I know this is a pretty vague question but I was wondering if there was a rule of thumb relation between the mean and the standard deviation of a given set of values by which you can say "Something's not right".

I'm thinking something like: If the standard deviation is half the value of the mean --> raise a flag.

  • $\begingroup$ Except for some very specific contexts (ones where you know a lot about the process and the sd can or should only be in a limited range of values), there's no such rule. $\endgroup$ – Glen_b -Reinstate Monica Jun 11 '13 at 2:06

No, there is no such rule. Even if all values are positive, it all depends on the distribution. Additive shifts will change the relationship between the mean and the SD, but not change anything essential. E.g. take IQ. Right now, it's scaled so it has a mean of 100 and an SD of 15. But if you subtracted 50 from every score, the mean would be 50 and the SD would still be 15, but the amount of spread would be the same.

Beyond that, suppose we are measuring the heights of humans.

if we measure a sample of ALL humans (all ages) vs. adults (men and women) vs. men vs. basketball players, the mean will certainly change (each will be bigger than the one before), but the SD will change more (each will be smaller, I would bet).

If any values are negative, all bets are off.


I imagine that you are focusing on situations in which a variable is necessarily positive.

But note that (e.g.) standard deviation proportional to the mean is diagnostic of a situation in which the coefficient of variation is constant, which suggests that a gamma or lognormal distribution might be appropriate.

The "problem" if there is one is that high SD/mean may imply that you don't have a normal distribution, so methods based on that may not work well, in which case you move to different methods.

A better practice is to plot the data, not rely on this ratio. (It might for example be pulled up by a single outlier.)

(LATER) I agree with Peter Flom: if the zero is arbitrary, the ratio is not a helpful thing to calculate.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.