Averages and measures of spread

A data set needs more than one number to describe it. A measure of location says where the data are centred. A measure of spread says how variable they are.

Two classes can have the same mean mark but very different consistency. For example,

$$4,5,5,5,6 \qquad\text{and}\qquad 1,2,5,8,9$$

both have mean $5$, but the second set is much more spread out. A useful comparison therefore pairs a measure of location with a measure of spread.

Prerequisites

You should be able to:

• order numerical data and use sigma notation $\sum$;
• calculate with squares and square roots;
• interpret frequency tables and class intervals;
• use the statistics functions on your calculator.

Review presenting and interpreting data if frequency tables or grouped data are unfamiliar.

Measures of location

For $n$ observations $x_1,x_2,\ldots,x_n$:

Measure	Definition	Particularly useful when
Mean	$\displaystyle \bar{x}=\frac{\sum x}{n}$	all values should influence the centre
Median	middle value after ordering	the distribution is skewed or contains extremes
Mode	most frequent value	the most common outcome matters

The mean uses every value, so it is sensitive to extreme observations. The median depends on position, not distance, so it is resistant to extremes. There can be no mode, one mode, or several modes.

The word average

In an exam, state which average you are using. “Average” can mean mean, median or mode, although an unqualified calculation usually intends the arithmetic mean.

Worked example 1: raw data

The journey times, in minutes, for seven journeys are

$$18,\ 21,\ 19,\ 20,\ 22,\ 19,\ 35.$$

First order the data:

$$18,\ 19,\ 19,\ 20,\ 21,\ 22,\ 35.$$

The fourth value is the median, so

$$\operatorname{median}=20.$$

The value $19$ occurs most often, so the mode is $19$. The mean is

$$\bar{x}=\frac{18+21+19+20+22+19+35}{7} =\frac{154}{7}=22.$$

The unusually long journey raises the mean above the median. For a typical journey time, the median is more representative here.

Finding the median position

For $n$ ordered observations, the median is at position

$$\frac{n+1}{2}.$$

If this position is an integer, use that observation. If it is halfway between two integers, average the two corresponding observations. For $n=8$, for example, the median is halfway between the fourth and fifth values.

Positions are not data values

The expression $(n+1)/2$ gives a position. It is not usually the median itself. Always order the data and then read the value at that position.

Measures of spread

Common measures of spread are:

$$\begin{aligned} \text{range}&=\text{maximum}-\text{minimum},\\ \text{interquartile range}&=Q_3-Q_1,\\ \text{variance}&=\frac{\sum (x-\bar{x})^2}{n},\\ \text{standard deviation}&=\sqrt{\frac{\sum (x-\bar{x})^2}{n}}. \end{aligned}$$

The range uses only the two extreme values. The interquartile range, or IQR, measures the spread of the middle $50\%$ of the ordered data. Both variance and standard deviation use every observation.

Variance has squared units. If $x$ is measured in centimetres, the variance is in $\text{cm}^2$. Standard deviation has the same units as the data, so it is usually easier to interpret in context.

Why deviations are squared

The deviations from the mean always sum to zero:

$$\sum(x-\bar{x})=\sum x-n\bar{x}=\sum x-n\left(\frac{\sum x}{n}\right)=0.$$

Simply averaging signed deviations would therefore give zero for every data set. Squaring makes all contributions non-negative and gives greater weight to observations far from the mean.

The efficient variance formula

Expanding $\sum(x-\bar{x})^2$ gives

$$\boxed{\operatorname{Var}(x)=\frac{\sum x^2}{n}-\bar{x}^{,2}}.$$

Hence

$$\boxed{\sigma_x=\sqrt{\frac{\sum x^2}{n}-\bar{x}^{,2}}}.$$

This is sometimes called the population standard deviation formula. In A level data questions, use division by $n$ unless the question or calculator convention specifies otherwise.

Two calculator standard deviations

Calculators often display $\sigma_x$, which divides by $n$, and $s_x$, which divides by $n-1$. They answer different questions. For the standard deviation of the listed data, select $\sigma_x$.

Worked example 2: variance and standard deviation

For the data $2,4,4,5,10$,

$$n=5,\qquad \sum x=25,\qquad \sum x^2=4+16+16+25+100=161.$$

Therefore

$$\bar{x}=\frac{25}{5}=5$$

and

$$\begin{aligned} \operatorname{Var}(x) &=\frac{161}{5}-5^2\\ &=7.2. \end{aligned}$$

The standard deviation is

$$\sigma_x=\sqrt{7.2}=2.683\ldots\approx2.68.$$

Do not round the variance before taking its square root. Retaining full calculator accuracy prevents avoidable error.

Self-check 1

Find the mean, variance and standard deviation of $3,3,7,7$.

Answer

Here $n=4$, $\sum x=20$ and $\sum x^2=116$. Thus

$$\bar{x}=5,$$ $$\operatorname{Var}(x)=\frac{116}{4}-5^2=4,$$

and $\sigma_x=\sqrt4=2$.

Frequency tables

If value $x$ occurs with frequency $f$, there are

$$n=\sum f$$

observations. Repeated values can be handled without writing them all out:

$$\bar{x}=\frac{\sum fx}{\sum f},$$ $$\operatorname{Var}(x)=\frac{\sum fx^2}{\sum f}-\bar{x}^{,2}.$$

Worked example 3: discrete frequency data

The number of goals scored by a team in $12$ matches is summarised below.

Goals, $x$	$0$	$1$	$2$	$3$	$4$
Frequency, $f$	$2$	$4$	$3$	$2$	$1$

Create the required totals:

$x$	$f$	$fx$	$fx^2$
$0$	$2$	$0$	$0$
$1$	$4$	$4$	$4$
$2$	$3$	$6$	$12$
$3$	$2$	$6$	$18$
$4$	$1$	$4$	$16$
Total	$12$	$20$	$50$

Then

$$\bar{x}=\frac{20}{12}=\frac53=1.666\ldots$$

and

$$\begin{aligned} \operatorname{Var}(x) &=\frac{50}{12}-\left(\frac53\right)^2\\ &=\frac{25}{18}. \end{aligned}$$

Therefore

$$\sigma_x=\sqrt{\frac{25}{18}}=1.178\ldots\approx1.18\text{ goals}.$$

For the median, use cumulative frequency. The sixth observation is $1$ and the seventh is $2$, so

$$\operatorname{median}=\frac{1+2}{2}=1.5.$$

The median need not itself be an observed value.

Grouped continuous data

When only class intervals and frequencies are known, the original values have been lost. Use each class midpoint $m$ as a representative value:

$$\bar{x}\approx\frac{\sum fm}{\sum f},$$ $$\operatorname{Var}(x)\approx\frac{\sum fm^2}{\sum f}-\bar{x}^{,2}.$$

These answers are estimates, because the observations are unlikely all to equal their class midpoints.

Worked example 4: estimating from grouped data

The masses of $20$ parcels are grouped as follows.

Mass $x$ in kg	Frequency $f$	Midpoint $m$	$fm$	$fm^2$
$0\leq x<2$	$3$	$1$	$3$	$3$
$2\leq x<4$	$8$	$3$	$24$	$72$
$4\leq x<6$	$6$	$5$	$30$	$150$
$6\leq x<10$	$3$	$8$	$24$	$192$
Total	$20$		$81$	$417$

Notice that the last midpoint is $8$, not $7$: the interval has width $4$.

The estimated mean is

$$\bar{x}\approx\frac{81}{20}=4.05\text{ kg}.$$

The estimated variance is

$$\operatorname{Var}(x)\approx\frac{417}{20}-4.05^2=4.4475,$$

so the estimated standard deviation is

$$\sigma_x\approx\sqrt{4.4475}=2.108\ldots\approx2.11\text{ kg}.$$

Frequency is not frequency density

In a histogram, bar area represents frequency. If class widths differ, bar height is frequency density, not frequency. Recover frequencies before calculating summary statistics. See histograms.

Self-check 2

Ten values lie in the classes below.

Class	$0\leq x<10$	$10\leq x<20$	$20\leq x<40$
Frequency	$2$	$5$	$3$

Estimate the mean.

Answer

The midpoints are $5$, $15$ and $30$. Therefore

$$\bar{x}\approx\frac{2(5)+5(15)+3(30)}{10} =\frac{175}{10}=17.5.$$

It is an estimate because the exact values within each class are unknown.

Quartiles and the interquartile range

After ordering the data, $Q_1$ marks approximately the $25\%$ position, the median $Q_2$ marks the $50\%$ position, and $Q_3$ marks the $75\%$ position. Then

$$\operatorname{IQR}=Q_3-Q_1.$$

Different accepted conventions can give slightly different quartiles for a small raw data set. In exam questions, follow the convention indicated by the data presentation or calculator guidance. For cumulative frequency data, quartiles are read at cumulative frequencies $n/4$, $n/2$ and $3n/4$, usually by interpolation from a graph.

Worked example 5: comparing distributions

Two machines fill packets. Their masses, in grams, have these summaries.

	Median	IQR	Mean	Standard deviation
Machine A	$501$	$8$	$503$	$7.1$
Machine B	$497$	$15$	$498$	$12.4$

Machine A has the larger typical packet mass because both its median and mean are larger. It is also more consistent because both its IQR and standard deviation are smaller.

A complete contextual comparison names the statistic, gives the direction, and interprets it:

Machine A has the smaller standard deviation, $7.1<12.4$, so its packet masses are less variable.

Do not say that Machine A is “better” without defining what better means. A larger centre could represent better performance in a test, but worse waiting times in a hospital.

Choosing suitable measures

Pair measures that respond similarly to extremes:

Shape or issue	Suitable pair	Reason
roughly symmetric data without extremes	mean and standard deviation	both use every observation
skewed data or possible outliers	median and IQR	both resist extreme observations
categorical or most common outcome	mode	mean and median may be meaningless

For the ordered salaries

$$22,\ 24,\ 25,\ 27,\ 29,\ 31,\ 180$$

in thousands of pounds, the mean is approximately $48.3$, which is not representative of most employees. The median is $27$. The large salary is genuine data, but it makes the median and IQR more informative summaries of a typical salary and its spread.

Read outliers and cleaning data before deciding that an unusual observation should be removed.

Linear transformations and coding

Suppose every observation is transformed by

$$y=a+bx.$$

Then

$$\bar{y}=a+b\bar{x},$$ $$\operatorname{Var}(y)=b^2\operatorname{Var}(x),$$

and

$$\sigma_y=|b|\sigma_x.$$

Adding $a$ shifts every value equally, so it changes location but not spread. Multiplying by $b$ scales all distances by $|b|$.

Worked example 6: changing units

Temperatures $x$ measured in degrees Celsius have mean $18$ and standard deviation $4$. Fahrenheit temperature is

$$F=32+\frac95x.$$

Therefore

$$\bar{F}=32+\frac95(18)=64.4$$

and

$$\sigma_F=\frac95(4)=7.2.$$

The $32$ changes the mean but has no effect on the standard deviation.

Worked example 7: reversing a code

Large values are coded using

$$y=\frac{x-100}{5}.$$

The coded data have mean $\bar y=3.2$ and standard deviation $\sigma_y=1.7$. Since $x=100+5y$,

$$\bar x=100+5(3.2)=116$$

and

$$\sigma_x=5(1.7)=8.5.$$

Effect on the IQR and range

Under $y=a+bx$, the range and IQR are also multiplied by $|b|$. Adding $a$ does not change either measure.

Self-check 3

The data $x$ have mean $12$ and variance $9$. Let $z=20-2x$. Find the mean, variance and standard deviation of $z$.

Answer

$$\bar z=20-2(12)=-4.$$

The multiplier is $-2$, so

$$\operatorname{Var}(z)=(-2)^2(9)=36$$

and

$$\sigma_z=|-2|\sqrt9=6.$$

A standard deviation cannot be negative.

Missing values from summary statistics

Summary statistics can be reversed to recover totals:

$$\sum x=n\bar{x}$$

and

$$\sum x^2=n\left(\operatorname{Var}(x)+\bar{x}^{,2}\right).$$

Worked example 8: combining two groups

Group A contains $12$ observations with mean $7.5$. Group B contains $8$ observations with mean $11$. The combined mean is not the simple average of $7.5$ and $11$, because the groups have different sizes.

The group totals are

$$\sum x_A=12(7.5)=90, \qquad \sum x_B=8(11)=88.$$

Hence

\bar{x}_{\text{combined}}=rac{90+88}{12+8}=8.9.

This is a weighted mean: the larger group contributes more.

Worked example 9: correcting an error

The mean of $20$ recorded values was calculated as $14.6$. One value had been entered as $42$ instead of $24$.

The recorded total was

$$20(14.6)=292.$$

Correcting the error gives

$$\sum x_{\text{correct}}=292-42+24=274.$$

Therefore the corrected mean is

$$\frac{274}{20}=13.7.$$

For a corrected variance, both $\sum x$ and $\sum x^2$ must be corrected. Replacing $42$ by $24$ changes the square total by $-42^2+24^2$.

Common misconceptions

• A larger mean does not imply a larger spread.
• Standard deviation is not the average value. It measures typical distance from the mean.
• $\sqrt{\operatorname{Var}(x)}$ is the standard deviation. Do not forget the square root.
• Adding a constant does not change standard deviation or IQR.
• Grouped-data calculations are estimates and should be described as such.
• A summary statistic should be interpreted in context, with units where appropriate.
• An outlier is not automatically an error and should not automatically be deleted.

Final self-check

A data set has $n=10$, $\sum x=80$ and $\sum x^2=730$.

1. Find the mean and standard deviation.
2. One further value, $20$, is added. Find the new mean.
3. Predict, without calculating the new standard deviation, whether the spread is likely to increase or decrease. Explain.

Answers

1. The mean is

   $$\bar{x}=\frac{80}{10}=8.$$

   The variance is

   $$\frac{730}{10}-8^2=9,$$

   so the standard deviation is $3$.

2. The new total is $80+20=100$ and the new number of observations is $11$. Thus

   $$\bar{x}_{\text{new}}=\frac{100}{11}=9.09\ldots$$

3. The spread is likely to increase. The added value $20$ lies far above both the old mean $8$ and the new mean $100/11$, so it contributes a large squared deviation. A full calculation confirms the increase.

Next steps

Use these summaries alongside presenting and interpreting data, then study histograms and outliers and cleaning data. Mean and standard deviation later become parameters of the normal distribution.