Year 10 Maths - 10A Statistics

Standard Deviation

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Theory

The standard deviation measures how far, on average, the scores are from the mean. A small value means the scores cluster near the mean; a large value means they are spread out. This Year 10A maths page shows the population standard deviation $\sigma$ by hand and on a calculator, from lists and frequency tables, and how it reacts to outliers and to changing every score.

The deviation of a score is $x - \bar{x}$, its distance from the mean (negative below the mean). Deviations always add to zero, so they are squared before averaging.

The population standard deviation $\sigma$ (sigma) is the square root of the mean of the squared deviations. It has the same units as the data.

Unlike the range, it uses every score. Because deviations are squared, a score far from the mean (an outlier) increases $\sigma$ a lot.

Calculators show two values: $\sigma_x$ (or $\sigma_n$), the population standard deviation used on this page, and $s_x$ (or $\sigma_{n-1}$), the sample standard deviation, which is slightly larger.

The mean and standard deviation together summarise a data set: the mean gives its centre and $\sigma$ its spread. Comparing both is a good way to compare two data sets.

Each score's deviation is its distance from the mean. Standard deviation is a kind of average of these distances.

Both sets have mean $10$. The standard deviation tells them apart: small when scores cluster near the mean, large when they are spread out.

\[\sigma = \sqrt{\dfrac{\sum (x - \bar{x})^2}{n}}\]

\[\sigma = \sqrt{\dfrac{\sum f(x - \bar{x})^2}{\sum f}}\]

σ = \sqrt{\frac{\sum {(x - \bar{x})}^{2}}{n}}

Change to every score	Mean	Standard deviation
add $c$	$+\,c$	unchanged
multiply by $k$ ($k > 0$)	$\times\,k$	$\times\,k$
multiply by $k$, then add $c$	$k\bar{x} + c$	$k\sigma$

Grouped data. Use each class centre as \(x\) (for \(20 \le t < 30\), use \(25\)) with the class frequency. The answer is an estimate.

Unknown scores. The mean gives the total. If \(\sigma\) is given, \(\sum (x - \bar{x})^2 = n\sigma^2\): for \(n = 4\) and \(\sigma = 3\) the squared deviations add to \(36\). Set up the equation and solve.

Adding a score equal to the mean adds \(0\) to \(\sum (x - \bar{x})^2\) but increases \(n\), so \(\sigma\) decreases.

By hand

Mean: $\bar{x} = \dfrac{\sum x}{n}$ (or $\dfrac{\sum fx}{\sum f}$).
Table: columns $x$, $x - \bar{x}$, $(x - \bar{x})^2$ (times $f$ for a frequency table).
Add the last column, divide by $n$, take the square root.

With a calculator

Statistics mode (1-variable). Turn the frequency column on for a table.
Enter each score (and its frequency). Read the data from stem-and-leaf plots, dot plots or histograms first.
Read $n$ and $\bar{x}$ to check the entry, then $\sigma_x$.

Interpreting

Compare spreads: the larger $\sigma$, the more spread out (less consistent) the data. Without calculating, look for scores far from the mean.
Within one standard deviation: count the scores from $\bar{x} - \sigma$ to $\bar{x} + \sigma$, using unrounded limits.
Changes: adding a constant leaves $\sigma$ alone; multiplying (including percentage changes) multiplies it.

Example 1 — Standard deviation by hand

Find the population standard deviation of $3,\ 5,\ 6,\ 9,\ 12$ without the statistics functions of a calculator. Give the exact value and the value correct to $2$ decimal places.

Solution

First the mean: $\bar{x} = \dfrac{35}{5} = 7$. Then square each deviation.

$x$	$x - \bar{x}$	$(x - \bar{x})^2$
$3$	$-4$	$16$
$5$	$-2$	$4$
$6$	$-1$	$1$
$9$	$2$	$4$
$12$	$5$	$25$
	Total	$50$

$\sigma$	$=$	$\sqrt{\dfrac{\sum (x - \bar{x})^2}{n}} = \sqrt{\dfrac{50}{5}}$
	$=$	$\sqrt{10} = 3.162\ldots \approx \mathbf{3.16}$

σ = \sqrt{\frac{50}{5}} = \sqrt{10}

The deviations add to $0$ ($-4 - 2 - 1 + 2 + 5 = 0$), a useful check on the mean.

Example 2 — From a frequency table

$25$ students recorded how many times they bought lunch at the canteen in one week: $0$ times $(6)$, $1$ $(9)$, $2$ $(7)$, $3$ $(2)$, $4$ $(1)$. Find the mean and the population standard deviation, correct to $2$ decimal places.

Solution

Each value counts $f$ times, so multiply by the frequency.

$\bar{x}$

$=$

$\dfrac{\sum fx}{\sum f} = \dfrac{33}{25} = \mathbf{1.32}$

$x$	$f$	$fx$	$f(x - \bar{x})^2$
$0$	$6$	$0$	$10.4544$
$1$	$9$	$9$	$0.9216$
$2$	$7$	$14$	$3.2368$
$3$	$2$	$6$	$5.6448$
$4$	$1$	$4$	$7.1824$
Total	$25$	$33$	$27.44$

$\sigma$	$=$	$\sqrt{\dfrac{\sum f(x - \bar{x})^2}{\sum f}} = \sqrt{\dfrac{27.44}{25}}$
	$=$	$1.047\ldots \approx \mathbf{1.05}$

On a calculator: statistics mode with a frequency column; enter $x$ and $f$, then read $\sigma_x$ (or $\sigma_n$), not $s_x$.

σ = \sqrt{\frac{27.44}{25}} \approx 1.05

Example 3 — Changing every score

The daily maximum temperatures in a town over one month had mean $18^\circ$C and population standard deviation $4^\circ$C.
(i) Find the mean and standard deviation in degrees Fahrenheit, using $F = 1.8C + 32$.
(ii) Explain which part of the formula changes each one.

Solution

(i) Multiplying by $1.8$ stretches every distance from the mean by $1.8$. Adding $32$ shifts every temperature, and the mean, but changes no distances.

$\text{mean}$	$=$	$1.8(18) + 32 = \mathbf{64.4^\circ F}$
$\sigma$	$=$	$1.8 \times 4 = \mathbf{7.2^\circ F}$

(ii) The mean is changed by both steps. The standard deviation is changed only by the $\times 1.8$; the $+32$ has no effect on spread.

σ_{F} = 1.8 \times σ_{C}

Example 4 — Within one standard deviation, and an outlier

Nine scores are $4,\ 6,\ 7,\ 8,\ 8,\ 9,\ 10,\ 12,\ 17$.
(i) Find the mean and the population standard deviation, correct to $2$ decimal places.
(ii) How many scores lie within one standard deviation of the mean?
(iii) The score $17$ is removed. Find the new standard deviation and explain the change.

Solution

(i) $\bar{x} = \dfrac{81}{9} = 9$. Squared deviations: $25, 9, 4, 1, 1, 0, 1, 9, 64$, total $114$.

$\sigma$

$=$

$\sqrt{\dfrac{114}{9}} = 3.559\ldots \approx \mathbf{3.56}$

(ii) Use the unrounded value for the limits:

$9 - 3.559\ldots$	$=$	$5.44\ldots$
$9 + 3.559\ldots$	$=$	$12.55\ldots$

Scores from $5.44$ to $12.55$: $6, 7, 8, 8, 9, 10, 12$, that is $\mathbf{7}$ of the $9$ scores.

(iii) Now $\bar{x} = \dfrac{64}{8} = 8$ and the squared deviations total $42$.

$\sigma$

$=$

$\sqrt{\dfrac{42}{8}} = 2.291\ldots \approx \mathbf{2.29}$

The standard deviation falls a lot: $17$ was far from the mean, and squaring its deviation ($64$) made it the biggest contributor.

σ = \sqrt{\frac{42}{8}} \approx 2.29

Common pitfalls

Use the population key. This course uses $\sigma_x$ ($\sigma_n$). The $s_x$ key divides by $n - 1$ and gives a larger answer.

Square the deviations before adding. The plain deviations always add to $0$.

Divide, then take the square root. $\sqrt{\dfrac{50}{5}}$, not $\dfrac{\sqrt{50}}{5}$.

Check n after entering data. A missed or doubled score changes $\sigma$. Compare $n$ with the count.

Adding to every score does not change spread. A $\$1$ rise on every price leaves $\sigma$ the same; a $10\%$ rise multiplies it by $1.1$.

Frequently asked questions

What does standard deviation measure?

It measures how spread out the scores are around the mean. A small standard deviation means the scores are close to the mean; a large one means they are spread out.

How do you calculate standard deviation by hand?

Find the mean, subtract it from each score, square each result, add the squares, divide by the number of scores, then take the square root.

Which standard deviation key should I use on the calculator?

Use the population standard deviation, labelled sigma x or sigma n. The s x or sigma n minus 1 key gives the sample standard deviation, which divides by n minus 1.

What does within one standard deviation of the mean mean?

It means from the mean minus one standard deviation up to the mean plus one standard deviation. Work out both limits with the unrounded standard deviation, then count the scores between them.

What happens to the standard deviation if you add 5 to every score?

Nothing. Every score and the mean move up by 5, so the distances from the mean, and the standard deviation, stay the same.

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Change to every score	Mean	Standard deviation
add \(c\)	\(+\,c\)	unchanged
multiply by \(k\) (\(k > 0\))	\(\times\,k\)	\(\times\,k\)
multiply by \(k\), then add \(c\)	\(k\bar{x} + c\)	\(k\sigma\)

\(x\)	\(x - \bar{x}\)	\((x - \bar{x})^2\)
\(3\)	\(-4\)	\(16\)
\(5\)	\(-2\)	\(4\)
\(6\)	\(-1\)	\(1\)
\(9\)	\(2\)	\(4\)
\(12\)	\(5\)	\(25\)
	Total	\(50\)

\(\sigma\)	\(=\)	\(\sqrt{\dfrac{\sum (x - \bar{x})^2}{n}} = \sqrt{\dfrac{50}{5}}\)
\(\)	\(=\)	\(\sqrt{10} = 3.162\ldots \approx \mathbf{3.16}\)

\(x\)	\(f\)	\(fx\)	\(f(x - \bar{x})^2\)
\(0\)	\(6\)	\(0\)	\(10.4544\)
\(1\)	\(9\)	\(9\)	\(0.9216\)
\(2\)	\(7\)	\(14\)	\(3.2368\)
\(3\)	\(2\)	\(6\)	\(5.6448\)
\(4\)	\(1\)	\(4\)	\(7.1824\)
Total	\(25\)	\(33\)	\(27.44\)

\(\sigma\)	\(=\)	\(\sqrt{\dfrac{\sum f(x - \bar{x})^2}{\sum f}} = \sqrt{\dfrac{27.44}{25}}\)
\(\)	\(=\)	\(1.047\ldots \approx \mathbf{1.05}\)

\(\text{mean}\)	\(=\)	\(1.8(18) + 32 = \mathbf{64.4^\circ F}\)
\(\sigma\)	\(=\)	\(1.8 \times 4 = \mathbf{7.2^\circ F}\)

\(9 - 3.559\ldots\)	\(=\)	\(5.44\ldots\)
\(9 + 3.559\ldots\)	\(=\)	\(12.55\ldots\)

Standard Deviation

📖 Theory

By hand

With a calculator

Interpreting

Common pitfalls

Frequently asked questions

Theory