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Gaussian - symmetric case:

consider a statistically independent sample of N observations from a normal distribution.The mean (loc) of the population is the results on which you would like to make a prediction. You can report the value of the sample mean and as an error on it the standard error (SD/$\sqrt{N}$), i.e. $mean \pm SD/\sqrt{N}$.

Asymmetric case:

consider the same situation as before but with an exponential distribution. It sounds to me wrong to report an estimate of the error of the result, i.e. the mean, which is a symmetric $ \pm $, as in the case before.

Does it make sense to report $ \pm \,SD/\sqrt{N}$ in the case of a mean of an asymmetric distribution. If not, what is a choice that make sense as general as possible?

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    $\begingroup$ The distribution of the sample mean is far less asymmetric than the distribution of a single observation. $\endgroup$
    – whuber ♦
    Mar 6, 2022 at 18:32

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The best course is to report a confidence interval for the parameter you care about. For a lognormal distribution (also asymmetrical), the parameter usually reported is the geometric mean. For values sampled from an exponential distribution, I don't know what parameter makes most sense but it may not be the mean.

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    $\begingroup$ Fitting in with Harvey's answer, if you compute a two-sided confidence interval using the standard error on the wrong scale, you will get a symmetric confidence interval that has poor coverage in both tails. Asymmetric sampling distributions call for asymmetric confidence intervals and make standard errors not very relevant. $\endgroup$
    – Frank Harrell
    Mar 7, 2022 at 11:42

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