Hey,
I'm looking for a webpage with simple worked examples of using the standard deviation equation. I can't find anything that explains all the math symbols in the equation.
Cheers.
If V is the variance, then V(X) = E(X)^2 - E(X^2) and the s.d. is the square root of the variance. If you have a probability function f(x) then E(g(x)) = integral f(x)g(x)dx, so E(x) = Integral f(x)x dx and E(x^2) = Integral f(x)x^2 dx.
If you are working with a discrete probability set, ie finitely many outcomes then the integrals become sums and f(x) becomes P(x), the probability of x occuring.
If you are working with a discrete probability set, ie finitely many outcomes then the integrals become sums and f(x) becomes P(x), the probability of x occuring.
If this isn't what you're looking for
http://en.wikipedia.org/wiki/Standard_deviation
at least it has links to other pages.
Always a good beginning, for any topic in which you aren't an expert.
http://en.wikipedia.org/wiki/Standard_deviation
at least it has links to other pages.
Always a good beginning, for any topic in which you aren't an expert.
I don't know about a website, but here's a basic, basic rundown.
The purpose of the standard deviation is to give us an estimate of how much a typical sample in a random distribution deviates from the mean (hence the name). Now, if one were to just go ahead and ask that question straight off, you might think that the thing to do would be the following:
dev = <|x - <x>|>
(where <> denotes an expectation value)
This would precisely be an estimate of the typical amount which any given sample deviates from the mean. But absolute value signs aren't very nice: they're discontinuous in the first derivative. So is there another way to arrive at a similar estimate without that ugly discontinuity? Certainly, we can just take the expectation value of the square instead:
variance = <(x - <x>)^2>
Of course, this has the wrong units, so we'll want to take the square root:
stddev = sqrt(<(x - <x>)^2>)
The question is, does this formula give what we want? Or should we weight it by some number? To answer that question, we turn to the most ubiquitous probability distribution around: the Gaussian distribution. I won't show the work here, but it's pretty easy to show that with a Gaussian distribution, the deviation and standard deviation are identical. So indeed, as long as our distribution is close to a Gaussian, the standard deviation measures how much your typical sample will deviate from the mean value.
What this means, fundamentally, is that when you make measurements, you should expect to get results which, on average, are one standard deviation away from the true value.
P.S. Note that the formula I wrote above isn't the way the standard deviation is usually written. This is because the above simplifies as follows:
<(x - <x>)^2>
= <x^2 - x<x> - <x>x + <x>^2>
= <x^2> + <x>^2 - 2<x>^2
= <x^2> - <x>^2
The purpose of the standard deviation is to give us an estimate of how much a typical sample in a random distribution deviates from the mean (hence the name). Now, if one were to just go ahead and ask that question straight off, you might think that the thing to do would be the following:
dev = <|x - <x>|>
(where <> denotes an expectation value)
This would precisely be an estimate of the typical amount which any given sample deviates from the mean. But absolute value signs aren't very nice: they're discontinuous in the first derivative. So is there another way to arrive at a similar estimate without that ugly discontinuity? Certainly, we can just take the expectation value of the square instead:
variance = <(x - <x>)^2>
Of course, this has the wrong units, so we'll want to take the square root:
stddev = sqrt(<(x - <x>)^2>)
The question is, does this formula give what we want? Or should we weight it by some number? To answer that question, we turn to the most ubiquitous probability distribution around: the Gaussian distribution. I won't show the work here, but it's pretty easy to show that with a Gaussian distribution, the deviation and standard deviation are identical. So indeed, as long as our distribution is close to a Gaussian, the standard deviation measures how much your typical sample will deviate from the mean value.
What this means, fundamentally, is that when you make measurements, you should expect to get results which, on average, are one standard deviation away from the true value.
P.S. Note that the formula I wrote above isn't the way the standard deviation is usually written. This is because the above simplifies as follows:
<(x - <x>)^2>
= <x^2 - x<x> - <x>x + <x>^2>
= <x^2> + <x>^2 - 2<x>^2
= <x^2> - <x>^2
Give this site a try:
standard-deviation.appspot.com/
standard-deviation.appspot.com/