The typical for a sample I collected = $-60.75$. Its linked standard deviation is $179.44$. My concern is, if the range of values for this details sample is minimal such that NO X-VALUE above $0$, how deserve to the conventional deviation be this large? From mine understanding, a M = $-60.75$ and a SD = $179.4$ show that around $68\%$ of this sample ranges from $-240.15$ come $+118.65$. However, given the restriction detailed above, how can this it is in accurate? Is my understanding of exactly how these two descriptives function incorrect? give thanks to you.

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If the PDF the the variable is$$f(x)=\frac4\pi\fracx1+x^4\,$$where $<\cdots>$ room Iverson Brackets.

The average is therefore$$\frac4\pi\int_0^\infty\fracx^21+x^4\,\urbanbreathnyc.comrmdx=\sqrt2$$Yet the variance, for this reason the conventional deviation, is boundless since$$\frac4\pi\int_0^\infty\fracx(x-\sqrt2)^21+x^4\,\urbanbreathnyc.comrmdx$$diverges.

As with the distribution above, this indicates that the circulation is non-symmetric, a many the variance is added by the tail to the appropriate of the mean.


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The sigma dominance only hold if the variable is generally distributed! In your case, the is probably since it all lies in the negative values? just imagine a high variance distribution and also then change to the left (take far a big value). Or it have the right to have fat tail. I suggest plot the data in some software program as a histogram, or density approximation and also it will be clear.


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Why the typical deviation approximated through the range rule of ignorance much different from the typical deviation calculated straight from sample data?
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