I would certainly indicate you to recontact the formula for standard deviation.For circumstances, when we take the corrected sample typical deviation into account we understand 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 must take the square root of the above expression in order to find the traditional deviation and we understand that we cannot have actually a negative number inside the square root.

In addition, the #N# means the dimension of the sample (team of world, animals and so on.) 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 with having actually either zero or positive number as you need to square the distinctions from the suppose.

Thus the inside of square root will certainly be greater than or equal to zero and also we will finish up with having actually a non negative number for typical deviation so it does not make any type of sense to talk about the square root of an unfavorable number.

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Sep 22, 2015

It have to always be positive because the calculation is based on the square of a distinction - making it positive no issue what the distinction is.

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Oct 8, 2015



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

#x^2 = 1#

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

The answer to this, is no. Conventionally once taking the square root we only take the positive worth. The idea that a negative value appears come from a generally omitted action and/or a not extremely recognized fact.

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

So far so excellent, yet you check out, the definition of the absolute value attribute is #sqrt(x^2)#, so we have

#|x| = sqrt(a)#

And since we currently have an equation taking care of a modulo, we have to put the plus minus sign

#x = +-sqrt(a)#

But you watch, despite using #s# or #sigma# for typical deviation and also #s^2# or #sigma^2#for the variance, they came to be the other means around!

Standard deviation was characterized as the square root of variance and square roots are by convention always positive. Because we"re not utilizing the typical deviation as an unrecognized value, that plus minus sign will not display up.