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Old 2012-01-25, 01:16   #1
roger
 
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Default Standard deviation for twin prime data

I tried to do a standard deviation list to see how the variance changes as does n, but I can't seem to get it right. It's been a long time since I did this high school math.

Maybe I've gone wrong somewhere:

I'm using the square root of the variance as the standard deviation.
Calculating the range of one standard deviation should be simply the average - std.dev, and average + std.dev, right?
Unfortunately, I'm getting lots of answers for std.dev that are well above the average, and so giving ranges of a single standard deviation starting well below zero.

I've been doing the calculations of the twin prime search twins in groups of 100 to give a somewhat accurate view of the range of variance and how it changes over time in relation to n.

Any suggestions? I've doublechecked my steps, and just don't know why the numbers don't make sense.
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Old 2012-01-25, 19:23   #2
ccorn
 
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I am not involved in the twin prime search and I am not familiar with its data. However...

Quote:
Originally Posted by roger View Post
Unfortunately, I'm getting lots of answers for std.dev that are well above the average, and so giving ranges of a single standard deviation starting well below zero.
This is generally possible, particularly if your data (i. e. the sampled random variable values) have a lower bound (e. g. zero) dictated by their very nature (e. g. positiveness) but no fundamental upper bound.

Of course, this then means that the probability distribution of your dataset cannot be expected to be direction-independent around its mean value, or around any other value for that matter.

In such cases, it might be more meaningful to compute stats on logarithms of the sampled values instead: Choose one reference value to your liking (it need not occur in the dataset), then divide all sampled values by that reference value, then take some logarithm with convenient base B (e.g. B=2, B=e, or B=10) of the resulting ratios. A negative result then just means that the original sampled value is less than the chosen reference value.
Of course, the interpretation changes accordingly: The arithmetic mean of the logarithms is not the logarithm of the arithmetic mean but the logarithm of the geometric mean, and adding/subtracting one standard deviation of the logarithms means multiplying/dividing the original values by some factor (namely by the base-B exponential of the standard deviation).

Last fiddled with by ccorn on 2012-01-25 at 19:43
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Old 2012-01-26, 04:45   #3
roger
 
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Thanks, I'll give it a try!
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