# Standard deviation and variance  (Page 3/3)

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 $\overline{X}$ Deviation $\left(X-\overline{X}\right)$ Deviation squared ${\left(X-\overline{X}\right)}^{2}$ 57 1 1 53 -3 9 58 2 4 65 9 81 48 -8 64 50 -6 36 66 10 100 51 -5 25 $\sum X=448$ $\sum x=0$ $\sum {\left(X-\overline{X}\right)}^{2}=320$

Note: The sum of the deviations of scores about their mean is zero. This always happens; that is $\left(X-\overline{X}\right)=0$ , for any set of data. Why is this? Find out.

Calculate the variance (add the squared results together and divide this total by the number of items).

$\begin{array}{ccc}\hfill \mathbf{Variance}& =& \frac{\sum {\left(X-\overline{X}\right)}^{2}}{n}\hfill \\ & =& \frac{320}{8}\hfill \\ & =& 40\hfill \end{array}$
$\begin{array}{ccc}\hfill \mathbf{Standard deviation}& =& \sqrt{\mathrm{variance}}\hfill \\ & =& \sqrt{\frac{\sum {\left(X-\overline{X}\right)}^{2}}{n}}\hfill \\ & =& \sqrt{\frac{320}{8}}\hfill \\ & =& \sqrt{40}\hfill \\ & =& 6.32\hfill \end{array}$

## Difference between population variance and sample variance

As with variance, there is a distinction between the standard deviation, $\sigma$ , of a whole population and the standard deviation, $s$ , of sample extracted from the population.

When dealing with the complete population the (population) standard deviation is a constant, a parameter which helps to describe the population. When dealing with a sample from the population the (sample) standard deviation varies from sample to sample.

In other words, the standard deviation can be calculated as follows:

1. Calculate the mean value $\overline{x}$ .
2. For each data value ${x}_{i}$ calculate the difference ${x}_{i}-\overline{x}$ between ${x}_{i}$ and the mean value $\overline{x}$ .
3. Calculate the squares of these differences.
4. Find the average of the squared differences. This quantity is the variance, ${\sigma }^{2}$ .
5. Take the square root of the variance to obtain the standard deviation, $\sigma$ .

What is the variance and standard deviation of the population of possibilities associated with rolling a fair die?

1. When rolling a fair die, the population consists of 6 possible outcomes. The data set is therefore $x=\left\{1,2,3,4,5,6\right\}$ . and n=6.

2. The population mean is calculated by:

$\begin{array}{ccc}\hfill \overline{x}& =& \frac{1}{6}\left(1+2+3+4+5+6\right)\hfill \\ & =& 3,5\hfill \end{array}$
3. The population variance is calculated by:

$\begin{array}{ccc}\hfill {\sigma }^{2}& =& \frac{\sum {\left(x-\overline{x}\right)}^{2}}{n}\hfill \\ & =& \frac{1}{6}\left(6,25+2,25+0,25+0,25+2,25+6,25\right)\hfill \\ & =& 2,917\hfill \end{array}$
4.  $\overline{X}$ $\left(X-\overline{X}\right)$ ${\left(X-\overline{X}\right)}^{2}$ 1 -2.5 6.25 2 -1.5 2.25 3 -0.5 0.25 4 0.5 0.25 5 1.5 2.25 6 2.5 6.25 $\sum X=21$ $\sum x=0$ $\sum {\left(X-\overline{X}\right)}^{2}=17.5$
5. The (population) standard deviation is calculated by:

$\begin{array}{ccc}\hfill \sigma & =& \sqrt{2,917}\hfill \\ & =& 1,708.\hfill \end{array}$

Notice how this standard deviation is somewhere in between the possible deviations.

## Interpretation and application

A large standard deviation indicates that the data values are far from the mean and a small standard deviation indicates that they are clustered closely around the mean.

For example, each of the three samples (0, 0, 14, 14), (0, 6, 8, 14), and (6, 6, 8, 8) has a mean of 7. Their standard deviations are 8.08, 5.77 and 1.15, respectively. The third set has a much smaller standard deviation than the other two because its values are all close to 7. The value of the standard deviation can be considered large' or small' only in relation to the sample that is being measured. In this case, a standard deviation of 7 may be considered large. Given a different sample, a standard deviation of 7 might be considered small.

Standard deviation may be thought of as a measure of uncertainty. In physical science for example, the reported standard deviation of a group of repeated measurements should give the precision of those measurements. When deciding whether measurements agree with a theoretical prediction, the standard deviation of those measurements is of crucial importance: if the mean of the measurements is too far away from the prediction (with the distance measured in standard deviations), then we consider the measurements as contradicting the prediction. This makes sense since they fall outside the range of values that could reasonably be expected to occur if the prediction were correct and the standard deviation appropriately quantified. (See prediction interval.)

## Relationship between standard deviation and the mean

The mean and the standard deviation of a set of data are usually reported together. In a certain sense, the standard deviation is a “natural” measure of statistical dispersion if the center of the data is measured about the mean.

## Means and standard deviations

1. Bridget surveyed the price of petrol at petrol stations in Cape Town and Durban. The raw data, in rands per litre, are given below:
 Cape Town 8,96 8,76 9,00 8,91 8,69 8,72 Durban 8,97 8,81 8,52 9,08 8,88 8,68
1. Find the mean price in each city and then state which city has the lowest mean.
2. Assuming that the data is a population find the standard deviation of each city's prices.
3. Assuming the data is a sample find the standard deviation of each city's prices.
4. Giving reasons which city has the more consistently priced petrol?
2. The following data represents the pocket money of a sample of teenagers. 150; 300; 250; 270; 130; 80; 700; 500; 200; 220; 110; 320; 420; 140.What is the standard deviation?
3. Consider a set of data that gives the weights of 50 cats at a cat show.
1. When is the data seen as a population?
2. When is the data seen as a sample?
4. Consider a set of data that gives the results of 20 pupils in a class.
1. When is the data seen as a population?
2. When is the data seen as a sample?

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