L3StatisticsCore

Measures of Spread

Two data sets can share a mean and be nothing alike. $\left\{49, 50, 51\right\}$ and $\left\{0, 50, 100\right\}$ both average 50, but one is tightly clustered and the other is not. A measure of $\textbf{spread}$ captures the difference, and no summary of data is complete without one.

35 min Data Presentation

The big picture

Standard deviation is the measure that matters most, because it is the one the rest of statistics is built on. The normal distribution is defined by a mean and a standard deviation; hypothesis tests on a mean use it; standardising a value to a $z$ -score divides by it. The reason it wins over simpler alternatives is that it uses every data point while producing a number in the original units — the variance gets you there by squaring, and the square root brings you back. Understanding why those two steps exist stops the formula feeling arbitrary and makes the later work far easier.

What you'll be able to do

Calculate range, interquartile range and interpercentile range
Calculate variance and standard deviation from raw and grouped data
Use the computational form $\sigma^{2} = \frac{\Sigma x^{2}}{n} - \bar{x}^{2}$
Apply coding to standard deviation and explain why it ignores the shift

Range and quartiles

The $\textbf{range}$ is the largest value minus the smallest. It is trivial to compute and almost useless as a summary, because it depends on exactly the two most extreme values and ignores everything between them.

The $\textbf{interquartile range}$ (IQR) is $Q_{3} - Q_{1}$ , the spread of the middle half of the data. Because it discards the top and bottom quarters, an outlier cannot affect it — which pairs it naturally with the median.

For $n$ values, use position $\tfrac{n}{4}$ for $Q_{1}$ and $\tfrac{3n}{4}$ for $Q_{3}$ with grouped data, interpolating within the class as you did for the median. The $\textbf{interpercentile range}$ generalises this: the 10th to 90th percentile range, for instance, trims the extreme tenth from each end.

\text{range} = \max - \min

Uses only two values.

\text{IQR} = Q_{3}-Q_{1}

Spread of the middle 50%.

Example 1Find the range and IQR of

3, 5, 6, 8, 9, 11, 14, 20

1Range:

20 - 3 = 17

2With

n = 8

listed values,

Q_{1}

is the mean of the 2nd and 3rd:

\tfrac{5+6}{2} = 5.5

Q_{3}

is the mean of the 6th and 7th:

\tfrac{11+14}{2} = 12.5

AnswerRange

17

; IQR

= 12.5 - 5.5 = 7

Why this works

The range of 17 is dominated by the single value 20. The IQR of 7 describes where most of the data actually sits — the same contrast as mean versus median, for the same reason.

Variance: why the squares

The natural idea is to average how far each value sits from the mean. But the deviations $\left(x - \bar{x}\right)$ always sum to zero — that is what being the mean means — so their average is always zero and tells you nothing.

Squaring fixes it. Squared deviations are never negative, so they cannot cancel, and larger deviations are weighted more heavily. The mean of the squared deviations is the $\textbf{variance}$ .

The cost is units: if the data is in metres, the variance is in square metres. Taking the square root gives the $\textbf{standard deviation}$ , back in the original units and therefore directly interpretable.

\sigma^{2} = \dfrac{\Sigma\left(x-\bar{x}\right)^{2}}{n}

Variance from the definition.

\sigma = \sqrt{\sigma^{2}}

Standard deviation — original units.

Why this works

Taking absolute values instead of squaring would also stop the cancellation, and that statistic exists. Squaring is preferred because it is differentiable and behaves predictably under addition, which is what makes the whole theory of the normal distribution possible.

The computational form

Computing every deviation is laborious. Expanding the definition gives an equivalent form that needs only two totals — the sum of the values and the sum of their squares.

For frequency data the same rearrangement applies with $f$ weighting each term. This is the version to use in practice, and the one the formula sheet provides.

\sigma^{2} = \dfrac{\Sigma x^{2}}{n} - \bar{x}^{2}

Mean of the squares minus the square of the mean.

\sigma^{2} = \dfrac{\Sigma fx^{2}}{\Sigma f} - \bar{x}^{2}

Frequency version.

Example 1Find the standard deviation of

2, 4, 4, 6, 9

\Sigma x = 25

, so

\bar{x} = 5

\Sigma x^{2} = 4 + 16 + 16 + 36 + 81 = 153

\sigma^{2} = \dfrac{153}{5} - 25 = 30.6 - 25 = 5.6

\sigma = \sqrt{5.6}

Answer

\sigma = 2.37

(3 s.f.)

Tip — "Mean of the squares minus the square of the mean" — in that order. Reversing it gives a negative number, which is impossible for a variance and is a useful self-check.

Coding, and comparing data sets

Under the coding $y = \dfrac{x-a}{b}$ , the standard deviation scales but does not shift: $\sigma_{x} = b\,\sigma_{y}$ . The subtraction of $a$ slides every value along equally and so changes nothing about how spread out they are.

That asymmetry with the mean — which does shift — is examined directly, and it makes sense the moment you picture it: adding 10 to every value moves the whole distribution without stretching it.

When comparing two data sets, always compare location and spread together, and say what the comparison means in context rather than merely stating which number is bigger.

\sigma_{x} = b\,\sigma_{y}

Coding scales spread; the shift

a

does not appear.

Example 1Coded data with

y = \dfrac{x-200}{10}

has

\bar{y} = 4.5

and

\sigma_{y} = 1.2

. Find

\bar{x}

and

\sigma_{x}

1Mean:

\bar{x} = 10(4.5) + 200 = 245

2Standard deviation:

\sigma_{x} = 10(1.2)

— the 200 plays no part.

Answer

\bar{x} = 245

\sigma_{x} = 12

Tip — A comparison needs both measures and a context sentence: "Set B has a higher mean and a smaller standard deviation, so its scores are both higher and more consistent."

Think like an examiner

Give standard deviation in the original units; variance is in squared units and is rarely the requested answer.
State clearly which formula you are using, and show $\Sigma x$ and $\Sigma x^{2}$ separately.
When comparing data sets, address location and spread, and interpret in context.
Remember coding: the mean shifts and scales, the standard deviation only scales.

Common misconceptions

Computing

\bar{x}^{2} - \dfrac{\Sigma x^{2}}{n}

The order is mean of squares first:

\dfrac{\Sigma x^{2}}{n} - \bar{x}^{2}

. A negative variance signals this error.

Quoting the variance when the standard deviation was asked for.

Take the square root. Check the units the question expects.

Adding the shift

a

when decoding a standard deviation.

\sigma_{x} = b\sigma_{y}

only. A translation does not change spread.

Comparing two data sets using the mean alone.

Spread carries half the information. Compare both, and interpret.

Spread

\text{IQR} = Q_{3}-Q_{1}

Resistant to outliers.

\sigma^{2} = \tfrac{\Sigma x^{2}}{n}-\bar{x}^{2}

Computational form.

\sigma^{2} = \tfrac{\Sigma fx^{2}}{\Sigma f}-\bar{x}^{2}

With frequencies.

\sigma = \sqrt{\sigma^{2}}

Back to original units.

\sigma_{x} = b\sigma_{y}

Coding scales only.

Stretch yourself

A frequency table gives $\Sigma f = 40$ , $\Sigma fx = 520$ and $\Sigma fx^{2} = 7360$ . Find the mean and standard deviation. Then state what happens to each if every value is increased by 15.

Hint — Use the frequency forms. For the second part, think about whether a shift stretches the data.

Reveal solution

SolutionMean:

\bar{x} = \dfrac{520}{40} = 13

. Variance:

\sigma^{2} = \dfrac{7360}{40} - 13^{2} = 184 - 169 = 15

, so

\sigma = \sqrt{15} = 3.87

(3 s.f.). Adding 15 to every value shifts the whole distribution 15 to the right: the mean becomes

13 + 15 = 28

, but the standard deviation is

\textbf{unchanged}

3.87

. Every value has moved by the same amount, so every deviation from the mean is exactly what it was before — the data has been translated, not stretched. This is the coding rule

\sigma_{x} = b\sigma_{y}

with

b = 1

: only a multiplier affects spread.

\blacksquare

Questions students ask

Why square the deviations rather than take absolute values?

Both stop the cancellation, but squares are differentiable and combine predictably when distributions are added. That algebraic convenience is what the normal distribution and all the later inference depend on.

When should I use IQR instead of standard deviation?

When the data is skewed or contains outliers, and whenever you are quoting the median as the average. IQR and median go together; mean and standard deviation go together.

Can standard deviation be negative?

No. It is the square root of an average of squares, so it is always zero or positive. Zero means every value is identical.

Why does adding a constant leave the standard deviation unchanged?

Because it moves every value and the mean by the same amount, so all deviations from the mean are unaltered. Spread measures relative position, not absolute position.

Key takeaways

Range uses only the extremes; IQR describes the middle half and resists outliers.
Deviations from the mean sum to zero, so variance squares them before averaging.
Standard deviation is the square root of variance, returning the original units.
Use $\sigma^{2} = \tfrac{\Sigma x^{2}}{n} - \bar{x}^{2}$ in practice.
Coding scales the standard deviation but a shift leaves it unchanged.