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November 12th, 2007, 12:20 PM   #1
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Radius of Gyration and Standard Deviation

In statics, the radius of gyration of a plane rectangular cross-section, for example, is the root-mean-square of the ratio of the distance, from the neutral axis, of each fiber to the distance of the extreme fiber, and in this case, ends up being 1.15 times as great as the mere mean. However, the meaning is not selected arbitrarily. The geometry of the section imposes that the distance to each fiber be factored in twice, hence the root-mean-square rather than the simple mean. So, anyway, it is the standard deviation of the distances.

I see the standard deviation used as a measure of dispersion in statistics, which I've never really studied, for example in IQ distributions. However, this lends added weight to the data points that are farther away from the mean. But why should the farther points carry more weight? I've never been able to fathom what the standard deviation tells me that the mere mean deviation wouldn't tell me. In other words, using the standard deviation in this context has always seemed to me an arbitrary practise. Can anyone explain to me any considerations that compel the choice, as they do in the case of the radius of gyration.
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November 12th, 2007, 05:44 PM   #2
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While the MAD is a useful statistic, it is difficult to use in analytical formulas describing probability distributions. I do not know how MAD could generate higher moments like continuing past squared to cubed deviations and so on. It is also debatable which better gives a proper weight to the tails of a distribution.
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November 14th, 2007, 06:06 AM   #3
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Thank you. You seem to be agreeing that the use of the standard deviation is a somewhat arbitrary choice, being perhaps preferable without being absolutely mandatory by the logic of the thing.
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February 23rd, 2017, 12:16 AM   #4
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When we ask for standard deviation, we need how close/far are the points located (squared average: standard deviation) from the mean and not how balanced (simple average) the whole system is!
Now why square and not cube/higher orders for that matter? That's simply because the distance formula mandates squares.
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February 23rd, 2017, 12:36 AM   #5
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When we ask for standard deviation, we need how close/far are the points located (squared average: standard deviation) from the mean and not how balanced (simple average) the whole system is!
Now why square and not cube/higher orders for that matter? That's simply because the distance formula mandates squares.
This post is 10 years old .
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February 23rd, 2017, 03:14 AM   #6
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This post is 10 years old .
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February 23rd, 2017, 03:21 AM   #7
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Nice!
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