Behavioural Genetic Interactive Modules



Variance is a key concept in the measurement of individual differences. The aim of much behavioural genetic research is to uncover the components of variance in a trait in order to understand what makes some individuals differ from other individuals. This module shows how the variance in a sample of observations is calculated and how it can be used to standardise measurements.


The Windows module, variance.exe, allows a number of data points on a single measure to be tabulated and for the variance calculated. Upon opening the module, you will see a window with several panels. The top left panel provides three different ways of entering data: inputing specific numbers by clicking on Enter values, generating random numbers, by clicking on Random values, or loading numbers from a file, Load from file.

The following pages contain example files of well different distributions that can be loaded into the demonstration:


Each value entered is recorded and tabulated as seen in the screenshot to the left. This shows 36 values have been entered (N=36), of which we can currently see the first nine.


These values are automatically plotted in the form of a bar chart. (Note that the scale of the chart is fixed from 0 to 100, so if only very small numbers are entered, the chart will look skewed).


At the bottom of the column of numbers, the sum of the variable, X, is calculated. Dividing this by the number of observations, N, gives the mean, which is shown below (the X with a bar above it).

The figures to the right represent the deviations from the mean, in the first column, and the squared deviations from the mean. Take, for example, the first row: the data point is 28. The mean is 51.39 for all 36 points, so the deviation from the mean for this point is 28 - 51.39 = -23.39. And -23.39 squared is 547. This calculation is repeated for every data point.


The deviations from the mean will always sum to zero (by definition). This can be seen in the figure to the left. The quantity of interest is the sum of the squared deviations. This is 29982.6.


By simply taking the average of the square deviations (i.e. dividing the sum by the number of observations) we obtain the variance. For technical reasons, it is common practice to subtract 1 from the number of observations in the sample (because the same sample was also used to calculate the mean, and this introduces a slight bias). So here we see that the variance is 29982.6 divided by 35 (36-1) which equals 856.64. The standard deviation is simply the square root of the variance, and is given below. Both these quantities are very useful in the further analysis of the trait.


Another use of these statistics is to standardise scores. Standard scores always have a mean of zero and a standard deviation of 1. We can see how standard scores are calculated here: each score is expressed as a deviation from the mean and then divided by the standard deviation for the sample. Twenty-eight therefore is re-expressed as -0.7991 ( (28 - 51.39) / 29.27 = -0.7991).








  1. Why will variances always be positive numbers?



  2. Which set of numbers have i) the greater mean, ii) the greater variance ?


    • A: 7,34,23,14,28,2,41,32,12
    • B: 67, 62, 65, 66, 62, 71, 67
    • C: 57, 52, 55, 56, 52, 61, 57











  1. Because variances are calculated from the squared deviations from the mean, they will always be positive. That is, the square of any number, whether positive or negative, will always be positive itself. The average of a set of positive numbers will also be positive itself.


  2. Set A has a smaller mean but a larger variance. Set A has a mean of 21.44 but a variance of 178.53. Set B has a mean of 65.71 but a variance of 9.9. As you might have noticed, Set C is similar to Set B, except each number is smaller by exactly 10. The mean is also 10 units smaller (55.71) - but the variance is the same (9.90). This makes sense: the variation about the set mean is the same. That is, looking at the values entered in the module, the deviations from the mean would have been identical in both cases (i.e. 57 - 55.71 = 67 - 65.71).

Please refer to the Appendix for further discussion of what variances represent and how they are used.

Site created by S.Purcell, last updated 20.05.2007