Can be derived (directly from the sample data sample average deviation of the average value),
Such an average value taken from the (expected value of the variance) is also not quite,
Imagine, for example, a sample refers to linear growth, may take a value to each of the entire value range,
So always have the same expectations and a sample of the overall sample,
After all the samples are then taken variance with the expectations of the overall sample,
Always have one of: ((sample a) and (expected value of the total sample) difference) is equal to 0,
Well, the last true variance can not contain the expected value of the items is 0,
Therefore, (the difference between the desired and the n samples of the total sample) is the number of non-zero values only :( n-1 th sample value in the case of linear growth)
We therefore expect only the sample variance ((n samples of the desired overall sample difference) squared) / (n-1);
This example just to help memory, and not wrong, in most cases can still clearly remember the concept and be able to correctly apply;
Focusing on (correct) understanding (concept), but also takes the case.