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Ok, for the switch-on voltage of a red LED I have the readings as follows, all in volts: $$ 1.45, 1.46, 1.46, 1.44, 1.45 $$ The mean of these readings, in volts, is $1.45$ (I rounded up to $2$ decimal places as my scale reading uncertainty was $\pm 0.01\,\mathrm{V}$, and my teacher told me to round them up since to state my scale reading uncertainty for the mean the mean will have to have the same number of decimal places as the scale reading uncertainty). Now, my random uncertainty for these values is $\pm 0.004\,\mathrm{V}$, which is not to $2$ decimal places (it's to $1$ significant figure). So, I was wondering if I would have to make my random uncertainty have $3$ significant figures ($\pm 0.00400\,\mathrm{V}$?) to express the random uncertainty in absolute form (Mean Value $\pm$ Random Uncertainty). And if I did so, would it be, in terms of physics, correct?

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  • $\begingroup$ How did you calculate "random uncertainty"? $\endgroup$ – user289661 Mar 29 '16 at 19:25
  • $\begingroup$ (Max.-Min)/No. of Results $\endgroup$ – user307397 Mar 29 '16 at 19:25
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    $\begingroup$ I believe it should be $\frac{max.-min.}{2}$, but this may not be what you are asked to do. $\endgroup$ – user289661 Mar 29 '16 at 19:27
  • $\begingroup$ Why 2? Is it not divided by the number of readings? That is the formula we have to use for our course anyway (the one I stated above). $\endgroup$ – user307397 Mar 29 '16 at 19:28
  • $\begingroup$ More on significant figures. $\endgroup$ – Qmechanic Apr 17 '16 at 14:13
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I believe in this case you should round your uncertainty to 0.01, in addition to the convention where it is better to give your results "the benefit of the doubt", if your digital measurement tool (your voltmeter) only had up to two decimal places, the smallest measurable value (0.01) would be uncertainty inherent in the measurement tool.

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  • $\begingroup$ So, are you saying I should round the random uncertainty to 0.01? Would that be, in physics terms, correct since obviously a 4 round down to a 0 and not to a 1. $\endgroup$ – user307397 Mar 29 '16 at 19:26
  • $\begingroup$ conventionally, you can not have an uncertainty of zero, so in order to make the decimal places consistent, that is to be done. $\endgroup$ – user289661 Mar 29 '16 at 19:28
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About significant figures in mesured values and uncertainties you can check chapter 2 of "An Introduction of Error Analysis" by John R. Taylor. It's a simple book about something any science student should know and that is not systematicly taught. Give it a look, it's worth the reading.

Two rules he sets for the writing of uncertainties are:

1) Rule for stating uncertainties:

Experimental uncertainties should amolst always be rounded to one significant figure

2) Rule for stating answers:

The last significant figure in any stated answer should usually beof the same order of magnitude (in the same decimal position) as the uncertainty

That is to say, without any reference to the measurment apparel, it's clearly unconvenient to write values such as (let's say for a speed I take this example from the book itself):$$90\ m.s^{-1}\pm30$$ which is not of any use except if you can explain why you can tolerate a $30\%$ error, or $$90\ m.s^{-1}\pm 0.234325$$ which is of no use as the error can be rounded to $0.2$ whithout any prejudice. In case your error is clearly inferior to the actual measured value, it's still convenient to use the scientific notation. In your case, I guess that $$1.45\ V\pm 4\cdot10^{-3}$$ would still be accepted, yet I would hold still to the rounding to $10^{-2}$.

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The point of doing multiple measurements is to get a value that is more accurate than any single measurement. In this case the mean of your five measurements is $1.452$. I didn't check the standard deviation, but assuming your value of $0.004$ is correct you should state your final result as:

$$ \text{V}_\text{on} = 1.452 \pm 0.004 \text{V} $$

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