WHAT DOES A NEGATIVE STANDARD DEVIATION MEAN

I would indicate you to recall the formula for standard deviation.For circumstances, when we take the corrected sample typical deviation right into account we know that;

#s = sqrt(1 /(N-1)sum_(i=1) ^N(x_i-bar x)^2 #

Standard Deviation

As you deserve to watch, you must take the square root of the over expression in order to find the typical deviation and also we recognize that we cannot have actually an unfavorable number inside the square root.

In addition, the #N# means the size of the sample (team of civilization, pets etc.) which is a positive number and if you expand the second part of the expression #sum_(i=1) ^N(x_i-bar x)^2# it is clear that you"ll finish up via having actually either zero or positive number as you have to square the distinctions from the mean.

Thus the inside of square root will be greater than or equal to zero and we will certainly end up through having a non negative number for traditional deviation so it does not make any kind of feeling to talk about the square root of a negative number.


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SCooke
Sep 22, 2015

It should constantly be positive because the calculation is based upon the square of a distinction - making it positive no issue what the distinction is.

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Lovecraft
Oct 8, 2015

No.

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Explanation:

I feel the others are going somewhere a little different below, in which they"re explaining why the variance deserve to never before be negative, but as we all know

#x^2 = 1#

Has 2 answers, #-1# and #1#, which have the right to raise a question much prefer your own, have the right to square roots be negative?

The answer to this, is no. Conventionally as soon as taking the square root we just take the positive value. The idea that a negative worth shows up come from a generally omitted step and/or a not exceptionally known fact.

#x^2 = a##sqrt(x^2) = sqrt(a)#

So much so great, however you check out, the meaning of the absolute value feature is #sqrt(x^2)#, so we have

#|x| = sqrt(a)#

And because we now have actually an equation handling a modulo, we need to put the plus minus sign

#x = +-sqrt(a)#

But you view, despite making use of #s# or #sigma# for traditional deviation and #s^2# or #sigma^2#for the variance, they happened the various other way around!

Standard deviation was identified as the square root of variance and square roots are by convention always positive. Because we"re not utilizing the traditional deviation as an unwell-known value, that plus minus authorize won"t show up.