Standard Deviation

**TigerTom** · May 29th 2007, 06:52 PM

I 'get' the standard deviation, and understand why it is useful, and *almost* why it is calculated like it is:

It is designed to show the average departure from the mean. But why is it not calculated thus:

My formula would be 'mean of the difference between x and mean(X)', which would logically be a good way to describe the spread of data, no?

I see the standard deviation comes out with a similar answer to my method, so what makes the real method more useful than mine? I just want to fully understand why it is done like it is.

Thanks in advance.

**ThePerfectHacker** · May 29th 2007, 07:17 PM

I think I have an excellent answer to this question.

One thing that used to bother me was something called the "quadradic mean".

As you know the "arithmetic mean" is computed as follows:
$\frac{x_1+x_2+...+x_n}{n} = \frac{\sum x }{n}$

Notice if all $x_1=x_2=x_3=...=x_n$ are all equal then the arithmetic mean is the same. As we expect. Because nothing changes.

Now there are many different ways to measure means, the arithmetic is a popular one. Another one is called "quadradic mean".
$\sqrt{\frac{x_1^2+x_2^2+...+x_n^2}{n}} = \sqrt{ \frac{\sum x^2}{n}}$ .

Look at what we are doing. We are squaring all the numbers. Then adding them together. Dividing them by how many numbers we had. And then finally taking the square root to undo the squares.

But you might ask, why not do this:
$\frac{\sqrt{x_1^2+x_2^2+...+x_n^2}}{n} = \frac{\sqrt{ \sum x}}{n}$ .

It also makes sense! You first square, then add then extract square root and then finially divide by the number of terms.

There is one main problem with the latter. Remember I said that in the arithmetic mean if all the numbers are the same then the result will be the same? The same with the quadradic mean. Here is a proof below.

Proof: If all the terms are non-negative and equal then the quadradic mean is the same. Simple:
$\sqrt{\frac{x^2+x^2+...+x^2}{n}} = \sqrt{ \frac{nx^2}{n}} = \sqrt{x^2} = x$ .

Exactly how we want it. Meaning if there is no change in the terms then that means should remain the same.

Now look what happens if we use your version of the quadradic mean. Will they all be the same?

$\frac{\sqrt{x^2+x^2+...+x^2}}{n} = \frac{\sqrt{nx^2}}{n} = \frac{x\sqrt{n}}{n} = \frac{x}{\sqrt{n}}$ .

Is that equal to $x$ . No! Not unless $x=0$ or $n=1$ .

My point is that if we use your version of the quadradic mean on numbers that are all the same we will end up with a smaller number. Which does not make so much sense because it should remain the same since all the numbers are unchanged.

This is how I remember the quadradic mean formula. That is whether the $n$ goes inside or outside the radical.

Same thing with the standard deviation. It looks almost like the quadradic mean thing.

**TigerTom** · May 29th 2007, 08:14 PM

Thanks - I learnt a lot from your post, and it allows me to express my question better (as you didn't quite answer it, or I missed it).

You mention this equation, then show it doesn't work to find the mean:

$ \frac{\sqrt{x_1^2+x_2^2+...+x_n^2}}{n} = \frac{\sqrt{ \sum x^2}}{n} $

But what I was proposing was more like this:

$ \frac{\sum \sqrt{x^2}}{n} $

But done on the difference of each x from the mean (all the squaring and rooting does is make the number positive):

$ \frac{\sum{\sqrt{({x} - \overline{x}})^2}}{n} $

So I am finding the arithmetic mean (positive) distance each value of x is from the mean of all values of x. Whereas the standard deviation seems to be using the 'quadratic mean' of these distances as it's measurement instead. What is the benefit of the standard deviation using the quadratic mean raher than the arithmetic mean?

**CaptainBlack** · May 29th 2007, 08:23 PM

Originally Posted by TigerTom

I 'get' the standard deviation, and understand why it is useful, and *almost* why it is calculated like it is:

It is designed to show the average departure from the mean. But why is it not calculated thus:

This last expression is the mean deviation, and is usualy written:

$ MD = \frac{1}{N}\sum_{i=1}^N |x_i-\bar{x}| $

and this is an important descriptive statistic that is not frequently used in
mathematical statistics. This is essentially because the introduction of the
absolute value makes analytical calculations using this statistic much more
complicated than the standard deviation.

RonL

**JakeD** · May 29th 2007, 08:32 PM

Originally Posted by TigerTom

I 'get' the standard deviation, and understand why it is useful, and *almost* why it is calculated like it is:

It is designed to show the average departure from the mean. But why is it not calculated thus:

My formula would be 'mean of the difference between x and mean(X)', which would logically be a good way to describe the spread of data, no?

I see the standard deviation comes out with a similar answer to my method, so what makes the real method more useful than mine? I just want to fully understand why it is done like it is.

Thanks in advance.

If I understand your formula, you're using

$\sqrt{(x - \bar{x})^2} = |x - \bar{x}|$

which is the absolute value of the difference $x - \bar{x} .$ That is a reasonable way to describe the spread. But the 'real' standard deviation has better statistical properties. It is one of the two parameters that specify a normal distribution (the other is the mean). And a big reason why the normal distribution is so important is the

Central Limit Theorem: Let $x_i,\i = 1,2,\ldots$ be a sequence of independent, identically distributed random variables each having mean $\mu$ and standard deviation $\sigma .$ Then for large $n$ , the sample mean $\bar{x}_n = \sum_{i=1}^n x_i /n$ has an approximate normal distribution with mean $\mu$ and standard deviation $\sigma / \sqrt{n}$ .

The Central Limit Theorem holds no matter what is the common distribution of the $x_i$ , so it is very general and the normal distribution applies to many settings. There is no such theorem that uses the mean absolute difference.

**TigerTom** · May 30th 2007, 06:12 AM

Originally Posted by CaptainBlack

This last expression is the mean deviation, and is usualy written:

$ MD = \frac{1}{N}\sum_{i=1}^N |x_i-\bar{x}| $

Ahhhh, ok. So it does exist, at least! I don't fully understand why it standard deviation is more helpful, but having read up on the mean absolute deviation, I've learnt a lot.

Originally Posted by JakeD

If I understand your formula, you're using

$\sqrt{(x - \bar{x})^2} = |x - \bar{x}|$

which is the absolute value of the difference $x - \bar{x} .$

Yes - I didn't know how to express it with mathematical notation. Thanks; getting stumped by notation is very frustating, especially when I would understand something if I knew what it meant.

Originally Posted by JakeD

That is a reasonable way to describe the spread. But the 'real' standard deviation has better statistical properties. It is one of the two parameters that specify a normal distribution (the other is the mean). And a big reason why the normal distribution is so important is the

Central Limit Theorem: <snip>

The Central Limit Theorem holds no matter what is the common distribution of the $x_i$ , so it is very general and the normal distribution applies to many settings. There is no such theorem that uses the mean absolute difference.

I only have a basic grasp of this part, but seeing there are properties the standard deviation has which empower it is useful and I think satisfies my question for the moment, thanks.

Thanks for all the help - it is really appreciated.

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