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If I wish to calculate the average deviation from a value in a list, can I use the standard deviation equation using the value in place of the mean?

If so, why? If not, what's a good alternative?

mino
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    If you want average deviation from $v$, why not compute that? $\frac 1n \sum |x_i-v|$. – lulu Mar 10 '17 at 21:15
  • @lulu Thank you. Is that not the same as the standard deviation equation but with v instead of mean? Or is there a difference? – mino Mar 10 '17 at 21:22
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    Standard deviation is the square root of the sum of the squared differences, not the same thing. To be sure there are some advantages to standard deviation...analytic tractability among them. And sometimes people want error functions that strengthen (or which tend to minimize) outliers. – lulu Mar 10 '17 at 21:27
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    I'm assuming that you have an actual list you are working from. If so, then there's no difficulty in computing the sum. If, instead, you just have summary data (like mean and variance) then it is probably easier to manipulate the standard expressions. – lulu Mar 10 '17 at 21:29
  • @lulu Hi. Thanks again. Yes, I have a list. E.g.: 0.1, 0.3, 0.34, 0.54, 0.6. And I will want to know the deviation from a value from the list, e.g.: 0.54 or a value not on the list but in the range such as 0.5. Can I use the standard deviation method you wrote with 'v' for this? Thank you! – mino Mar 10 '17 at 21:36
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    Yes, you can. I wouldn't call it standard deviation though...that already has a definition. Average absolute deviation is good, if you need to call it something. – lulu Mar 10 '17 at 21:39
  • @lulu Thank you for the advice! :-) – mino Mar 10 '17 at 21:40

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A starting point may be something such as

$$\sqrt{ (100a - 50) ^ 4 } \times 4$$

This formula assumes that $a$ is the ratio of coin flips resulting in heads to total coin flips, or $\frac{h}{T})$. It uses the principle that in perfect conditions the probability of either heads or tails is 50%. And so this measures the absolute deviation from $0.5$.

In the context of randomness I found a score of 9 is an acceptable threshold for detecting anomalies or biases.

There may be more standard or optimal solutions for this however.

bryc
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