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#21
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It doesn't tell me anything specific, and I already know it;s a measure of the spread. It's like anwering "what's a flush" with "It's a type of poker hand"; or my question with "it's a statistical measurement". My point is I understand how mean deviation is measuring the spread - what exactly it's telling me. [/ QUOTE ] In what sense? You know how to compute it. What else do you know about it that makes you say that you "know exactly what it is telling you." eastbay |
#22
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You know how to compute it. What else do you know about it that makes you say that you "know exactly what it is telling you." [/ QUOTE ] I've been off so far, but I think what he is saying is that mean deviation is in some sense more natural or intuitive. The thing is, with every mathematical step you take, you are in danger of going beyond what is intuitive or natural. Start with the counting numbers. Probably these seem natural and intuitive enough. Add zero. This may seem natural too, but actually the idea didn't appear until maybe a few thousand years ago. Add negative numbers. Remember having trouble with these? Probably they now seem natural too. But then take the square root of -1. Does an imaginary number feel natural? Well, probably yes, if you are an electrical engineer. But sooner or later, you have to give up on intuition. Even for the most brilliant physicists, the tortured mathematical descriptions of reality they are forced to resort to in their grand unification schemes must seem far from intuitive. Maybe we can come up with a description of standard deviation that would make it feel natural. But maybe we just have to let it be unintuitive until we get used to it. |
#23
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"You know how to compute it. What else do you know about it that makes you say that you "know exactly what it is telling you."
I don't understand why you are doing what you do to compute it. |
#24
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"You know how to compute it. What else do you know about it that makes you say that you "know exactly what it is telling you." I don't understand why you are doing what you do to compute it. [/ QUOTE ] I'll give it a shot... First, we have to know what we are looking for before we can come up with a caculation for it. With something like the mean (average) we know we want ONE number that best represents the whole distribution of scores. The mean, or average score, is the best example of the distribution of scores if the sample size is large and normally distributed. When we ask for the standard deviation, we are asking for ONE number that represents "on average, how much does each score in this distribution deviate from the mean of the distribution?" So...we want a number that tells us the average deviation from the mean. Well, if the formula for the mean is equal to the sum of scores divided by sample size M = (SUM X / N) then the formula for the average (standard) deviation would intuitively be the sum of the deviations (from the mean) divided by the sample size, SD = (SUM dev / N). Well, that doesn't work because the sum of the deviations is zero. So we square the deviations to get rid of the zeros (SD = (SUM dev)sqrd / N). But that formula doesn't work because squaring the deviations will produce a standard devation that can be larger than any of the individual deviations that produced it. It doesn't make sense to say the average deviation from the mean is 6.4 when each score only deviated from the mean about 1 to 3 points. That is why we take the square root, which produces the final, correct, formula for standard deviation. SD = SQRT of ((SUM dev)sqrd / N) In short, you are calculating an average so you use the same basic formula for doing so....sum the scores and divide by the number of scores. But because we are calculating an average deviation from the mean, we have to manipulate the numbers a couple of times in order to get around a characterstic of those deviations (that they sum to zero). I have a feeling that you already know/knew all that [img]/images/graemlins/blush.gif[/img], but I don't know how else to answer the question. [img]/images/graemlins/confused.gif[/img] Griff |
#25
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All i know is i studied it for 3 months and got an 87% in the exam on it and i still cant remember how the hell to do it in a poker game. I think it is good if you know the principles of frequency distribution for poker becasue if u can understand all of that the standard deviation becomes much simpler.
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#26
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Thanks for your explanation. One of my questions was that when finding mean deviation you solve the negative problem by simply removing the negative signs, and why not just do that rather than squaring. Several people here have given some helpful answers tot hat. One of my friends who's a big maths boffin told me that it had something to do with pythagoras's theorem and some other stuff I didn;t really understand; anyone heard anything along those lines?
Anyway here's what I'm telling myself from now on: SD is a similar measure to MD but is calculated in a different way that gives it some more useful properties. Does that make sense? |
#27
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I also took a stat course, in one ear out the other, but got a B+ overall... so maybe not as aloof as I thought I was during those classes. SD and SE are not easy concepts to grasp in a practical sense. You'd be better off reading a book on this subject or even a textbook might help through it's examples and such.
I'm not sure how much it mentions it but I think "Getting the Best of it" (in the books section) has a section on probability in gambling, maybe go check that out. |
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