---> It indicates the amount of clustering around the mean; showing how much the values are clustered allows the analysis of data to be taken much further than simply measuring central tendency.
In a normal tendency:-
- 68% of the values lie within +/- 1 Standard Deviation (SD) of the mean
- 95% of the values lie within +/- 2 SD of the mean
- 99% of the values lie within +/- 3 SD of the mean
---> A low SD indicates a more clustered distribution. A higher SD indicates a more spread-out/dispersed distribution.
Worked Example using Column B in AIB (Population density- persons per hectare)
4.42, 18.52, 16.73, 22.02, 38.44, 24.39, 36.64, 37.99, 23.04, 53.24, 33.66, 37.07, 34.97, 20.61, 27.76, 25.61, 26.04, 14.87
n=18
First thing to do is work out the mean, in the same way mentioned above: 495.9/18 = 27.55
Then, you need to subtract the mean, individually, from each of the values in the data set and then square the result. This can take a while but I find it easiest and quickest to do it in a little table....
x
|
4.42
|
18.52
|
16.73
|
22.02
|
38.44
|
24.29
|
36.64
|
37.99
|
23.04
|
(x-x-)2
|
535.00
|
81.54
|
117.07
|
30.57
|
118.59
|
10.63
|
82.63
|
109.00
|
20.34
|
x
|
53.24
|
33.66
|
37.07
|
34.97
|
20.61
|
27.76.
|
25.61
|
26.04
|
14.87
|
(x-x-)2
|
659.98
|
37.33
|
90.63
|
55.06
|
48.16
|
0.004
|
3.76
|
2.28
|
160.78
|
Next, you need to add up all the values on the bottom row of the table - the (x- xbar) squared results - and this gives you sigma (S)
S = 2163.4
Then divide this value by (n-1), remembering than n = number of samples/values
2163.4/18-1 = 127.26
Then square root the result to give you your SD value,
Square root of 127.26 = 11.28
---> This result indicates that 68% of the population densities for the different regions of Poole lie within +/- 11.28 of the mean (27.55), illustrating large disparity across Poole in terms of population density.
---> Ideally when doing Standard Deviation you should complete one for two sets of data and compare the results (like you do for the Rivers fieldwork where you complete standard deviation for sediment size in pool's and riffles then compare the results, before using knowledge of Hjulstrom and velocity changes you recorded to explain why results deviate accordingly). So, as an example for Poole, if you calculate the standard deviation for 'owner occupied dwelling as % of total dwellings' (Column C) and 'rented from council or Housing Association as % of total dwellings' (Column D) you would see that Column C displays greater deviation than Column D, suggesting that across Poole, houses rented from the council are move evenly distributed than those privately owned and occupied, then using the background knowledge of Poole you have, you could try to explain why this trend occurs.
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