# Uncertainty in a given value

I have read that the uncertainty in the charge of a given value, e.g. charge of an electron, is $\pm1$ in the last significant figure.

So $e=1.6\times 10^{-19}$ and the uncertainty would be $\pm 0.1\times 10^{-19}$, but to me it seems wrong. Shouldn't it be $\pm0.05\times 10^{-19}$ as this is the upper and lower bound of what the charge can be to 3 significant figures?

I have read that the uncertainty in the charge of a given value, e.g. charge of an electron, is ±1 in the last significant figure.

"Given" doesn't really mean anything useful. This is just the previously measured value of $$e$$.

Measurements don't have to have any particular error bars on the last digit. Wikipedia gives the current best value of $$e$$ as $$−1.6021766208(98)×10^{−19}$$ C, with the source being CODATA, who probably did a weighted average of various measurements. So the error bars in this example are 98 times greater than $$\pm 1$$ in the last digit, and almost 200 times greater than $$\pm 0.5$$ in the last digit. If the error bars on a piece of published data are big enough to be relevant in your calculation, then use the published error bars.

If you're doing a calculation, and you find that the precision of your result is limited by the precision of the 2 sig fig value you were using for $$e$$, then you simply made a mistake by not looking up a more accurate value. Don't do that!

$$\pm 0.5$$ in the last digit would be the maximum error (not standard deviation) due to rounding. For the reasons discussed above, this would never be relevant when using some piece of published data. This would be more likely to be relevant, for example, if you're doing your own measurement using a digital balance.

You are probably referring to least count error of the measurement. Generally, the least count error is taken to be the least count of measuring instrument.

In this case, the measurement was $1.6\times10^{-19}$, so the least count would be $0.1 \times10^{-19}$. So this would be maximum least count error.

Suppose the least count of an instrument was $1$ unit and reading taken by a person was $1$. It means the true value could be anywhere between $4$ and $5$. The true value might be $4.99$ or $4.00$, but the reading would not be taken as $5$ due to convention and would be taken as $4$ only. So we see maximum least count error cannot be greater than the least count of the instrument.

Ben Crowell's answer is right. Here is an extension to deal with the use of instruments with a digital readout.

An instrument with a digital readout might be either a sensor (e.g. voltmeter) or a source (e.g. signal generator). If it is a sensor and it is well designed, then it ought to show sufficient digits that the accuracy of its reported measurement is not limited by its own digital display. In this case you must consult the manufacturer's reference to find the precision, and you should also doubt that because the manufacturer was probably a bit optimistic and the instrument will have aged etc.

If it is a source, then there are two possible designs.

1. It reports $$n$$ digits, and the precision of the signal is better than $$n$$ digits.

2. It reports $$n$$ digits, and the precision of the signal is less than $$n$$ digits.

Again, you have to check to find which it is. In neither case do the digits on the diaplay tell you the precision!

Case 1 is not impossible, by the way. For example, I have a signal generator which, whem I dial up "1.000 kHz" then puts out a signal whose frequency is is $$1 \pm 10^{-5}$$ kHz. It only lets me increment the signal in steps of 1 Hz, but it is good enough to give the indicated frequency precise to 10 mHz.

Finally, for everyday measurements I'll admit that we do often find ourselves using a measuring instrument such as a digital voltmeter in which the precision is indicated by the digits it shows. In this case, assuming the instrument is designed to do rounding in the appropriate way, then you could attach a notional $$\pm 0.5$$ to the last digit, but this would not be a standard deviation. A flat probability distribution of width $$1$$ has a standard deviation of about $$0.3$$.