Measures of Location and Spread

What you'll learn

How to describe the “centre” of a data set using the mean, median and quartiles.
How to describe how spread out data are using range, interquartile range and standard deviation.
How to estimate averages and standard deviation from grouped data.
How coding affects the median, mean and standard deviation.

1. What are measures of location and spread?

In statistics, a long list of data is often hard to interpret directly. We use summary measures to describe the data quickly.

Definition

Measures of location and spread

A measure of location tells you where the data are centred. Common examples are the mean, median and quartiles.
A measure of spread tells you how varied the data are. Common examples are the range, interquartile range and standard deviation.

Key Idea

The big picture

A good statistical summary usually needs both a measure of location and a measure of spread: one tells you the typical value, the other tells you how consistent the data are.

2. Mean, median and quartiles for raw data

The mean is the arithmetic average. For data values called $x$ , the mean is written as $xˉ\bar{x}$ .

xˉ=∑xn\bar{x}=\frac{\sum x}{n}

Here, $n$ is the number of data values, and $∑x\sum x$ means “add all the data values”.

The median is the middle value when the data are in order.

The quartiles split ordered data into quarters:

An ordered data strip showing how the median, lower quartile and upper quartile divide a data set into four parts.

The lower quartile, usually written $Q_1$ , is around one quarter of the way through the data.
The upper quartile, usually written $Q_3$ , is around three quarters of the way through the data.
The interquartile range is $Q_3-Q_1$ .

Example

Finding the mean, median and quartiles from raw times

Twelve runners record these 400 m times, in seconds:

46.3, 45.0, 45.8, 47.1, 45.6, 46.0, 45.4, 46.8, 45.9, 45.5, 46.2, 45.7

Add the times and divide by 12 to find the mean.
$xˉ=551.312=45.941…\bar{x}=\frac{551.3}{12}=45.941\ldots$
Round sensibly, so the mean time is about 45.9 s.
Put the data in ascending order.
$47.145.0,\ 45.4,\ 45.5,\ 45.6,\ 45.7,\ 45.8,\ 45.9,\ 46.0,\ 46.2,\ 46.3,\ 46.8,\ 47.1$
Since there are 12 values, the median is halfway between the 6th and 7th values.
$median=45.8+45.92=45.85\text{median}=\frac{45.8+45.9}{2}=45.85$
The lower half is the first 6 values, so $Q_1$ is halfway between the 3rd and 4th values.
$Q1=45.5+45.62=45.55Q_1=\frac{45.5+45.6}{2}=45.55$
The upper half is the last 6 values, so $Q_3$ is halfway between the 9th and 10th values.
$Q3=46.2+46.32=46.25Q_3=\frac{46.2+46.3}{2}=46.25$

Common Mistake

Forgetting to order the data

You must order the data before finding the median or quartiles. The middle value in the original list is usually meaningless.

3. Standard deviation

The standard deviation measures how far the data values typically are from the mean. A small standard deviation means the values are tightly clustered; a large standard deviation means they are more spread out.

Two dot plots with the same mean but different spreads illustrate small and large standard deviation.

Definition

Standard deviation formula

For AS-Level statistics questions using a whole data set or summary statistics, use

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

The value $∑x2\sum x^2$ means “square each data value, then add the squares”.

Example

Using summary statistics to find the mean and standard deviation

For a group of 80 employees, the travel times to work have summary statistics:

n=80,∑x=3440,∑x2=163200n=80,\qquad \sum x=3440,\qquad \sum x^2=163200

Find the mean and standard deviation.

Find the mean using $xˉ=∑xn\bar{x}=\frac{\sum x}{n}$ .
$xˉ=344080=43\bar{x}=\frac{3440}{80}=43$
Substitute into the standard deviation formula.
$sd=16320080−432\text{sd}=\sqrt{\frac{163200}{80}-43^2}$
Simplify inside the square root.
$sd=2040−1849\text{sd}=\sqrt{2040-1849}$
Calculate the standard deviation.
$sd=191=13.820…\text{sd}=\sqrt{191}=13.820\ldots$
Give a sensible rounded answer: the mean is 43 minutes and the standard deviation is 13.8 minutes.

Tip

Calculator check

On your calculator, the standard deviation with divisor $n$ is usually labelled $σx\sigma_x$ . Avoid using the sample standard deviation unless the question specifically asks for it.

4. Using $S_{xx}$

Sometimes you are given $S_{xx}$ instead of $∑x2\sum x^2$ .

Definition

What $S_{xx}$ means

The statistic $S_{xx}$ measures the total squared variation from the mean:

Sxx=∑x2−(∑x)2nS_{xx}=\sum x^2-\frac{(\sum x)^2}{n}

So the standard deviation is

sd=Sxxn\text{sd}=\sqrt{\frac{S_{xx}}{n}}

Example

Finding standard deviation from $S_{xx}$

A survey of 70 staff gives

n=70,∑x=1890,Sxx=5670n=70,\qquad \sum x=1890,\qquad S_{xx}=5670

Find the mean and standard deviation.

Use $xˉ=∑xn\bar{x}=\frac{\sum x}{n}$ .
$xˉ=189070=27\bar{x}=\frac{1890}{70}=27$
Use the formula involving $S_{xx}$ .
$sd=567070\text{sd}=\sqrt{\frac{5670}{70}}$
Calculate.
$sd=81=9\text{sd}=\sqrt{81}=9$
The mean is 27 and the standard deviation is 9.

5. Grouped data and midpoints

When data are grouped into classes, you usually do not know the exact original values. So the mean and standard deviation are estimates.

For each class, use the class midpoint as a representative value.

A class interval is represented by its midpoint when estimating from grouped data.

For example, the midpoint of $140<h≤150140<h\le150$ is

140+1502=145\frac{140+150}{2}=145

Example

Estimating the mean and standard deviation from grouped heights

A gardener records the heights of 50 plants:

A frequency diagram of the grouped plant-height data shows why midpoints are used for estimating the mean and standard deviation.

$120<h≤130120<h\le130$ : frequency 5
$130<h≤140130<h\le140$ : frequency 12
$140<h≤150140<h\le150$ : frequency 20
$150<h≤160150<h\le160$ : frequency 9
$160<h≤180160<h\le180$ : frequency 4

Estimate the mean and standard deviation.

Find the class midpoints: 125, 135, 145, 155 and 170.
Multiply each midpoint by its frequency, then add.
$∑fx=5(125)+12(135)+20(145)+9(155)+4(170)=7220\sum fx=5(125)+12(135)+20(145)+9(155)+4(170)=7220$
Divide by the total frequency.
$xˉ=722050=144.4\bar{x}=\frac{7220}{50}=144.4$
Now find $∑fx2\sum fx^2$ by squaring each midpoint first.
$∑fx2=5(1252)+12(1352)+20(1452)+9(1552)+4(1702)=1049150\sum fx^2=5(125^2)+12(135^2)+20(145^2)+9(155^2)+4(170^2)=1049150$
Substitute into the grouped standard deviation formula.
$sd=104915050−144.42\text{sd}=\sqrt{\frac{1049150}{50}-144.4^2}$
Calculate.
$sd=11.47…\text{sd}=11.47\ldots$
The estimated mean is 144.4 cm and the estimated standard deviation is about 11.5 cm.

Common Mistake

Squaring the frequency

For $∑fx2\sum fx^2$ , square the midpoint, not the frequency. The calculation is frequency times midpoint squared.

6. Linear interpolation for the median

For grouped data, the median is also an estimate. We assume the data are evenly spread within the median class. This method is called linear interpolation.

The picture below shows the idea: you locate the halfway position, then move proportionally through the median class.

Linear interpolation estimates the median by moving proportionally through the median class from the previous cumulative frequency.

Linear interpolation diagram for estimating the median from grouped data

Definition

Linear interpolation formula

For grouped data,

median=L+n2−cf×w\text{median}=L+\frac{\frac{n}{2}-c}{f}\times w

where:

$L$ is the lower boundary of the median class.
$n$ is the total frequency.
$c$ is the cumulative frequency before the median class.
$f$ is the frequency of the median class.
$w$ is the class width.

Example

Estimating a grouped median

Use the plant-height data from before:

$120<h≤130120<h\le130$ : frequency 5
$130<h≤140130<h\le140$ : frequency 12
$140<h≤150140<h\le150$ : frequency 20
$150<h≤160150<h\le160$ : frequency 9
$160<h≤180160<h\le180$ : frequency 4

Estimate the median height.

Find the total frequency.
$n = 5 + 12 + 20 + 9 + 4 = 50$
The median is at position $n2=25\frac{n}{2}=25$ .
Find cumulative frequencies: 5, 17, 37, 46, 50.
The 25th value lies in the class $140<h≤150140<h\le150$ , because the cumulative frequency goes from 17 to 37 there.
Identify the interpolation values: $L = 140$ , $c = 17$ , $f = 20$ , and $w = 10$ .
Substitute into the formula.
$median=140+25−1720×10\text{median}=140+\frac{25-17}{20}\times10$
Calculate.
$median=144\text{median}=144$
The estimated median height is 144 cm.

Common Mistake

Class boundaries matter

If data are recorded to the nearest mile, a class labelled 20–29 actually runs from 19.5 to 29.5 for interpolation. Use the real class boundaries, not just the printed labels.

7. Coding data

Sometimes data are coded to make the numbers easier to handle. For example,

x=t−202x=\frac{t-20}{2}

means the original value $t$ has had 20 subtracted, then been divided by 2.

To undo the coding, rearrange the formula:

t = 2 x + 20

Key Idea

How coding affects measures

If $t = a x + b$ , then measures of location, such as the mean and median, become $a$ times as large and then have $b$ added. Standard deviation is only multiplied by $∣ a ∣$ because adding a constant does not change spread.

Coding stretches data by the scale factor and shifts location, but only the stretch changes standard deviation.

Example

Undoing coding for median and standard deviation

For coded travel times $x$ , the estimated median is 14.2 and the estimated standard deviation is 8.1. The coding used was

x=t−202x=\frac{t-20}{2}

Find the median and standard deviation of the original times $t$ .

Rearrange the coding formula.
$t = 2 x + 20$
Transform the median using the full formula.
$t=2(14.2)+20=48.4\text{median of }t=2(14.2)+20=48.4$
Transform the standard deviation by multiplying by 2 only.
$t=2(8.1)=16.2\text{sd of }t=2(8.1)=16.2$
The original median is 48.4 minutes and the original standard deviation is 16.2 minutes.

Common Mistake

Adding to the standard deviation

Do not add 20 to the standard deviation. Adding a constant shifts every value by the same amount, so the spread stays the same.

Exam technique

In the exam

For raw data, order the values before finding the median or quartiles.
For grouped means and standard deviations, use midpoints and remember your answers are estimates.
For interpolation, identify $L$ , $c$ , $f$ , $w$ and $n2\frac{n}{2}$ before substituting.
For coding, undo the formula carefully; multiply standard deviation by the scale factor only.

Self review

Check yourself

Can you explain the difference between the mean and the median?
When estimating a grouped mean, why do you use class midpoints?
If $x=t−105x=\frac{t-10}{5}$ , what happens to the standard deviation when you convert from $x$ back to $t$ ?

Measures of Location and Spread

What you'll learn

How to describe the “centre” of a data set using the mean, median and quartiles.
How to describe how spread out data are using range, interquartile range and standard deviation.
How to estimate averages and standard deviation from grouped data.
How coding affects the median, mean and standard deviation.

1. What are measures of location and spread?

In statistics, a long list of data is often hard to interpret directly. We use summary measures to describe the data quickly.

Definition

Measures of location and spread

A measure of location tells you where the data are centred. Common examples are the mean, median and quartiles.
A measure of spread tells you how varied the data are. Common examples are the range, interquartile range and standard deviation.

Key Idea

The big picture

A good statistical summary usually needs both a measure of location and a measure of spread: one tells you the typical value, the other tells you how consistent the data are.

2. Mean, median and quartiles for raw data

The mean is the arithmetic average. For data values called $x$ , the mean is written as $xˉ\bar{x}$ .

xˉ=∑xn\bar{x}=\frac{\sum x}{n}

Here, $n$ is the number of data values, and $∑x\sum x$ means “add all the data values”.

The median is the middle value when the data are in order.

The quartiles split ordered data into quarters:

An ordered data strip showing how the median, lower quartile and upper quartile divide a data set into four parts.

The lower quartile, usually written $Q_1$ , is around one quarter of the way through the data.
The upper quartile, usually written $Q_3$ , is around three quarters of the way through the data.
The interquartile range is $Q_3-Q_1$ .

Example

Finding the mean, median and quartiles from raw times

Twelve runners record these 400 m times, in seconds:

46.3, 45.0, 45.8, 47.1, 45.6, 46.0, 45.4, 46.8, 45.9, 45.5, 46.2, 45.7

Add the times and divide by 12 to find the mean.
$xˉ=551.312=45.941…\bar{x}=\frac{551.3}{12}=45.941\ldots$
Round sensibly, so the mean time is about 45.9 s.
Put the data in ascending order.
$47.145.0,\ 45.4,\ 45.5,\ 45.6,\ 45.7,\ 45.8,\ 45.9,\ 46.0,\ 46.2,\ 46.3,\ 46.8,\ 47.1$
Since there are 12 values, the median is halfway between the 6th and 7th values.
$median=45.8+45.92=45.85\text{median}=\frac{45.8+45.9}{2}=45.85$
The lower half is the first 6 values, so $Q_1$ is halfway between the 3rd and 4th values.
$Q1=45.5+45.62=45.55Q_1=\frac{45.5+45.6}{2}=45.55$
The upper half is the last 6 values, so $Q_3$ is halfway between the 9th and 10th values.
$Q3=46.2+46.32=46.25Q_3=\frac{46.2+46.3}{2}=46.25$

Common Mistake

Forgetting to order the data

You must order the data before finding the median or quartiles. The middle value in the original list is usually meaningless.

3. Standard deviation

Two dot plots with the same mean but different spreads illustrate small and large standard deviation.

Definition

Standard deviation formula

For AS-Level statistics questions using a whole data set or summary statistics, use

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

The value $∑x2\sum x^2$ means “square each data value, then add the squares”.

Example

Using summary statistics to find the mean and standard deviation

For a group of 80 employees, the travel times to work have summary statistics:

n=80,∑x=3440,∑x2=163200n=80,\qquad \sum x=3440,\qquad \sum x^2=163200

Find the mean and standard deviation.

Find the mean using $xˉ=∑xn\bar{x}=\frac{\sum x}{n}$ .
$xˉ=344080=43\bar{x}=\frac{3440}{80}=43$
Substitute into the standard deviation formula.
$sd=16320080−432\text{sd}=\sqrt{\frac{163200}{80}-43^2}$
Simplify inside the square root.
$sd=2040−1849\text{sd}=\sqrt{2040-1849}$
Calculate the standard deviation.
$sd=191=13.820…\text{sd}=\sqrt{191}=13.820\ldots$
Give a sensible rounded answer: the mean is 43 minutes and the standard deviation is 13.8 minutes.

Tip

Calculator check

On your calculator, the standard deviation with divisor $n$ is usually labelled $σx\sigma_x$ . Avoid using the sample standard deviation unless the question specifically asks for it.

4. Using $S_{xx}$

Sometimes you are given $S_{xx}$ instead of $∑x2\sum x^2$ .

Definition

What $S_{xx}$ means

The statistic $S_{xx}$ measures the total squared variation from the mean:

Sxx=∑x2−(∑x)2nS_{xx}=\sum x^2-\frac{(\sum x)^2}{n}

So the standard deviation is

sd=Sxxn\text{sd}=\sqrt{\frac{S_{xx}}{n}}

Example

Finding standard deviation from $S_{xx}$

A survey of 70 staff gives

n=70,∑x=1890,Sxx=5670n=70,\qquad \sum x=1890,\qquad S_{xx}=5670

Find the mean and standard deviation.

Use $xˉ=∑xn\bar{x}=\frac{\sum x}{n}$ .
$xˉ=189070=27\bar{x}=\frac{1890}{70}=27$
Use the formula involving $S_{xx}$ .
$sd=567070\text{sd}=\sqrt{\frac{5670}{70}}$
Calculate.
$sd=81=9\text{sd}=\sqrt{81}=9$
The mean is 27 and the standard deviation is 9.

5. Grouped data and midpoints

When data are grouped into classes, you usually do not know the exact original values. So the mean and standard deviation are estimates.

For each class, use the class midpoint as a representative value.

A class interval is represented by its midpoint when estimating from grouped data.

For example, the midpoint of $140<h≤150140<h\le150$ is

140+1502=145\frac{140+150}{2}=145

Example

Estimating the mean and standard deviation from grouped heights

A gardener records the heights of 50 plants:

A frequency diagram of the grouped plant-height data shows why midpoints are used for estimating the mean and standard deviation.

$120<h≤130120<h\le130$ : frequency 5
$130<h≤140130<h\le140$ : frequency 12
$140<h≤150140<h\le150$ : frequency 20
$150<h≤160150<h\le160$ : frequency 9
$160<h≤180160<h\le180$ : frequency 4

Estimate the mean and standard deviation.

Find the class midpoints: 125, 135, 145, 155 and 170.
Multiply each midpoint by its frequency, then add.
$∑fx=5(125)+12(135)+20(145)+9(155)+4(170)=7220\sum fx=5(125)+12(135)+20(145)+9(155)+4(170)=7220$
Divide by the total frequency.
$xˉ=722050=144.4\bar{x}=\frac{7220}{50}=144.4$
Now find $∑fx2\sum fx^2$ by squaring each midpoint first.
$∑fx2=5(1252)+12(1352)+20(1452)+9(1552)+4(1702)=1049150\sum fx^2=5(125^2)+12(135^2)+20(145^2)+9(155^2)+4(170^2)=1049150$
Substitute into the grouped standard deviation formula.
$sd=104915050−144.42\text{sd}=\sqrt{\frac{1049150}{50}-144.4^2}$
Calculate.
$sd=11.47…\text{sd}=11.47\ldots$
The estimated mean is 144.4 cm and the estimated standard deviation is about 11.5 cm.

Common Mistake

Squaring the frequency

For $∑fx2\sum fx^2$ , square the midpoint, not the frequency. The calculation is frequency times midpoint squared.

6. Linear interpolation for the median

For grouped data, the median is also an estimate. We assume the data are evenly spread within the median class. This method is called linear interpolation.

The picture below shows the idea: you locate the halfway position, then move proportionally through the median class.

Linear interpolation estimates the median by moving proportionally through the median class from the previous cumulative frequency.

Linear interpolation diagram for estimating the median from grouped data

Definition

Linear interpolation formula

For grouped data,

median=L+n2−cf×w\text{median}=L+\frac{\frac{n}{2}-c}{f}\times w

where:

$L$ is the lower boundary of the median class.
$n$ is the total frequency.
$c$ is the cumulative frequency before the median class.
$f$ is the frequency of the median class.
$w$ is the class width.

Example

Estimating a grouped median

Use the plant-height data from before:

$120<h≤130120<h\le130$ : frequency 5
$130<h≤140130<h\le140$ : frequency 12
$140<h≤150140<h\le150$ : frequency 20
$150<h≤160150<h\le160$ : frequency 9
$160<h≤180160<h\le180$ : frequency 4

Estimate the median height.

Find the total frequency.
$n = 5 + 12 + 20 + 9 + 4 = 50$
The median is at position $n2=25\frac{n}{2}=25$ .
Find cumulative frequencies: 5, 17, 37, 46, 50.
The 25th value lies in the class $140<h≤150140<h\le150$ , because the cumulative frequency goes from 17 to 37 there.
Identify the interpolation values: $L = 140$ , $c = 17$ , $f = 20$ , and $w = 10$ .
Substitute into the formula.
$median=140+25−1720×10\text{median}=140+\frac{25-17}{20}\times10$
Calculate.
$median=144\text{median}=144$
The estimated median height is 144 cm.

Common Mistake

Class boundaries matter

If data are recorded to the nearest mile, a class labelled 20–29 actually runs from 19.5 to 29.5 for interpolation. Use the real class boundaries, not just the printed labels.

7. Coding data

Sometimes data are coded to make the numbers easier to handle. For example,

x=t−202x=\frac{t-20}{2}

means the original value $t$ has had 20 subtracted, then been divided by 2.

To undo the coding, rearrange the formula:

t = 2 x + 20

Key Idea

How coding affects measures

Coding stretches data by the scale factor and shifts location, but only the stretch changes standard deviation.

Example

Undoing coding for median and standard deviation

For coded travel times $x$ , the estimated median is 14.2 and the estimated standard deviation is 8.1. The coding used was

x=t−202x=\frac{t-20}{2}

Find the median and standard deviation of the original times $t$ .

Rearrange the coding formula.
$t = 2 x + 20$
Transform the median using the full formula.
$t=2(14.2)+20=48.4\text{median of }t=2(14.2)+20=48.4$
Transform the standard deviation by multiplying by 2 only.
$t=2(8.1)=16.2\text{sd of }t=2(8.1)=16.2$
The original median is 48.4 minutes and the original standard deviation is 16.2 minutes.

Common Mistake

Adding to the standard deviation

Do not add 20 to the standard deviation. Adding a constant shifts every value by the same amount, so the spread stays the same.

Exam technique

In the exam

For raw data, order the values before finding the median or quartiles.
For grouped means and standard deviations, use midpoints and remember your answers are estimates.
For interpolation, identify $L$ , $c$ , $f$ , $w$ and $n2\frac{n}{2}$ before substituting.
For coding, undo the formula carefully; multiply standard deviation by the scale factor only.

Self review

Check yourself

Can you explain the difference between the mean and the median?
When estimating a grouped mean, why do you use class midpoints?
If $x=t−105x=\frac{t-10}{5}$ , what happens to the standard deviation when you convert from $x$ back to $t$ ?

Measures of Location and Spread

What you'll learn

1. What are measures of location and spread?

2. Mean, median and quartiles for raw data

3. Standard deviation

4. Using $S_{xx}$

5. Grouped data and midpoints

6. Linear interpolation for the median

7. Coding data

Recap questions

You've reached the end

Measures of Location and Spread

What you'll learn

1. What are measures of location and spread?

2. Mean, median and quartiles for raw data

3. Standard deviation

4. Using $S_{xx}$

5. Grouped data and midpoints

6. Linear interpolation for the median

7. Coding data

Recap questions

You've reached the end

Measures of Location and Spread

What you'll learn

1. What are measures of location and spread?

2. Mean, median and quartiles for raw data

3. Standard deviation

4. Using SxxS_{xx}Sxx​

5. Grouped data and midpoints

6. Linear interpolation for the median

7. Coding data

Recap questions

You've reached the end

Measures of Location and Spread Revision Guide

Measures of Location and Spread

What you'll learn

1. What are measures of location and spread?

2. Mean, median and quartiles for raw data

3. Standard deviation

4. Using SxxS_{xx}Sxx​

5. Grouped data and midpoints

6. Linear interpolation for the median

7. Coding data

Recap questions

You've reached the end

Measures of Location and Spread Revision Guide

4. Using $S_{xx}$

4. Using $S_{xx}$