I would suggest you to recall the formula for typical deviation.For instance, once we take the corrected sample traditional 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 have the right to see, you have to take the square root of the above expression in order to uncover the traditional deviation and also we know that us cannot have actually a negative number inside the square root.

In addition, the #N# means the size of the sample (group that people, pets etc.) i beg your pardon is a hopeful number and also if you expand the second component of the expression #sum_(i=1) ^N(x_i-bar x)^2# it is clear the you"ll end up with having either zero or hopeful number as you have to square the distinctions from the mean.

Thus the within of square root will certainly be higher than or equal to zero and we will finish up with having actually a non negative number for typical deviation so the doesn"t make any type of sense come talk around the square source of a an adverse 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 - make it optimistic no issue what the difference is.

You are watching: Can the standard deviation ever be negative Lovecraft
Oct 8, 2015

No.

Explanation:

I feel the others room going what a bit different here, in i beg your pardon they"re explaining why the variance can never it is in negative, yet as us all know

#x^2 = 1#

Has two answers, #-1# and also #1#, which deserve to raise a question much like her own, deserve to square roots be negative?

The answer to this, is no. Traditionally when acquisition the square root we just take the positive value. The concept that a an unfavorable value appears come native a generally omitted action and/or a not very known fact.

See more: Which Of The Following Is An Advantage Of Cmos Logic Devices Over Ttl Devices?

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

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

#|x| = sqrt(a)#

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

#x = +-sqrt(a)#

But girlfriend see, in spite of using #s# or #sigma# for standard deviation and also #s^2# or #sigma^2#for the variance, they came to be the other method around!

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