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Hello guys I hope everyone is having a great day

So I was working on my physics homework on uncertainties for school and as soon as I completed it I noticed that the uncertainty of an average is smaller than the uncertainty of an individual readings. So basically I want to know why that is and I was hoping you guys could help me out.

Thanks in advance

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  • $\begingroup$ Hi, welcome to PSE. It would be more benefitting for us to answer your question if you could also provide more context to the question. For e.g., what was being measured, what were the values and what is the average? $\endgroup$
    – Someone
    Commented Mar 20, 2019 at 22:03
  • $\begingroup$ Consider, if measuring multiple times didn't reduce your uncertainty about the thing being measured, why would you ever bother doing it? Or another way of looking at it: Uncertainty is a lack of information. If I measure more times I have more information, and therefore less uncertainty. $\endgroup$
    – The Photon
    Commented Mar 21, 2019 at 0:32
  • $\begingroup$ That is the motivation to average in the first place. $\endgroup$
    – my2cts
    Commented Mar 12, 2021 at 15:09

3 Answers 3

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Remember that when you take an average of a quantity, you're essentially dividing all the measures of that quantity over the number of measures, that is,

$$\overline{x}=\frac{\sum_i x_i}{N}$$

where in case you're not familiar, the $\sum$ is just a short way of indicating the sum of all the measures you took, and $N$ is the total number of measures. We usually represent averages as either $<x>$ or $\overline{x}$.

Suppose that a measure has an uncertainty $\Delta x_i$, so the your real value $x_{i,R}$ must lie between,

$$x_i-\Delta x_i<x_{i,R}<x_i +\Delta x_i$$

Now, when you take the average, you're essentially dividing the above quantities by $N$, including the very uncertainty, so you get an average uncertainty with the result,

$$\overline{\Delta x}=\frac{\sum_i \Delta x_i}{N}$$

Since $N>1$, then it is obvious that the average uncertainty will be smaller than the other uncertainties. Note that the more measures you take (the bigger $N$), the more you can be sure about the real value, since the average uncertainty will tend to zero.

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  • $\begingroup$ Please note that the uncertainty of the mean, as you defined it does not decrease with $N$, because when $N$ increases the numerator should increase roughly by the same ratio of the increase in $N$. However, it is unlikely that all the measurement errors are similar, if the measurements were independent, thus, the standard deviation is more useful as a measure of error. $\endgroup$
    – Ertxiem - reinstate Monica
    Commented Mar 21, 2019 at 3:17
  • $\begingroup$ You're right, it would be more useful in this case to express it in terms of the standard deviation, as explained by your answer. I'll upvote that. $\endgroup$
    – Charlie
    Commented Mar 21, 2019 at 4:51
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The uncertainty that decreases is related with the variance, which is the square of the standard deviation.

The total variance is the sum of the individual variances (so, it increases with $N$), but the total standard deviation is not the sum of the individual standard deviations (we need to compute it as the square root of the total variance).

When we study the mean, we need to divide the sum of the measurements by $N$, and the variance will be divided by $N^2$, therefore, the variance of the mean will decrease with $N$ and the standard deviation of the mean will decrease with $\sqrt{N}$.

Consequently, if we want to be 10 times more precise (in the sense of a standard deviation of the mean 10 times smaller), we would need 100 times more measurements (provided everything else is unchanged).

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Measuring something repeatedly will build our confidence in the result, i.e. the variability will decrease (or at least won't increase) and be as good or better than any one individual measurement's variability $-$ even that of the most precise measurement. Suppose that I measure quite precisely that the length of a bar, $\eta$, is definitely somewhere between 23.4 and 23.6 cm. Then in a second experiment I independently measure that $\eta$ is between 21 and 24 cm. I can hold on to the first, very precise measurement, but (thanks to the second measurement) supplement it by saying that $\eta$ is probably closer to 23.4 than 23.6 cm. So we've improved the situation, beyond even the most precise measurement. In physics, more information is always a good thing.

If you want to go deeper into the maths: the equation for the mean of your measurements is an average (sometimes a weighted one), and our natural intuition about averages not being lower than the lowest example will apply there. However, the equation for overall variance turns out not to be an average. The equations for overall variance can be found on Wikipedia and will depend on the circumstances of your experiment (e.g. how independent your measurements are, etc.).

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