Measures of Variability (part 2)

In the last post I introduced the simplest of measures of variability, that of range. Range has its problems, though, so it isn’t the method of choice in most cases.

Mean deviation (or average deviation) is another type of variability measure. This one takes into consideration the distance the scores are from their own mean. A deviation is the mathematical difference (representing the distance) between any score and the mean of a data set, so taking the average of those distances should give us the mean deviation. But wait a minute! For the mean to even be the mean, half the distances would be below that mean while half the distances would be above it, so when we add up those distances we will wind up with a value of zero since those above and those below would cancel each other out. And we all know what happens when we divide the sum (which in this case is zero) by the number of scores to get the average. We’ll just wind up with zero again, which would indicate that our scores vary an average of zero. This is a problem, of course, since scores in a data set are going to vary at least a little.


Xi (scores)                 Xi – M  (deviations)  

10                               1.6

8                                  -0.4                    Mean Deviation =   S (Xi – M)   = 0 / 5 = 0

16                               7.6                                                                 n

5                                  -3.4                      This can’t be correct…

3                                  -5.4

42                               0                     

M = 8.4


One way around this is to add up the absolute values of the deviations instead, which gets rid of those pesky negative signs, and gives us the absolute total distance the scores are from their mean. Then, when we divide by the number of scores we get a reasonable value.


Xi                |Xi – M | 

10                   1.6

8                      0.4                     Mean Deviation =   S | Xi – M |   = 18.4 / 5 = 3.68

16                   7.6                                                                n

5                      3.4

3                      5.4

42                   18.4               

M = 8.4


Now we can see that these scores deviate from the mean an average of 3.68. But who cares, right?   Well, you would if your main goal is comparing the variability of scores between two sets of similar data. In fact, this is all the mean deviation is good for because the deletion of the negative signs changes things in such a way as to make the value useless in more complicated type of analyses. If you need to use the average distance the scores are from the mean in actual analyses, a much better measure of variability is standard deviation. I will explain this one next time.


Comments are closed