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.

You are watching: Can the mean be negative  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. 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. Thanks for contributing solution to urbanbreathnyc.comematics ridge Exchange!

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