I would imply you come recall the formula for traditional deviation.For instance, as soon as we take it the corrected sample traditional deviation into account we know that;

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

Standard Deviation

As you have the right to see, you should take the square root of the over expression in stimulate to discover the standard deviation and we recognize that us cannot have a negative number within the square root.

In addition, the #N# means the dimension of the sample (group of people, animals etc.) which is a positive number and also if you broaden the second part of the expression #sum_(i=1) ^N(x_i-bar x)^2# it is clear the you"ll finish up with having actually either zero or hopeful number as you have to square the distinctions from the mean.

Thus the within of square root will certainly be greater than or equal to zero and we will finish up with having a non an adverse number for conventional deviation so the doesn"t make any kind of sense to talk around the square root of a negative number. SCooke
Sep 22, 2015

It must constantly be positive due to the fact that the calculation is based upon the square of a difference - making it hopeful no issue what the difference is.

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Oct 8, 2015

No.

Explanation:

I feel the others are going what a bit different here, in which they"re explaining why the variance have the right to never be negative, however as we all know

#x^2 = 1#

Has two answers, #-1# and #1#, which can raise a question much like her own, have the right to square roots be negative?

The answer come this, is no. Conventionally when taking the square source we only take the confident value. The principle that a negative value shows up come from a frequently omitted action and/or a not very known fact.

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

So far so good, however you see, the an interpretation of the absolute value role is #sqrt(x^2)#, so us have

#|x| = sqrt(a)#

And since we now have actually an equation managing a modulo, we must put the plus minus sign

#x = +-sqrt(a)#

But you see, regardless of using #s# or #sigma# for conventional deviation and also #s^2# or #sigma^2#for the variance, they came to be the other way around!

Standard deviation was characterized as the square root of variance and also square roots room by convention always positive. Since we"re not making use of the traditional deviation together an unknown value, the plus minus authorize won"t show up.