Standard Deviation & Variance Revision Notes for AQA A-Level Mathematics

2.1.3 Standard Deviation & Variance

Standard Deviation

This measure of spread measures, on average, how far each data point deviates from the mean of a set.

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Example: Find the standard deviation of $(2, 1, 3, 6)$

Calculate the mean:

\bar{x} = \frac{2 + 1 + 3 + 6}{4} = 3

Calculate how far each point deviates from the mean (best in tabular form):

\begin{array}{|c|c|c|} \hline x & x - \bar{x} & (x - \bar{x})^2 \\ \hline 2 & -1 & 1 \\ 1 & -2 & 4 \\ 3 & 0 & 0 \\ 6 & 3 & 9 \\ \hline \end{array}

Add to the table in $(x - \bar{x})^2$ column to make these deviations positive (see above in red).

To find the mean square deviation, we calculate:

\sigma^2 = \frac{\Sigma (x - \bar{x})^2}{n}\\

Where $\Sigma (x - \bar{x})^2$ is the sum of all squared deviations.
$n$ is the number of items. This quantity, which measures squared spread, is called the variance.

\sigma^2 = \frac{1 + 4 + 0 + 9}{4} = \frac{14}{4} = 3.5

Standard deviation is the square root of the variance: $\sigma = \sqrt{\sigma^2} = \sqrt{\frac{\Sigma(x - \bar{x})^2}{n}} = \sqrt{3.5} \approx 1.871 \, \text{(4sf)}$

This quantity, on its own, does not mean much. We need another standard deviation to compare it to, for it to make sense.

Summary of Formulae

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

$\overline{x^2}$ is the mean of all squared data.
$\bar{x}^2$ is the mean of all data then squared. Verifying formula (2) for our previous example (2, 1, 3, 6)****:

\begin{array}{|c|c|} \hline &x & x^2 \\ \hline &2 & 4 \\ &1 & 1 \\ &3 & 9 \\ &6 & 36 \\ \hline \Sigma & 12 & 50 \\ \hline \end{array}

\sigma^2 = \overline{x^2} - (\bar{x})^2 = \frac{50}{4} - \left(\frac{12}{4}\right)^2 = 3.5

\Rightarrow \sigma = \sqrt{3.5} = 1.871 \, \text{(4sf)}

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Question 2: (AQA SS1B, Jan 2007, Q1) The times, in seconds, taken by 20 people to solve a simple numerical puzzle were:

17, \quad 19, \quad 22, \quad 26, \quad 28, \quad 31, \quad 34, \quad 36, \quad 39, \quad 39 \\ 41, \quad 42, \quad 43, \quad 47, \quad 50, \quad 51, \quad 53, \quad 55, \quad 57, \quad 58

(a) Calculate the mean and the standard deviation of these times.

\bar{x} = 39.35, \quad \sigma = 12.36

(b) In fact, 23 people solved the puzzle. However, 3 of them failed to solve it within the allotted time of 60 seconds.

Calculate the median and the interquartile range of the times taken by all 23 people.

(Inputting 3 extra values bigger than 60):

\text{Median} = 42,\\ \quad \text{IQR} = 55 - 31 = 24

(c) For the times taken by all 23 people, explain why:

(i) The mode is not an appropriate numerical measure;

There is no mode.

(ii) The range is not an appropriate numerical measure.

We don't know the highest value, so the range is not possible to calculate.

Range	IQR	$\sigma$
Uses two data points	Uses two data points	Uses all data
Includes outliers	Ignores outliers	Includes extreme outliers
Easy to calculate	Relatively easy to calculate	More difficult to calculate

Standard Deviation & Variance (AQA A-Level Mathematics): Revision Notes

2.1.3 Standard Deviation & Variance