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Topic: Standard deviation  (Read 3413 times)

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gizmo_attacks

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Standard deviation
« on: September 27, 2010, 01:17:02 PM »
Hello all, I have to figure out the standard deviation and coefficient of variation for several equations.  The first one is,

y = 5.75 (±0.03) + 0.833 (±0.001)-8.021 (±0.001) = -1.438

I understand the concept of standard deviation, but I have never figured it out from an equation like this, just from a set of values, I am utterly lost and have not found anything online where they show you how to figure it from an equation such as this.  Any help would be much appreciated, the sooner the better. Thanks

Offline JGK

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Re: Standard deviation
« Reply #1 on: September 27, 2010, 03:54:44 PM »
Repeat the calculation using all variations of the values (eg for the first parameter use 5.78 and 5.72)

this will give you a range of end values for which you can calulate a mean and SD
Experience is something you don't get until just after you need it.

Offline Stepan

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Re: Standard deviation
« Reply #2 on: September 27, 2010, 05:01:42 PM »
Theoretically CV^2=CV1^2+CV2^2+CV3^2+... and if you know individual CV's you calculate total.

Practically for your example it will be +/-0.03 as it >> than +/-0.001

gizmo_attacks

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Re: Standard deviation
« Reply #3 on: September 28, 2010, 03:06:09 PM »
Thanks, would I also have to include 5.75 for the first value, or just the two adjusted values?

Offline Stepan

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Re: Standard deviation
« Reply #4 on: September 28, 2010, 04:58:06 PM »
Sorry my fault: CV was for Coefficient of Variations.

In case of your formula, the standard deviation for Y is 0.03  (I assume 0.03 is standard deviation for 5.75) and CV(y)=0.03/1.438 - it is basically a relative error.

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