Negative Mean with Positive Standard Deviation
The mean for a sample I collected = $-60.75$. Its associated standard deviation is $179.44$. My question is, if the range of values for this particular sample is restricted such that NO X-VALUE exceeds $0$, how can the standard deviation be this large? From my understanding, a M = $-60.75$ and a SD = $179.4$ indicate that approximately $68\%$ of this sample ranges from $-240.15$ to $+118.65$. However, given the restriction noted above, how can this be accurate? Is my understanding of how these two descriptives function incorrect? Thank you.
$\endgroup$ 12 Answers
$\begingroup$If the PDF of the variable is $$ f(x)=\frac4\pi\frac{x}{1+x^4}\,[x\ge0] $$ where $[\cdots]$ are Iverson Brackets.
The mean is therefore $$ \frac4\pi\int_0^\infty\frac{x^2}{1+x^4}\,\mathrm{d}x=\sqrt2 $$ Yet the variance, hence the standard deviation, is infinite since $$ \frac4\pi\int_0^\infty\frac{x(x-\sqrt2)^2}{1+x^4}\,\mathrm{d}x $$ diverges.
As with the distribution above, this implies that the distribution is non-symmetric, a lot of the variance is contributed by the tail to the right of the mean.
$\endgroup$ $\begingroup$The sigma rule only holds if the variable is normally distributed! In your case, it is probably because it all lies in the negative values? Just imagine a high variance distribution and then shifted to the left (take away a large value). Or it can have fat tail. I suggest plot the data in some software as a histogram, or density approximation and it will be clear.
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