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I have a distribution of data with a positive skew, shown in the image. The standard deviation is 1.34 and the mean is 2.01. I want to illustrate on the graph the range of values that are within one standard devaiton of the mean with vertical lines. I believe that for a normal distribution I could simply minus the standard deviation from the mean and add the standard deviation to the mean to show the range, however I am unsure how to show this for a skewed distribution as it will not be equally distributed around the mean. Thanks

Data plotted in a scatter graph

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  • $\begingroup$ Correction: the standard deviation is 1.10 and the mean is 1.70 $\endgroup$
    – Jason Ron
    Oct 11, 2018 at 18:47
  • $\begingroup$ Is this a Poisson distribution? If it is not a normal distribution at all then it may not make sense to even discuss standard deviation. More info on calculating probabilities with Poisson in case it helps: intmath.com/counting-probability/… $\endgroup$ Oct 11, 2018 at 19:00
  • $\begingroup$ Could you please explain how this graph works? How, exactly, does this scatterplot represent a "distribution of data"? What do its axes represent? $\endgroup$
    – whuber
    Oct 11, 2018 at 19:07

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This probably isn't a very sensible thing to do with skewed distributions. In the Gaussian case, showing the $\pm \sigma$ range is the same (technically only very close) to showing the range which represents a $\frac{2}{3}$ probability of any given measurement falling in. Similarly, showing the $\pm 2\sigma$ range is quite close to showing a 95% confidence interval, it's a nice quirk of the Normal Distribution.

I would suggest that you might want to do things like shade an area either side of the mean, which carry equal probability (i.e. same area), and together sum to 66%, 90%, 95% etc, those are the more meaningful quantities.

Obviously this is harder to do for skewed distributions and will require some grid search over integrations usually. If you know the mean, and you want to plot the 66% confidence interval, you have to solve

$\int _{\mu}^{a}f(x)dx=0.33$

and

$\int_{b}^{\mu}f(x)fx=0.33$

numerically, essentially calculating these integrals loads of times over a grid-search.

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  • $\begingroup$ Thank you, your response makes sense and I'll instead show percentiles from the median $\endgroup$
    – Jason Ron
    Oct 11, 2018 at 19:15

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