Representations of Data

What you'll learn

How to turn grouped data into correct class widths.
How histograms use area, not just height, to represent frequency.
How to find quartiles, outliers and draw box plots.
How to compare distributions using centre, spread and outliers.

1. Grouped data and class widths

In statistics, data are often collected in groups called classes or class intervals, such as 10–14 miles or 20–24 minutes.

The frequency is the number of observations in a class.

For histograms, the width of a class matters. If data are measured to the nearest whole number, the true class boundaries go halfway between the whole numbers.

For example, a class written as 10–14 actually covers values from 9.5 up to 14.5, so its class width is 5.

A number line showing how the rounded class 10–14 corresponds to true boundaries 9.5 and 14.5 with class width 5.

Definition

Class width

The class width is the size of the interval covered by a class, using the true class boundaries if the data have been rounded.

Example

Finding true class widths

A variable is measured to the nearest whole number. Two classes are 6–10 and 11–18. Find their true class widths.

The class 6–10 includes values that round to 6, 7, 8, 9 or 10, so its true boundaries are 5.5 and 10.5.
Find the width of the first class:
$10.5 - 5.5 = 5$
The class 11–18 has true boundaries 10.5 and 18.5.
Find the width of the second class:
$18.5 - 10.5 = 8$

Common Mistake

Counting the labels instead of the interval

For a class like 14–20 measured to the nearest whole number, the class width is 7, not 6. Use boundaries 13.5 and 20.5.

2. Histograms: area represents frequency

A histogram is used for grouped continuous data. Unlike a bar chart, the bars can have different widths, and the key idea is:

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

So:

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

This is why the area of each bar represents the frequency.

Two histogram bars with equal area but different widths and heights, illustrating that frequency is represented by area.

Here is the main idea visually: wider bars need a smaller height if they are to represent the same frequency.

Annotated histogram showing frequency as area

Key Idea

Histogram rule

In a histogram, compare areas, not just heights. A tall narrow bar and a short wide bar can represent the same frequency.

Example

Completing a missing frequency from a histogram

A histogram shows completion times, measured to the nearest second. The class 6–10 has frequency 20 and its bar is 5 grid squares high. The class 11–15 has a bar 3 grid squares high. Find the frequency for 11–15.

The given histogram setup for using one known bar to find the missing frequency of another equal-width class.

The class 6–10 has true width 5. Its frequency density is:
$205=4\frac{20}{5} = 4$
A height of 5 grid squares represents a frequency density of 4, so one grid square represents:
$45=0.8\frac{4}{5} = 0.8$
The bar for 11–15 is 3 grid squares high, so its frequency density is:
$\times 0.8 = 2.4$
The class 11–15 also has width 5, so its frequency is:
$2.4 \times 5 = 12$

3. Finding bar widths and heights from one given bar

Sometimes you are told the actual width and height of one bar on a drawn histogram. Use it to find the horizontal and vertical scales.

The horizontal scale links class width to centimetres.
The vertical scale links frequency density to centimetres.

Example

Finding the width and height of another histogram bar

A variable is measured to the nearest whole number. The class 9–13 has frequency 12 and is drawn with width 2 cm and height 3 cm. Find the width and height of the bar for class 14–20, which has frequency 21.

A scale diagram showing how one histogram bar fixes separate horizontal and vertical scales for another bar.

Find the true width of 9–13. Its boundaries are 8.5 and 13.5:
$13.5 - 8.5 = 5$
Since width 5 is drawn as 2 cm, the horizontal scale is:
$unit\frac{2}{5} = 0.4 \text{ cm per unit}$
The class 14–20 has boundaries 13.5 and 20.5, so its true width is:
$20.5 - 13.5 = 7$
Therefore its drawn width is:
$\times 0.4 = 2.8 \text{ cm}$
Now find the frequency density of 9–13:
$125=2.4\frac{12}{5} = 2.4$
A frequency density of 2.4 is drawn as height 3 cm, so the vertical scale is:
$unit\frac{3}{2.4} = 1.25 \text{ cm per frequency-density unit}$
The frequency density of 14–20 is:
$217=3\frac{21}{7} = 3$
So its drawn height is:
$\times 1.25 = 3.75 \text{ cm}$

Tip

Two separate scales

Histogram width and height scales are independent. Do not use the horizontal scale to find a height.

4. Estimating probabilities and percentages from a histogram

If one person is selected at random, a probability can be estimated using frequency:

frequency\text{probability} = \frac{\text{frequency wanted}}{\text{total frequency}}

From a histogram, frequencies come from areas. If the vertical scale is not fully labelled, use proportional areas.

Example

Estimating a percentage from histogram areas

A scaled histogram has intervals of width 20 seconds. The bar heights for 0–20, 20–40, 40–60, 60–80, 80–100, 100–120 and 120–140 seconds are 2, 3, 6, 8, 5, 4 and 2. The number of people taking more than 100 seconds is 60. Estimate the percentage taking less than 60 seconds.

The histogram for the percentage estimation example, with the known region above 100 seconds and the target region below 60 seconds marked.

Find the area for more than 100 seconds:
$20 \times 4 + 20 \times 2 = 120$
Since 120 area units represents 60 people, each area unit represents:
$60120=0.5\frac{60}{120} = 0.5$
Find the area for less than 60 seconds:
$20 (2 + 3 + 6) = 220$
Estimate the number of people taking less than 60 seconds:
$220 \times 0.5 = 110$
Find the total area:
$20 (2 + 3 + 6 + 8 + 5 + 4 + 2) = 600$
Estimate the total number of people:
$600 \times 0.5 = 300$
Find the percentage:
$110300×100≈36.7%\frac{110}{300} \times 100 \approx 36.7\%$

5. Box plots and quartiles

A box plot summarises data using five key values:

minimum
lower quartile, written $Q_1$
median
upper quartile, written $Q_3$
maximum

The median is the middle value when the data are in order. The quartiles split the data into quarters. The interquartile range, or IQR, measures the spread of the middle 50% of the data:

IQR=Q3−Q1\text{IQR} = Q_3 - Q_1

A box plot shows the middle 50% of the data as the box, with the median marked inside.

A labelled generic box plot showing minimum, quartiles, median, maximum, IQR and range.

Annotated box plot showing quartiles, whiskers and outliers

Example

Finding quartiles from a list

The ordered data are:

8, 12, 15, 18, 20, 22, 24, 25, 29, 33, 56

Find the median, quartiles and IQR.

There are 11 values, so the median is the 6th value: 22.
The lower half is 8, 12, 15, 18, 20, so $Q_1$ is 15.
The upper half is 24, 25, 29, 33, 56, so $Q_3$ is 29.
Find the IQR:
$29 - 15 = 14$

Common Mistake

Including the median twice

When there is an odd number of data values, do not include the median in both halves when finding $Q_1$ and $Q_3$ .

6. Outliers and drawing box plots

An outlier is a value that is unusually far from the rest of the data. At AS Level, you are usually given this rule:

fence=Q1−1.5×IQR\text{lower fence} = Q_1 - 1.5 \times \text{IQR}

fence=Q3+1.5×IQR\text{upper fence} = Q_3 + 1.5 \times \text{IQR}

Values below the lower fence or above the upper fence are outliers.

Example

Identifying an outlier and describing the box plot

Use the data from the previous example, where $Q_1 = 15$ , $Q_3 = 29$ and IQR is 14.

A box plot for the example showing whiskers to non-outliers and the outlier plotted separately.

Find the lower and upper fences:
$fence=29+1.5×14=50\begin{aligned} \text{lower fence} &= 15 - 1.5 \times 14 = -6 \\ \text{upper fence} &= 29 + 1.5 \times 14 = 50 \end{aligned}$
Compare the data values with the fences. The value 56 is greater than 50, so 56 is an outlier.
The smallest non-outlier is 8 and the largest non-outlier is 33.
The box plot would have a box from 15 to 29, a median line at 22, whiskers to 8 and 33, and a cross at 56.

Common Mistake

Whiskers do not go to outliers

If there are outliers, the whiskers stop at the smallest and largest values that are not outliers. The outliers are plotted separately, usually with crosses.

7. Comparing distributions

When comparing two box plots, write about:

location: usually compare medians
spread: compare IQRs and ranges
outliers: mention any unusual values in context

Example

Comparing two sets of completion times

Group A has minimum 7, $Q_1 = 10$ , median 13, $Q_3 = 16$ , largest non-outlier 21, with outliers at 25 and 27 minutes. Group B has minimum 6, $Q_1 = 11$ , median 14, $Q_3 = 17$ and maximum 22. Determine whether Group B has outliers, then compare the groups.

Parallel box plots for Groups A and B, supporting comparison of median, spread and outliers.

For Group A, 75% of people completed the task by $Q_3$ , so by 16 minutes.
Find the IQR for Group B:
$17 - 11 = 6$
Find the outlier fences for Group B:
$fence=17+1.5×6=26\begin{aligned} \text{lower fence} &= 11 - 1.5 \times 6 = 2 \\ \text{upper fence} &= 17 + 1.5 \times 6 = 26 \end{aligned}$
Group B has no outliers because 6 and 22 are both within the fences.
Group A has the lower median, so a typical person in Group A was faster. The IQRs are both 6, so the middle 50% of times have similar spread. However, Group A has two high outliers, meaning a few people in Group A took unusually long.

8. Mean and standard deviation, or median and IQR?

The mean and standard deviation are useful when data are fairly symmetrical with no major outliers.

The median and IQR are better when the data are skewed or contain outliers, because they are less affected by extreme values.

For summary data:

xˉ=∑xn\bar{x} = \frac{\sum x}{n}

and

deviation=∑x2n−xˉ2\text{standard deviation} = \sqrt{\frac{\sum x^2}{n} - \bar{x}^2}

Example

Choosing suitable summary statistics

For 50 journey times, $∑x=425\sum x = 425$ and $∑x2=4312.5\sum x^2 = 4312.5$ . A box plot shows two large high outliers. Calculate the mean and standard deviation, then choose suitable summary statistics.

Calculate the mean:
$xˉ=42550=8.5\bar{x} = \frac{425}{50} = 8.5$
Calculate the standard deviation:
$4312.550−8.52=14≈3.74\sqrt{\frac{4312.5}{50} - 8.5^2} = \sqrt{14} \approx 3.74$
Since the box plot shows high outliers, the median and IQR would usually be more suitable for describing these data.

Exam technique

In the exam

For histograms, write down class widths first, then use frequency density or area.
For rounded data, use true class boundaries such as 9.5 to 14.5.
For box plots, calculate IQR and outlier fences before deciding where the whiskers go.
When comparing distributions, mention median, IQR or range, and outliers in context.

Self review

Check yourself

Why can two histogram bars with different heights represent the same frequency?
How do you decide whether a value is an outlier using $Q_1$ , $Q_3$ and the IQR?
In what situation would median and IQR be better than mean and standard deviation?

What you'll learn

How to turn grouped data into correct class widths.
How histograms use area, not just height, to represent frequency.
How to find quartiles, outliers and draw box plots.
How to compare distributions using centre, spread and outliers.

1. Grouped data and class widths

In statistics, data are often collected in groups called classes or class intervals, such as 10–14 miles or 20–24 minutes.

The frequency is the number of observations in a class.

For histograms, the width of a class matters. If data are measured to the nearest whole number, the true class boundaries go halfway between the whole numbers.

For example, a class written as 10–14 actually covers values from 9.5 up to 14.5, so its class width is 5.

A number line showing how the rounded class 10–14 corresponds to true boundaries 9.5 and 14.5 with class width 5.

Definition

Class width

The class width is the size of the interval covered by a class, using the true class boundaries if the data have been rounded.

Example

Finding true class widths

A variable is measured to the nearest whole number. Two classes are 6–10 and 11–18. Find their true class widths.

The class 6–10 includes values that round to 6, 7, 8, 9 or 10, so its true boundaries are 5.5 and 10.5.
Find the width of the first class:
$10.5 - 5.5 = 5$
The class 11–18 has true boundaries 10.5 and 18.5.
Find the width of the second class:
$18.5 - 10.5 = 8$

Common Mistake

Counting the labels instead of the interval

For a class like 14–20 measured to the nearest whole number, the class width is 7, not 6. Use boundaries 13.5 and 20.5.

2. Histograms: area represents frequency

A histogram is used for grouped continuous data. Unlike a bar chart, the bars can have different widths, and the key idea is:

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

So:

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

This is why the area of each bar represents the frequency.

Two histogram bars with equal area but different widths and heights, illustrating that frequency is represented by area.

Here is the main idea visually: wider bars need a smaller height if they are to represent the same frequency.

Annotated histogram showing frequency as area

Key Idea

Histogram rule

In a histogram, compare areas, not just heights. A tall narrow bar and a short wide bar can represent the same frequency.

Example

Completing a missing frequency from a histogram

The given histogram setup for using one known bar to find the missing frequency of another equal-width class.

The class 6–10 has true width 5. Its frequency density is:
$205=4\frac{20}{5} = 4$
A height of 5 grid squares represents a frequency density of 4, so one grid square represents:
$45=0.8\frac{4}{5} = 0.8$
The bar for 11–15 is 3 grid squares high, so its frequency density is:
$\times 0.8 = 2.4$
The class 11–15 also has width 5, so its frequency is:
$2.4 \times 5 = 12$

3. Finding bar widths and heights from one given bar

Sometimes you are told the actual width and height of one bar on a drawn histogram. Use it to find the horizontal and vertical scales.

The horizontal scale links class width to centimetres.
The vertical scale links frequency density to centimetres.

Example

Finding the width and height of another histogram bar

A scale diagram showing how one histogram bar fixes separate horizontal and vertical scales for another bar.

Find the true width of 9–13. Its boundaries are 8.5 and 13.5:
$13.5 - 8.5 = 5$
Since width 5 is drawn as 2 cm, the horizontal scale is:
$unit\frac{2}{5} = 0.4 \text{ cm per unit}$
The class 14–20 has boundaries 13.5 and 20.5, so its true width is:
$20.5 - 13.5 = 7$
Therefore its drawn width is:
$\times 0.4 = 2.8 \text{ cm}$
Now find the frequency density of 9–13:
$125=2.4\frac{12}{5} = 2.4$
A frequency density of 2.4 is drawn as height 3 cm, so the vertical scale is:
$unit\frac{3}{2.4} = 1.25 \text{ cm per frequency-density unit}$
The frequency density of 14–20 is:
$217=3\frac{21}{7} = 3$
So its drawn height is:
$\times 1.25 = 3.75 \text{ cm}$

Tip

Two separate scales

Histogram width and height scales are independent. Do not use the horizontal scale to find a height.

4. Estimating probabilities and percentages from a histogram

If one person is selected at random, a probability can be estimated using frequency:

frequency\text{probability} = \frac{\text{frequency wanted}}{\text{total frequency}}

From a histogram, frequencies come from areas. If the vertical scale is not fully labelled, use proportional areas.

Example

Estimating a percentage from histogram areas

The histogram for the percentage estimation example, with the known region above 100 seconds and the target region below 60 seconds marked.

Find the area for more than 100 seconds:
$20 \times 4 + 20 \times 2 = 120$
Since 120 area units represents 60 people, each area unit represents:
$60120=0.5\frac{60}{120} = 0.5$
Find the area for less than 60 seconds:
$20 (2 + 3 + 6) = 220$
Estimate the number of people taking less than 60 seconds:
$220 \times 0.5 = 110$
Find the total area:
$20 (2 + 3 + 6 + 8 + 5 + 4 + 2) = 600$
Estimate the total number of people:
$600 \times 0.5 = 300$
Find the percentage:
$110300×100≈36.7%\frac{110}{300} \times 100 \approx 36.7\%$

5. Box plots and quartiles

A box plot summarises data using five key values:

minimum
lower quartile, written $Q_1$
median
upper quartile, written $Q_3$
maximum

The median is the middle value when the data are in order. The quartiles split the data into quarters. The interquartile range, or IQR, measures the spread of the middle 50% of the data:

IQR=Q3−Q1\text{IQR} = Q_3 - Q_1

A box plot shows the middle 50% of the data as the box, with the median marked inside.

A labelled generic box plot showing minimum, quartiles, median, maximum, IQR and range.

Annotated box plot showing quartiles, whiskers and outliers

Example

Finding quartiles from a list

The ordered data are:

8, 12, 15, 18, 20, 22, 24, 25, 29, 33, 56

Find the median, quartiles and IQR.

There are 11 values, so the median is the 6th value: 22.
The lower half is 8, 12, 15, 18, 20, so $Q_1$ is 15.
The upper half is 24, 25, 29, 33, 56, so $Q_3$ is 29.
Find the IQR:
$29 - 15 = 14$

Common Mistake

Including the median twice

When there is an odd number of data values, do not include the median in both halves when finding $Q_1$ and $Q_3$ .

6. Outliers and drawing box plots

An outlier is a value that is unusually far from the rest of the data. At AS Level, you are usually given this rule:

fence=Q1−1.5×IQR\text{lower fence} = Q_1 - 1.5 \times \text{IQR}

fence=Q3+1.5×IQR\text{upper fence} = Q_3 + 1.5 \times \text{IQR}

Values below the lower fence or above the upper fence are outliers.

Example

Identifying an outlier and describing the box plot

Use the data from the previous example, where $Q_1 = 15$ , $Q_3 = 29$ and IQR is 14.

A box plot for the example showing whiskers to non-outliers and the outlier plotted separately.

Find the lower and upper fences:
$fence=29+1.5×14=50\begin{aligned} \text{lower fence} &= 15 - 1.5 \times 14 = -6 \\ \text{upper fence} &= 29 + 1.5 \times 14 = 50 \end{aligned}$
Compare the data values with the fences. The value 56 is greater than 50, so 56 is an outlier.
The smallest non-outlier is 8 and the largest non-outlier is 33.
The box plot would have a box from 15 to 29, a median line at 22, whiskers to 8 and 33, and a cross at 56.

Common Mistake

Whiskers do not go to outliers

If there are outliers, the whiskers stop at the smallest and largest values that are not outliers. The outliers are plotted separately, usually with crosses.

7. Comparing distributions

When comparing two box plots, write about:

location: usually compare medians
spread: compare IQRs and ranges
outliers: mention any unusual values in context

Example

Comparing two sets of completion times

Parallel box plots for Groups A and B, supporting comparison of median, spread and outliers.

For Group A, 75% of people completed the task by $Q_3$ , so by 16 minutes.
Find the IQR for Group B:
$17 - 11 = 6$
Find the outlier fences for Group B:
$fence=17+1.5×6=26\begin{aligned} \text{lower fence} &= 11 - 1.5 \times 6 = 2 \\ \text{upper fence} &= 17 + 1.5 \times 6 = 26 \end{aligned}$
Group B has no outliers because 6 and 22 are both within the fences.
Group A has the lower median, so a typical person in Group A was faster. The IQRs are both 6, so the middle 50% of times have similar spread. However, Group A has two high outliers, meaning a few people in Group A took unusually long.

8. Mean and standard deviation, or median and IQR?

The mean and standard deviation are useful when data are fairly symmetrical with no major outliers.

The median and IQR are better when the data are skewed or contain outliers, because they are less affected by extreme values.

For summary data:

xˉ=∑xn\bar{x} = \frac{\sum x}{n}

and

deviation=∑x2n−xˉ2\text{standard deviation} = \sqrt{\frac{\sum x^2}{n} - \bar{x}^2}

Example

Choosing suitable summary statistics

Calculate the mean:
$xˉ=42550=8.5\bar{x} = \frac{425}{50} = 8.5$
Calculate the standard deviation:
$4312.550−8.52=14≈3.74\sqrt{\frac{4312.5}{50} - 8.5^2} = \sqrt{14} \approx 3.74$
Since the box plot shows high outliers, the median and IQR would usually be more suitable for describing these data.

Exam technique

In the exam

For histograms, write down class widths first, then use frequency density or area.
For rounded data, use true class boundaries such as 9.5 to 14.5.
For box plots, calculate IQR and outlier fences before deciding where the whiskers go.
When comparing distributions, mention median, IQR or range, and outliers in context.

Self review

Check yourself

Why can two histogram bars with different heights represent the same frequency?
How do you decide whether a value is an outlier using $Q_1$ , $Q_3$ and the IQR?
In what situation would median and IQR be better than mean and standard deviation?

What you'll learn

1. Grouped data and class widths

2. Histograms: area represents frequency

3. Finding bar widths and heights from one given bar

4. Estimating probabilities and percentages from a histogram

5. Box plots and quartiles

6. Outliers and drawing box plots

7. Comparing distributions

8. Mean and standard deviation, or median and IQR?

Recap questions

Review Representations of Data by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Representations of Data

What you'll learn

1. Grouped data and class widths

2. Histograms: area represents frequency

3. Finding bar widths and heights from one given bar

4. Estimating probabilities and percentages from a histogram

5. Box plots and quartiles

6. Outliers and drawing box plots

7. Comparing distributions

8. Mean and standard deviation, or median and IQR?

Recap questions

Review Representations of Data by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Representations of Data

What you'll learn

1. Grouped data and class widths

2. Histograms: area represents frequency

3. Finding bar widths and heights from one given bar

4. Estimating probabilities and percentages from a histogram

5. Box plots and quartiles

6. Outliers and drawing box plots

7. Comparing distributions

8. Mean and standard deviation, or median and IQR?

Recap questions

Review Representations of Data by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Representations of Data Revision Guide

Representations of Data

What you'll learn

1. Grouped data and class widths

2. Histograms: area represents frequency

3. Finding bar widths and heights from one given bar

4. Estimating probabilities and percentages from a histogram

5. Box plots and quartiles

6. Outliers and drawing box plots

7. Comparing distributions

8. Mean and standard deviation, or median and IQR?

Recap questions

Review Representations of Data by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Representations of Data Revision Guide