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### #1

Posted 14 March 2009 - 01:39 AM

### #3

Posted 14 March 2009 - 10:54 AM

Doing the calculation is not that hard.

If you understand what standard deviation is all about, the calculation even makes sense!

So let's see what it is we're doing in the calculation.

Start with a

**collection of numbers**.

You'd like to describe them somehow, perhaps to compare them with other collections of numbers.

It could be the heights of your classmates, or how much money they have in their pockets.

First thing you could do is find the average amount, called the

**mean**.

Add all the numbers, and get say $60.95.

There are

**23**classmates, say, so their average coin is $2.65. The mean is

**$2.65**.

What else might you want to know?

Does every one have the same amount? Probably not.

But if everyone doesn't have exactly $2.65,

**how much more or less do they have?**

Since some have more and some have less, the average of the differences is 0. Not helpful.

But if you square the differences first, you get positive numbers, and the average won't be zero.

So that's the idea.

The average of the

**square**of the differences from the mean gives you something called the

**variance**.

And its

**square root**gives you the

**standard deviation**.

There's only one catch:

When you take that last average, you don't use the number [23] of classmates, you use

**one less**: 22.

I won't go into why you do that, just believe: it gives a more meaningful result.

So here's the deal:

Standard Deviation:

Standard Deviation:

- Find the mean.
- Find the differences from the mean.
- Square the differences.
- Find the average of the squares [but use N-1 instead of N]
- Take the square root.

The

**mean**tells you the average amount of money owned by your classmates.

The

**standard deviation**tells you a kind of average of how much what they have differs from the mean.

For example, if the standard deviation were zero, everyone would have exactly $2.65.

If the standard deviation were huge, some of your classmates would have very little, and others quite a lot.

Hope that helps.

- bn

*Vidi vici veni.*

### #4

Posted 05 April 2009 - 06:15 PM

### #5

Posted 05 April 2009 - 07:26 PM

when I was in statistics I always wondered why we didn't use the mean of the absolute values of the differences from the mean, rather than the convoluted sdev formula. Does anyone know if there's some special advantage with the standard dev. formula that averaging the absolute-value-differences from the mean wouldn't accomplish?

A thought shared by me also... unfortunately its Easter holidays and I wont be able to ask any math teachers for a few weeks but Ill see what i can find out

### #6

Posted 07 April 2009 - 08:56 AM

when I was in statistics I always wondered why we didn't use the mean of the absolute values of the differences from the mean, rather than the convoluted sdev formula. Does anyone know if there's some special advantage with the standard dev. formula that averaging the absolute-value-differences from the mean wouldn't accomplish?

Just my thought, may not be correct.

I think the standard deviation is calculated that way because the statisticians believes most of the time the distribution, when the sample size getting bigger, will falls into the form of Normal Distribution, which is in a Symetrical Bell-Curved form, with mean at the center. Since curve is involved, the square and square root sure has to be involved in the calculation. So that 1 standard deviation give you 68% of the population, 2 sd gives you 95% and so on.

I am not sure but I believe if we use the mean of the absolute values of the differences from the mean, will result as treating the distribution in a triangle form instead of a bell curve form.

### #7

Posted 07 April 2009 - 10:54 AM

Just my thought, may not be correct.

I think the standard deviation is calculated that way because the statisticians believes most of the time the distribution, when the sample size getting bigger, will falls into the form of Normal Distribution, which is in a Symetrical Bell-Curved form, with mean at the center. Since curve is involved, the square and square root sure has to be involved in the calculation. So that 1 standard deviation give you 68% of the population, 2 sd gives you 95% and so on.

I am not sure but I believe if we use the mean of the absolute values of the differences from the mean, will result as treating the distribution in a triangle form instead of a bell curve form.

Im pretty sure this is what woon is talking about

Yeah, if we dont have the square and square root involved then it will be a triangle. I don't know when someone would actually use the curve, but ive just done an exam with SD in and UR's way is much easier (mostly because I was bored so did the SD questions in my head...)

### #8

Posted 07 April 2009 - 05:19 PM

### #9

Posted 07 April 2009 - 05:32 PM

Imagine you have a dice. If this is a fair dice, the chance of you to throw any number (1, 2,3, 4, 5 or 6) is equal = 1/6. So if you plot a bar graph with probability at y-axis, it is a 6 bar with the same height.

Let's say now we throw it twice and get the average of the number. You will then notice that number that you get is now 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5, 5.5, 6. Does all these number having equal probability anymore? The answer is no. For getting 1 and 6 the chance is 1/36 each, 1.5 and 5.5 are 1/18 each, 2 and 5 are 1/12, and so on, with 3.5 having the highest chance = 1/2. As you can see, the bar graph now change from flat one become triangle,

now try to throw it 3 times and get the average, what do you see the bar graph?

how about 4 times, 5 times, 10 times, 20 times and so on...

You will notice, as the sampling size getting bigger (for this case is the number of throws), the bar graph will becoming a bell-curve shape. That's the way it approaching normal distribution.

### #10

Posted 09 April 2009 - 12:56 AM

Excellent explanation, bonanova. I'm sure you know why we use n-1 but simply chose not to include it because an understanding of that is not necessary for a standard deviation problem. We use n-1 instead of n because we wish to know theWhen you take that last average, you don't use the number [23] of classmates, you use

one less: 22.

I won't go into why you do that, just believe: it gives a more meaningful result.

**degrees of freedom**.

The degrees of freedom the term representing how much of the data is changeable. If you know n-1 sets of data and you also know the mean, there is only one possible value for the last piece of data so there are n-1 degrees of freedom. For example, if three of four test scores were 92, 88, 95, and the mean test score was 92, we could vary any of those three test scores but the fourth test score must be 93. That test score is not free to be changed so instead of

**4**degrees of freedom, there are

**n-1 = 4-1 = 3**degrees of freedom.

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