I don't get your problem. It seems (but you do not explicitly say this) that you think of adding and subtracting the SD to/from the mean to get some "upper end" and "lower end". But it's unclear why you would (or should) do this. Often, when the variable is assumed to have a normal distribution, the range of mean ± SD is an approximate 64% prediction interval, and mean ± 2SD is an approximate 95% prediction interval. But this only makes sense when the distribution is (approximately) normal.

Very obviously, the normal distribution is a very bad model to describe the distribution of your variable. Therefore, thinking about mean ± SD as some reasonable prediction intreval makes no sense (as you correctly noted will such an interval cover negative values that are impossible).

This is not the fault of or a problem with the SD. The SD is the square-root of the variance. This is it. Nothing less and nothing more. It is just not a very useful measure of dispersion for your variable, because the normal distribution is not a very useful model for the distribution of your data.

You may still report the mean and the SD, but this will give a wrong impression about the distribution of your variable and likely be misinterpreted by readers. If you want to give a summary of the values you observed, better give the median and the interquartile range (the first and the third quartile).

If you want to make some inferential statistics, be aware that the variable is not normal distributed. Use a model assuming a more reasonable distribution (e.g. gamma or log-normal) or transform the data so that the transformed data is (approximately) normal distributed.

I believe that there is an error in calculation, in using data, please repeat your calculation precisely, and you probably need some one to check your work with you, Regards.

Hello Elseline Smith

I agree with Dr Jochen Wilhelm that often we assume the data has a normal distribution but often it does not. I have also run into this problem with small sample sizes.

Medhat Elsahookie how can you write this, aber Jochen Wilhelm precisely described the issue?

Rainer Duesing , I am sure that some people writing answers never read the other answers than have been already given :(

The standard deviation is a heartsink statistic. Most people don't look at it because they know they don't understand it, and among the people who think they do, it is clear that most don't. Their understanding is merely repeating something about ±2SD.

If you want to describe the dispersion of your data, I recommend quantiles such as quartiles. Better still, since an actual numeric value is seldom crucial to understand dispersion, graphical methods are often the best choice, allowing easy comparison of groups. Plotting the actual data with overlaid boxplots is useful, or even raincloud plots for added density bling!

Ronán Michael Conroy , I already used them, but i didn't know that they are called "raincloud plots" :) Thank you!

Note that "violin plots", which are essentially "mirrored raincloud plots", are a sometimes better alternative to boxplots. And there are also ridgeline-density plots do show and compare a couple of densities (see https://r-charts.com/en/distribution/ggridges_files/figure-html/ggridges.png *).

And I advocate not to overlay or add summary indicators like boxplots, more simply mean/SD od median/IQR or similar, and only show the data wherever possible, like with plain an simple "beeswarm plots" (I really like this name).

I'll take this opportunity to warn against using box plots, violin plots or raincloud plots (or similar) for small samples (n

Jochen Wilhelm – point well made about tiny samples, which is why I always try to show the data, jittered or, in the case of integer data, possibly stacked. The trouble is that in health research we often have to work with the data available, and some indication of distribution shape is better than none. I used to work on one of the world's largest studies of a rare ovarian condition, N=18! I'd still be waiting if we held out for another 82 patients to give us a margin of error of ±10%!

Here is an example for 10 independent samples from an exponential distribution (lambda=1), with n=18 each, represented as ridge density plot. The impressions are anything, from "almost normal" over "unimodal skewed", "bimodal" to "multimodal", with or without "outliers". Pick any of them as "your sample" - it gives you a wrong impression in any case. None of the densities hints that the distribution is actually exponential.

Jochen, in your simulation above that is not the distribution of the means from the specific exponential distribution right?

That's correct. It's showing the kernel density estimates of the sampled data.

I see, thanks for the clarification. But I thought because of the Central Limit Theorem, even if your data's distribution is not normally distributed, its means will be normally distributed. I see in your graphs (which are very cool btw) that this doesn't seem to be case, can you clarify what am I missing?

The data are from an exponential distribution. The plots show the (kernel-density estimates of the) data. Nothing is or needs to be normal distributed here. At no point I mentioned samples means. Sorry if I didn't make myself clear in the last post.

Maybe a part of your confusion comes from the "kind-of-bell-shape" looks of these density curves. This is because a Gaussian kernel was used to build the density estimates (other smooth kernels would produce similar looking smooth shapes). I used this because this is the standard when making raincloud, ridged density, or violin plots. Please note that the curves in the figure all start at negative values (at around -1) - but we know that this is outside the support of an exponential random variable.

Using an exponential kernel is not possible (to my knowledge) because it has a "hard" lower bound. However, such kernel density estimates are used to get an impression about what the distribution could be - not to apply a rule for a distribution that is already known.

It is possible to have a standard deviation that is bigger than the mean when the data is negatively skewed. Negative skewness means that the tail of the distribution is longer on the left side than on the right side. This can lead to negative values in the data.

In a normal distribution, standard deviations larger than the means do not cause any problem in and of themselves. However, in a normal distribution, this would imply many negative values. As soon as you have at least two numbers in the data set which are not exactly equal to one another, standard deviation has to be greater than zero – positive 1.