Derivation of standard deviation formula

[Bharti Singla]

Hello everyone!
Was just looking at the derivation of standard deviation formula. It has been derived very wisely; as the sum of all deviations always become zero, we got a way to deal with it is to squaring all the deviations and averaging them out and finally square root the result to get the 'average deviation'. But looking at the given steps below cause a little confusion.
In the second step, we have squared the term (x-xbar) only but in the fourth step(to undo the effect of squaring), we are taking the whole term summation(x-xbar)^2/n under square root i.e.we haven't squared the denominator but taking it for rooting. Why? Although I got it 'logically' that we are first calculating the 'average of squared deviations' and then by doing square root of it, we can get average deviation. But, not able to digest it mathematically!
If we are squaring a quantity, we should do square root of that quantity only to undo the effect.
Is there any way or any other proof to show it mathematically?

Any help will be much appreciated.

 

[Hemant Rupani]

I will try to explain with a mathematical example:-
Suppose, you have sample data with 4 outcomes.
Let their absolute deviation from mean are a,b,c,d.

We take sd ((a^2+b^2+c^2+d^2)/4)^0.5 ... (A)
You are thinking why not ((a^2+b^2+c^2+d^2)/4^2)^0.5 i.e. ((a^2+b^2+c^2+d^2)^0.5)/4 ... (B)

Let a=b=c=d. Now, ideally sd should be around same value as any deviance as all deviance are same.

Now, if you solve (A) & (B),
You'll get 'a' from (A) & 'a/2' from (B)

Hence, there is underestimation of sd from (B).
The sample size increases, so the underestimation from (B)