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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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$$e/m=\color{red}{1.758\,820\,1}\color{blue}{50}\,\color{magenta}{(44)}\color{green}{×10^{11}} \ \mathrm{C/kg}=\left(\color{red}{1.758\,820\,1}\color{blue}{50}\color{green}{×10^{11}} \pm \color{red}{0.000\,000\,0}\color{blue}{44}\color{green}{×10^{11}}\right) \ \mathrm{C/kg}$$

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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
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This uncertainty is standard deviation (often written $\sigma$).

It means you are $\sim95\%$ sure that your number $m$ is in $[m_e-2\sigma,m_e+2\sigma]$ where $m_e$ is the measure given.

I think it's only $\sim65\%$ when you use $\pm\sigma$

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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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