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Let be two samples collections A and B. Mean of the first is 5 and the second is 15. Standard deviation of the first is 2 and the second is 5.

Can we conclude something even though the two datasets have a different mean ?

Can we consider the ratio (standard deviation / mean) and conclude that A (0.4) is more widespread than B (0.33) ?

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    As long as the distributions have the same shape, I guess it makes sense...2012-09-17

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Your proposed ratio (std / mean) is called the coefficient of variation (http://en.wikipedia.org/wiki/Coefficient_of_variation). Since this is independent of the unit of measure of a sample/distribution, it is perfect for comparison of the dispersion.

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    Okay then $i$f you can assume a paramtric form you can compare the sample estimates of variance. If both distributions are normal there is the F test for equality of variance. In other situations there are robust and nonparametric tests for determining if there is a difference in scale between two distributions, This was discussed recently in another question.2012-09-18
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You can think of them the same way you normally would, unless the standard deviation is linked to the mean in some way (such as with the poisson or exponential distribution). For more intuition, note that if you center a variable by its mean, you haven't changed anything fundamentally about the variable, and the standard deviation is unchanged.