogram, we plot frequency density on the y-axis rather than frequency. The frequency density for each group is found using the formula:

\text{frequency density} = \dfrac{\text{frequency}}{\text{class width}}

In this topic we will at both drawing and interpreting histograms.

Example: Below is a grouped frequency table of the lengths of 71 pieces of string. Construct a histogram of the data.

To construct a histogram, we will need the frequency density for each class. Dividing the frequency of the first class by its width, we get

 

\text{frequency density } =\dfrac{8}{20-0} = 0.4

 

Then, for the second class we get

 

\text{frequency density }=\dfrac{13}{30-20}=1.3

 

Once we have calculated the frequency density with the remaining groups, then it is good to add a third column to the table containing the frequency density values, see: below.

Once this new column is completed, all that remains is to plot the histogram.

 

With lengths on the x-axis frequency density on the y-axis, each bar that we draw will have width equal to its class width, and height equal to the relevant frequency density.

 

The resulting histogram is shown below

The other part of this topic is interpreting histograms when they’re given to you pre-drawn. You might see them without the frequency density axis labelled, like the example below.

 

Example: Below is a histogram times taken to complete a quiz. 44 people took between 0 and 1.5 minutes. Work out how many people took between 3 and 4 minutes.

It’s imperative here to remember that it’s the area that matters. So, to answer this question, we’re going to use the information to work out how much 1 small square of area is worth.

 

Between 0 and 1.5 minutes includes all of the first bar and some of second. From 0 to 1 minutes there are 10\times 12 =120 small squares, and from 1 to 1.5 there are 5\times 20=100 small squares (marked on the graph below for clarity).

So, in total there are 100+120=220 small squares between 0 and 1.5 minutes, and the question tells us that this accounts for 44 people. Therefore, 1 person is equal to

 

220 \div 44=5\text{ small squares}.

 

Now, reading from the graph we get that there are 11 \times 10 = 110 small squares between 3 and 4 minutes, so given that 5 small squares is one person, there must be

 

110 \div 5 = 22

 

who took between 3 and 4 minutes to do the quiz. Note: as 110 is half of 220, the answer must be half of 44 – this is a nice time-saver answer, but it won’t always be that nice unfortunately.

Example Questions

1) Below is a grouped frequency table of the heights of plants growing in a garden. Construct a histogram of the data.

 

Answer

We must calculate the frequency density for each class by dividing each frequency by its associated class width. Writing the frequency density in its own column, we get

 

 

Now we need to plot the histogram with height on the x-axis and frequency density on the y-axis. The result should look like

 

2) Below is a histogram of data on how long people could hold their breath. There were 54 people who could hold it for at least 1 minute. Work out how many could hold it for between 20 and 40 seconds.

 

Answer

We want to see how many people one small square of area is worth.

The question tells us that there were 54 people who could hold their breath for 1 minute or more, so that means 54 people account for the whole of the last bar and part of the second-to-last bar. The number of small squares after 60 seconds is

(5 \times 15) + (15 \times 4) = 75 + 60 = 135

As seen on the graph below if you’re unsure.

 

Then, if this area constitutes 54 people, we must have that

 

\text{1 person } = \dfrac{135}{54} = 2.5\text{ small squares}

 

The number of small squares between 20 and 40 is

 

(5 \times 32) + (5 \times 20) = 160 + 100 = 260

 

Also shown on the graph below. Therefore, the number of people who held their breath for between 20 and 40 seconds is

 

\dfrac{260}{2.5} = 104\text{ people}.

 

Histograms and Frequency density questions are a large topic in the GCSE Maths syllabus. Whether you are a GCSE Maths tutor in Harrogate or you teach Maths in another area of the country, the Histogram questions and Frequency density revision materials will be of use in lessons and are suitable to set as homework. All the Histogram questions are appropriate for AQA, OCR, WJEC and Edexcel as well as all other major exam boards.