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Standard Deviation

\displaystyle \sigma ^2 = \frac{1}{N} \sum_{i=1}^N (x_i - \overline{x})^2.


Or, more simply:

\displaystyle \sigma ^2 = \frac{1}{N} \sum (x - \overline{x})^2.


  • \sigma = 'standard deviation'.
  • \sum mean 'the sum of'.
  • \overline{x} = 'the mean' (said 'x bar').
  • N is the number of peices of data you have.
  • i runs from 1 to N as a label for each piece of data.


The standard deviation measures the spread of the data about the mean value. It is useful in comparing sets of data which may have the same mean but a different range. For example, the mean of the following two is the same: 15, 15, 15, 14, 16 and 2, 7, 14, 22, 30. However, the second is clearly more spread out. If a set has a low standard deviation, the values are not spread out too much.

Example

Find the standard deviation of 4, 9, 11, 12, 17, 5, 8, 12, 14 First work out the mean: 10.222 Now, subtract the mean individually from each of the numbers in the question and square the result. This is equivalent to the (x - \overline{x})^2 step. x refers to the values in the question.


x 4 9 11 12 17 5 8 12 14
(x - \overline{x})^2 38.7 1.49 0.60 3.16 45.9 27.3 4.94 3.16 14.3


Now add up these results (this is the 'sigma' in the formula): \sum (x - \overline{x})^2 = 139.55

Divide by N. N is the number of values, so in this case is N = 9: \frac{1}{N} \sum (x - \overline{x})^2 = 15.51

And finally, square root this: \sigma = 3.94


The standard deviation can usually be calculated much more easily with a calculator and this is usually acceptable in exams. With some calculators, you go into the standard deviation mode (often mode '.'). Then type in the first value, press 'data', type in the second value, press 'data'. Do this until you have typed in all the values, then press the standard deviation button (it will probably have a lower case sigma on it). Check your calculator's manual to see how to calculate it on yours.


NB: If you have a set of numbers (e.g. 1, 5, 2, 7, 3, 5 and 3), if each number is increased by the same amount (e.g. to 3, 7, 4, 9, 5, 7 and 5), the standard deviation will be the same and the mean will have increased by the amount each of the numbers were increased by (2 in this case).


Frequcied Data

When dealing with data such as the following:

x f
4 9
5 14
6 22
7 11
8 17


the formula for standard deviation becomes:


\displaystyle \sigma ^2 = \frac{\displaystyle \sum_{i=1}^N f_{x_i}(x_i - \overline{x})^2}{\displaystyle \sum_{i=1}^N f_{x_i}}}.


Or, more simply:

\displaystyle \sigma ^2 = \frac{\sum (x - \overline{x})^2}{\sum f}.


Where:

  • f_{x_i} = frequency of x_i.
  • N = number of different values the data takes.
  • \displaystyle \sum f = \sum_{i=1}^{N} f_{x_i} = the total frequency = total number of pieces of data.

Example

Try working out the standard deviation of the above data. You should get an answer of 1.32 .


Comments

I'm unsure of the level of these notes - are they needed for Higher GCSE or just GCSE statistics? Probably both.

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