What is standard deviation? (was: Small question I would like answered?)

CubingDuck

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I suppose i sound like a noon, but I always see cube videos, and it is says standard deviation: something or other

What does that mean?
 
@Lord Voldemort: That's not technically correct. Conventionally, the individual distances form the mean are squared, then averaged, then the square root is taken of the end result.

In your example: sd=sqrt[((31-30)^2+(27-30)^2+(32-30)^2)/3]=2.16

The standard deviation is not the average distance from the mean. It is a measure of the average distance from the mean.
 
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If you want to be REALLY technical, the standard deviation is the root mean square of the deviations from the mean.

In statistics, "average" can mean any measure designed to give the "middle" of a set of values - median, mode, arithmetic mean (what we usually imply when we just say "mean"), geometric mean, root mean square, etc. So the standard deviation is one possible average of the deviations from the mean, but there are others.
 
Sorry about that. I deleted it as to not confuse anyone.
I have never formally studied statistics, just what I've heard before and what I could deduce from the name. Just curios though, would any problems arise if you were to calculate SD how I did it?
 
Easier definition than qqwref:

It just measures how far apart your data is, or the variability
Ex: a set of data that has numbers like 1 100000000 will have a big standard deviation while a set with elements 1 1 1 1 1 will have 0 standard deviation. Basically, the lower the number, the more consistent the data is.

@Lord Voldemort
It would help if you told us what way you did it.
This is the real method though:

Find the mean of the set of data.
Subtract the mean from each element of the data and square it.
Add all of the squared numbers up.
Divide by the number of elements
Take the square root
 
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*snip*
Find the mean of the set of data.
Subtract the mean from each element of the data and square it.
Add all of the squared numbers up.
Divide by the number of elements
Take the square root
And for the sample standard deviation (i.e. an estimate of the standard deviation of the entire populace based on a smaller cross-section of that population) Divide by one less than the number of elements in step 4.

Which is fun.

Just as a fun little experiment, I took the mean and standard deviation of my last 400 times in CalCube. I turn out to be an extremely normal cuber. Which is the first time I've ever been called normal, but at least I'm not fishy. Oh god... too many statistics puns in one post.
 
Standard Deviation is a way of measuring how much data points differ from each other. It can be used to verify scientific experiments. In cubing it tells you how consistent you are.

If you know the Standard Deviation and the average of your times, you can use that data to plot the probability of a time occurring. In other words, you can plot a theoretical distribution of your times.
 
@ moogra -
Instead of squaring, adding, and square rooting, I just added the absolute value of the difference, and obviously divided by the number of terms.
 
a measure of the dispersion of observations about the mean, sometimes limited to apply only to normal distributions, but more commonly used to refer to any distribution, in which case it is the same as root mean square
 
95% of all data (theoretically) will fall 3 Standard Deviations above and below the mean.

For example, if your average is 20.00 seconds and your standard deviation is .5 then 95% of all solves in your average will be 3(.5) seconds above or below 20.00 seconds.
 
The standard deviation is a measure of the variability of a random variable (i.e. your times). Your true standard deviation cannot ever be known, and in any case it should vary over time, but it can be estimated by the procedures already outlined. I would like to add, as one already has, that for a sample size of N, you should sum the errors squared and divide by N-1, rather than N, and then take the square root. This estimate will be unbiased, whereas division by N will yield an artificially low estimate.

Edit: Whoops, sorry. A small correction.

The estimate for SD given with N-1 instead of N is not unbiased, but its square is an unbiased estimator of the variance (which is the square of the SD). The one with N is the maximum likelihood estimator for SD. Nitpicking...

@ moogra -
Instead of squaring, adding, and square rooting, I just added the absolute value of the difference, and obviously divided by the number of terms.
Well, that's wrong unfortunately.

95% of all data (theoretically) will fall 3 Standard Deviations above and below the mean.
This depends on the distribution. It's about right for normal variables (if you add "within" before "3" and replace "3" with "2"), but solving times are only approximately normal.
 
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Those last two posts were enlightening.
@ TheBB - I was asking moogra why that method is wrong, but I think the problem was the way I defined SD. As I said, I've never formally studied statistics, that was something of an educated guess at best.
 
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