What is Standard Deviation Measuring?

jillst24

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Joined
Jul 7, 2010
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28
I have a set of data.
The Median is 214 days. The mean is 344.
I calculated the STDEVPA at 624 but don't really understand what it is telling me.

HELP?
 

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I am actually trying to understand what it is telling in me English...I understand the excel formula.

I am trying to figure out how long soemthing is.....i have the median days and the mean days and the ST DEV now...i am trying to summarize the data.
 
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If you have an engineering or applied maths background the (statistical) standard deviation is equivalent to the positive square root of the Moment of Inertia.

Otherwise its usual statistical meaning is to measure the "spread" of a set of data in the same units of measurement (kilograms, miles etc,) in which the original data are measured.

It is calculated by taking the (positive) square root of the average of the squared differences between each data value and the mean (or average) value of all data values in the relevant dataset.
 
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Standard Deviation is basically a measure of how much your data is varying. If every point of data you have was the exact same number, say all 26's, then you would have a standard deviation of 0 because there is no variance between the numbers. The more your data varies the larger the standard deviation will be.
 
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My rule of thumb is "Average distance from the average".

That is to say, your data set will have an average. There may or may not be any values that are equal to that average, so essentially you take the difference between that value and the average, and average those differences. That's your standard deviation, in very non pure mathematical terms.
 
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My rule of thumb is "Average distance from the average".

That is to say, your data set will have an average. There may or may not be any values that are equal to that average, so essentially you take the difference between that value and the average, and average those differences. That's your standard deviation, in very non pure mathematical terms.
Unfortunately this isn't correct. Not even very close.

If you modify it to:
"There may or may not be any values that are equal to that average, so essentially you take the squared difference between that value and the average, and average those squared differences, then take the positive square root of the result. That's your standard deviation ..."
then you'll be much closer to the mark.

Your rule of thumb could also do with a similar modification.

Check with Wikipedia or any statistical text if you have questions on this.
 
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