short cut to Standard Deviation

[spaminator101]

now that i think about it isnt standard deviation just the average deviation from the mean
so shouldn't you be able to find SD by averaging the absolute values of the deviation from the mean
instead of finding the sqrt of your variance

i have tried this out and it didn't work
i was wondering if anyone here had tried it and similar/different results

[BruceZ]

[ QUOTE ]
now that i think about it isnt standard deviation just the average deviation from the mean

[/ QUOTE ]

No, variance is the average squared deviation from the mean. Standard deviation is the square root of the variance. No getting around the square root.

Var(x) = (1/N)*sum [i = 1 to N] (Xi - u)^2

where Xi are the data samples and u = the average.

SD(x) = sqrt[Var(x)]

SD is a root mean square average, or rms average, since it is the root of a mean of squares. You've heard of watts rms in connection with stereo amps and speakers? Same idea, except rms watts is a misnomer. There is no rms power, but there is rms voltage and rms current, which is what they really mean.


[ QUOTE ]
so shouldn't you be able to find SD by averaging the absolute values of the deviation from the mean
instead of finding the sqrt of your variance

[/ QUOTE ]

No.

If you want a shortcut, you can find the variance by averaging the squares of the results, and then subtracting the square of the average.

Var(x) = E(x^2) - [E(x)]^2

= (1/N)*sum[i = 1 to N] Xi^2 - {(1/N)*sum [ i = 1 to N] Xi}^2

SD(x) = sqrt[Var(x)]

Be careful of roundoff error if you use this one.

[AaronBrown]

To add to BruceZ's excellent answer, the average deviation from the mean is known as the Mean Absolute Deviation. If all the deviations are the same size, it is the same as the standard deviation. For example, if you flip a coin, heads you win $1, tails you lose $1, the standard deviation and the mean absolute deviation and every individual deviation are all $1.

But if the outcomes do not all have the same deviation, the standard deviation will be larger than the mean absolute deviation. That's because squaring puts more weight on the extreme deviations. For example, if you bet $100 on a single number in American roulette, your expected value is $3,500*(1/38) - $100*(37/38) = -$5.26. Your mean absolute deviation is $3,505.26*(1/38) + $94.74*(37/38) = $184.49. But your standard deviation is [$3,505.26^2*(1/38) + $94.74^2*(37/38)]^0.5 = $576.26. It's much higher, because the square weights the large deviation much more than the small one.

[spaminator101]

thanks to both of the posters here brucez and aaron brown
this had been bothering me for some time