A-level Mathematics

11.3 Diagrams for single-variable data and the histogram — Learn

Prerequisites: reading a table, finding a fraction of a total, and drawing and reading axes at GCSE.

By the end of this section you can: choose a diagram for single-variable data; explain why the area of a histogram bar represents frequency; calculate a frequency density; and use a histogram to find a frequency or a probability.

1. Diagrams for discrete and categorical data

A bar chart shows the frequency of each category. The bars have gaps between them, and the height of each bar is the frequency. A bar chart is used for categories, such as eye colour, or for discrete values, such as the number of siblings.

A pie chart shows each category as a share of a whole. The angle of each sector is

angle=frequencytotal×360∘.\text{angle} = \frac{\text{frequency}}{\text{total}} \times 360^\circ.

A pictogram uses a symbol to stand for a number of items. Each symbol must have a key.

For these diagrams, the frequency is read from the height, the angle or the number of symbols.

2. Grouped continuous data and the histogram

Continuous data, such as time or mass, are grouped into classes. A histogram shows grouped continuous data. Its bars have no gaps, because the variable passes smoothly from one class to the next.

The important difference from a bar chart is this: the area of a bar represents the frequency, not its height. When the classes all have the same width, the heights are proportional to the frequencies, so a histogram looks like a bar chart. When the classes have different widths, the heights alone mislead.

The height of a bar is called the frequency density:

frequency density=frequencyclass width.\text{frequency density} = \frac{\text{frequency}}{\text{class width}}.

So the frequency of a class is

frequency=frequency density×class width.\text{frequency} = \text{frequency density} \times \text{class width}.

3. Completing a frequency table

Example 1. The table shows some grouped data. Complete the frequency density column.

Class Class width Frequency Frequency density
00–1010 10 20 2
1010–2020 10 35 3.5
2020–4040 20 30 1.5

For the first row, 2010=2\dfrac{20}{10} = 2. For the second, 3510=3.5\dfrac{35}{10} = 3.5. For the third, 3020=1.5\dfrac{30}{20} = 1.5.

Notice that the third class has a larger frequency than the first, but a smaller frequency density, because the class is twice as wide.

4. Finding a frequency from a histogram

Example 2. A histogram has frequency density 44 for the class 55–1515. How many values lie between 55 and 1515?

frequency=4×10=40.\text{frequency} = 4 \times 10 = 40.

Example 3. The same histogram has frequency density 44 for the class 55–1515. How many values lie between 55 and 1010?

The part of the class from 55 to 1010 has width 55, so

frequency=4×5=20.\text{frequency} = 4 \times 5 = 20.

The area over part of a bar gives the frequency in that part.

5. Estimating a probability from a histogram

A histogram of a sample can be used to estimate probabilities. The probability that a value lies in a class is the frequency in that class divided by the total frequency.

Example 4. A histogram of 200 values has frequency density 66 in the class 1010–2020. Estimate the probability that a value lies in that class.

The frequency is

6×10=60.6 \times 10 = 60.

So the probability is

60200=310.\frac{60}{200} = \frac{3}{10}.

This is an estimate from the sample, and it connects the histogram to the probability distributions of Chapter 13.

6. Comparing distributions

Two distributions can be compared using their histograms. Describe:

A distribution with a long tail on the right is positively skewed. Waiting times are often positively skewed, because most waits are short and a few are long.

7. Try it: why the area matters

Predict first. The figure draws a histogram of the same 60 values with different class widths. Before you move the control, predict what happens to the bar heights when the class width doubles. Write down whether the total area of all the bars will change.

Test. Change the class width and watch the bars. Read the total frequency each time.

Explain. Write two sentences. The first explains why the bars become shorter when the classes become wider. The second explains why the total area stays the same.

Watch out. Three mistakes are common.

Reading histogram bar heights as frequencies. The height is a frequency density. Multiply it by the class width to get the frequency.

Leaving gaps between histogram bars. A histogram has no gaps, because the variable is continuous. Gaps belong on a bar chart.

Using the frequency as the height when class widths differ. This makes a wide class look far more common than it is. The area, not the height, is the frequency.

Re-check: multiply every frequency density by its class width and add the results. The total must equal the number of values in the data set.

You should now be able to

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