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I was doing some exercises on error propagation when I came across this problem:

$4\pi^2/(0,034 \pm 0,004 \space s^2/cm)$

I calculated my uncertainty to be $\Delta$ = 136 $cm/s^2$

And so the hole thing gives me: $(1200 \pm 136) \space cm/s^2$.

Now going into my question I rounded $4\pi^2/0,034 = 1161.12993$ to $1200$ which makes sense since my uncertainty is 136 but in the solution to this problem $136$ is rounded to $100$ and that's not making much sense to me.

So should I always round my uncertainty to the nearest hundred (assuming my uncertainty has 3 digits)?

user2550495
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1 Answers1

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You should round the uncertainty of the result to at most two significant figures (see this document, §7.2.6; you can use more figures in intermediate calculations, though).

Then, you have to round the quantity value to a number of figures compatible with the uncertainty.

So, in your case, the result would be

$$(1160\pm 140)\,\mathrm{cm/s^2}.$$

If you're unsure about the figures, rewrite the values in scientific notation.

Wojciech Morawiec
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Massimo Ortolano
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  • This is wrong. It's perfectly possible to have an uncertainty with 3 or more sig figs. It just doesn't happen very often in real life. –  Sep 21 '17 at 22:40
  • @BenCrowell Given the meaning of uncertainty, it's useless to report it with more than two significant digits. As I said, more digits can be used in intermediate calculations though. – Massimo Ortolano Sep 22 '17 at 05:05