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I have some data that I strongly suspect are lognormally distributed, and I'd like to summarize the distribution using the mean and standard deviation. I've read that with lognormal distributions the goemetric mean and standard deviation should be used, but using them produces slightly strange results.

Specifically, when using the sample mean and standard deviation I calculate that 86% of my data lie within +/- 1 standard deviation about the mean.

However, when I use the geometric mean and standard deviation I calculate that only ~5% of my data lie within +/- 1 standard deviation about the mean.

I have over 30,000 pieces of data, and the distribution looks strongly lognormal, but I don't really understand these results. Is it valid to use the geometric mean and standard deviation here?

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    $\begingroup$ This paper is helpful, regarding concentration and dispersion measures for the log normal: "Log-Normal Distributions across the Sciences: Keys and Clues Author(s): Eckhard Limpert, Werner A. Stahel, Markus Abbt Source: BioScience, Vol. 51, No. 5 (May, 2001), pp. 341-352 at jstor.org/stable/1314039 $\endgroup$
    – Alecos Papadopoulos
    Commented Sep 2, 2014 at 16:08

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Chebyshev and similar +/- one sigma intervals refer to arithmetic mean and std, not to geometric ones. If you think your $X$ is lognormal, then construct an "arithmetic" -/+ one $\sigma$ interval for $log(X)$, then exponentiate it to get the CI for $X$.

In other words, you have zero chance of constructing an interval with the expected coverage the way you did because for lognormal such interval should not be symmetric around the mean. After you exponentiate, you'll see that the CI became asymmetric wrt the arithmetic mean of $X$.

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  • $\begingroup$ I got it wrong, let me edit the answer. $\endgroup$
    – James
    Commented Sep 2, 2014 at 17:25
  • $\begingroup$ Yep, that was it - I was doing exp(mean(log(data))) + exp(sd(log(data))) instead of exp(mean(log(data)) + sd(log(data))). $\endgroup$
    – DLaw
    Commented Sep 2, 2014 at 17:30

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