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I was doing a propagation of uncertainty on the following numbers, and I got a lower error from the calculated quantity than the measured quantities. So why is it that my compounded error smaller than the individual errors it is made up of?

The two measured quantities are $A = 1.0 \pm 0.2 \hspace{0.1cm} m$ and $ B = 2.0 \pm 0.2 \hspace{0.1cm} m$. I am supposed to find $ \sqrt{AB}$. I calculated it to be $1.41 \pm 0.158 \hspace{0.1cm} m$. I somehow end up getting a lower error than my beginning errors!

I'm pretty sure my work is right (I verified it here). Am I getting a lower error because I'm using too many digits to report the calculation (i.e. it's a sig fig mistake)? Even if it is, why is it theoretically possible to get a lower error from a function having the same dimensions than the errors I began with?

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  • $\begingroup$ The standard deviation is usually too small, if we estimate it. This is a known fact -- it called bias. Hence, your observation is not special. However, instead of letting us do the calc for your, you should present your calculation. Especially, because I believe that you just have a typo in your result and multiplied the error by a factor of ten. $\endgroup$
    – Semoi
    Jul 26, 2020 at 21:16
  • $\begingroup$ @Semoi the link I had put in the question does the calculations, and it had the same result, so I didn't bother showing my calculations (I just used the partial derivatives in a quadrature formula). I did mistype the calculated error, as you pointed out - by a factor of 10. But my question is why is my compounded error smaller than the individual errors it is made up of. $\endgroup$
    – GLS
    Jul 26, 2020 at 21:34

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You get a smaller error because $\sqrt{AB}$ is the geometric mean of the two measurements, i.e. a form of averaging.

When you take the average of a set of numbers, the uncertainty on the average is smaller than the uncertainty in the individual measurements (if the measurements have similar uncertainties), and so it is here.

$\sqrt{AB}$ will always be in between $A$ and $B$. It is this centralising tendency that results in a smaller standard deviation.

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  • $\begingroup$ That's a perfect explanation, thanks! $\endgroup$
    – GLS
    Jul 27, 2020 at 22:11

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