I hope no question is too simple. I am just learning, and I came across an expression that compared standard deviation and mean of a distribution. I thought that it makes no sense to compare them, meaning that no useful information can be obtained from it. Am I wrong?


closed as off-topic by Sycorax, Alexis, mdewey, Peter Flom Mar 27 at 10:45

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    $\begingroup$ Could you give the context in which they were being compared, and the exact nature of the comparison? A quote, and if possible a reference or link would help, if there is one. $\endgroup$ – Glen_b Mar 21 '13 at 22:15
  • $\begingroup$ I'm voting to close this question as off-topic because the OP has not answered questions asked. $\endgroup$ – Peter Flom Mar 27 at 10:45

You can look at $\sigma / \mu$ which is the coefficient of variation.

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    $\begingroup$ Could you help answer the question by explaining what "useful information" might be obtained from this ratio and what assumptions are needed for that to be the case? $\endgroup$ – whuber Mar 21 '13 at 19:01
  • $\begingroup$ In one of your comments you mention that this is about a Normal population. But even that is not enough to know exactly what use the coefficient of variation would be. For example, if the mean is close to 0, the CV will be huge. But so? If you then added (say) 100 to every value, the standard deviation would stay the same and the mean would be the old mean + 100 (that is, approximately 100) and, of course, the CV would be much smaller. So, you need to think about context to be able to say what the CV tells you. $\endgroup$ – Peter Flom Mar 21 '13 at 22:31

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