Histograms

Why histograms use frequency density Extended

A histogram shows grouped continuous data, such as times, lengths or masses. It looks like a bar chart with no gaps, but there is one big difference: in a histogram the area of each bar represents the frequency, not its height.

This matters when the classes have different widths. Suppose 24 people took between 40 and 60 seconds to solve a puzzle and 22 took between 30 and 40 seconds. If both bars were drawn 24 and 22 units tall, the 40 to 60 bar, being twice as wide, would look far bigger than the other, as if many more people were in it. In fact its 24 people are spread over 20 seconds, while the 22 people are packed into just 10 seconds. The data is more concentrated in the 30 to 40 class.

To show this fairly, the height of each bar is the frequency density: the frequency per unit of class width.

$$\text{frequency density}=\text{frequency}÷\text{class width}$$

Then each bar's area is $\text{class width}\times \text{frequency density}=\text{frequency}$, so the area tells the truth about how many values are in each class. On a histogram the vertical axis is labelled Frequency density, never "Frequency".

Calculating frequency density Extended

For each class:

1. Find the class width: upper bound minus lower bound. For $20<t⩽30$ the width is $30-20=10$.
2. Divide the frequency by the class width.

Add two columns to the table, one for class width and one for frequency density, and work across each row. The units of frequency density are "per unit of the variable", for example people per second, though the axis label is simply "Frequency density".

Watch the class widths carefully. They are differences between bounds, not the bounds themselves: dividing 24 by 60 for the class $40<t⩽60$ instead of by 20 is a common slip. Widths can also be decimals: the class $1.5<m⩽1.8$ has width $0.3$.

Drawing a histogram Extended

To draw a histogram from a table:

1. Calculate the frequency density of every class.
2. Draw a continuous scale for the variable on the horizontal axis. Every class bound must be in the right place, so a class of width 20 is twice as wide as a class of width 10.
3. Label the vertical axis Frequency density and choose a scale that fits the largest frequency density.
4. Draw each bar from its lower bound to its upper bound, with height equal to its frequency density. There are no gaps between bars, because the data is continuous.

Use a ruler and check each bar against the table. The commonest errors are a bar of the wrong width (drawn on a class number instead of the scale) and a bar drawn to the height of the frequency.

Reading frequencies from a histogram Extended

Turn the formula round. Since frequency density = frequency $÷$ class width,

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

which is the area of the bar. To find the frequency of a class from a histogram, read the height of its bar on the frequency density axis, find its width from the horizontal scale, and multiply.

Read the vertical scale carefully: each small square is often worth 0.1 or 0.2, not 1. And it is easy to read the height of a bar and give that as the frequency. A frequency density is not a frequency unless the class width happens to be 1.

Estimating within part of a class Extended

Questions often ask how many values lie above or below a value that is inside a class, such as "how many people took more than 50 seconds?" when the classes are $40<t⩽60$ and $60<t⩽100$.

We do not know where in its class each value lies, so we assume the values are spread evenly across the class. Then the number of values in part of a class is in proportion to the width of that part:

$$\text{number}=\text{frequency density}\times \text{width of the part}$$

This is the area of the part of the bar you want. From 50 to 60 is half of the class $40<t⩽60$, so it holds about half of its 24 people, 12. Add the whole classes above: 12 more in $60<t⩽100$. About $12+12=24$ people took more than 50 seconds.

The answer is an estimate, because of the even-spread assumption. It can be a non-integer, and if so leave it as it is unless the question asks you to round.

Completing tables and histograms Extended

Exam questions often give you half the information in the table and half in the histogram, and ask you to complete both. Work class by class:

• a bar is drawn but its frequency is missing: frequency = height $\times$ width;
• a frequency is given but its bar is missing: height = frequency $÷$ width, then draw it.

If one frequency is missing from both and the total is given, subtract the known frequencies from the total first.

Sometimes the vertical axis has no scale and you are told the frequency of one class instead. Use that class to find the scale: its frequency $÷$ its width gives the frequency density its bar represents, and the rest of the axis follows.

The modal class Extended

The modal class is the class where the data is most concentrated. On a histogram with unequal widths, that is the class with the highest frequency density, the tallest bar, which is not necessarily the class with the largest frequency.

In the puzzle data the class $40<t⩽60$ has the largest frequency, 24, but the tallest bar is $30<t⩽40$, with frequency density $2.2$. Its 22 people are packed into 10 seconds, so that is where the times are most concentrated.

When every class has the same width, frequency density is just frequency divided by the same number, so the bars have the same shape as a frequency diagram and the class with the largest frequency is the tallest.