Does calculating relative uncertainty not account for the extreme values or am I wrong?

Question

Assume two values, $A = 4 \pm 0.5$ and $B = 10 \pm 1$. When calculating $AB$, we must convert them into their percentage uncertainties.

$A = \frac{0.5}{4}\times 100 = 4\pm12.5\%$

$B = \frac{1}{10}\times 100 = 10\pm 10\%$

$AB = (4\times10)\pm(12.5\% + 10\%) = 40 \pm22.5\%$

Converting back into absolute uncertainties: $a\Delta (AB) = \frac{22.5}{100}\times 40$ = 9

Hence, $AB = 40 \pm 9 $

However, I find that this method doesn't account for the extremes possible. The largest possible value indicated by the initial uncertainty is $4.5\times 11 = 49.5$, whereas my calculated uncertainty suggests that $49$ is the largest possible number.

Is this the way it is, or am I doing something wrong?

Answer 1

For one thing, I believe your formula is incorrect. The real uncertainty is about 6.4, shown So actually even smaller than 9. And more importantly, the uncertainty number is, depending on the case, the standard deviation, or maybe two or three std deviations, assuming the error in the numbers is Gaussian distributed. So it's not meant to encompass all possible error which is technically infinite, just a certain statistical band.

The uncertainty in a measurement is different from a tolerance given on an engineering drawing, which mandates that the real number fall between those two extremes. Even though they look the same.