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In a physics text book I read the following:

$$e/m=1.758820150(44) ×10^{11} \mathrm{C/kg} $$ In this expression, $(44)$ indicates the likely uncertainty in the last two digits, $50$.

How should I understand this uncertainty? Does it mean $\pm 44$ on the last two digits?

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The digits in parentheses are the uncertainty, to the precision of the same number of least significant digits. (The meaning of the uncertainty is context-dependent but generally represents a standard deviation, or a 95% confidence interval.) So: $$e/m=1.758\,820\,1\color{blue}{50}\,\color{magenta}{(44)}×10^{11} \ \mathrm{C/kg}=\left(1.758\,820\,1\color{blue}{50}×10^{11} \pm 0.000\,000\,0\color{magenta}{44}×10^{11}\right) \ \mathrm{C/kg}$$

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    $\begingroup$ Colors are rather dull for physics expressions if you ask me. IMO, it clouds the beauty of latex! $\endgroup$ – Waffle's Crazy Peanut Jan 10 '15 at 11:14
  • $\begingroup$ I think the green part could be better turned to normal black. Even red could be turned to black IMO. Only the blue and magenta parts really need highlighting, colorfulness of the others just distracts. $\endgroup$ – Ruslan Jul 6 '20 at 16:09
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The uncertainty is equal to the standard deviation $\sigma = 0.000000044×10^{11} C/kg = 4400 C/kg$. It means you are

  • approx. 68% confident that the true value $m_e$ lies within the interval $[\hat m_e-\sigma, \hat m_e+\sigma]$, where $\hat m_e$ is the measure value.
  • and approx. $95\%$ confident that the true value $m_e$ lies within the interval $[\hat m_e-2\sigma, \hat m_e+2\sigma]$.
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  • $\begingroup$ So, since it is +/- $2 \sigma $, then the uncertainty is +/- 88 of the last two digits? And not +/- 44? $\endgroup$ – Steeven Jan 10 '15 at 11:11
  • $\begingroup$ Well $\pm$ and standard deviation aren't the same thing and you can't just translate. , it depends on the certainty you want. Did you ever worked with confidence interval? $\endgroup$ – servabat Jan 10 '15 at 11:13
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    $\begingroup$ NO! The (44) is the one sigma value, not the two sigma value. And it's best not to think in terms of ±. It's just uncertainty. $\endgroup$ – David Hammen Jan 10 '15 at 13:27

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