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How would I calculate the uncertainty for the average of this set?

$32.5 \pm 0.1$

$32.0 \pm 0.1$

$32.3 \pm 0.1$

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closed as off topic by Colin K, David Z Sep 3 '12 at 18:35

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Would someone please fix the formatting? – Nyx Jun 9 '12 at 17:40
up vote -1 down vote accepted

You can callculate the standart deviation as show in this link

The standart deviation $\sigma=\sqrt{\frac{\sum_{i=1}^n a_i^2}{n}-\left(\frac{\sum_{i=1}^n a_i}{n}\right)^2}$, where $a_i$ is the $i$-th number in your set and $n$ is the number of numbers you have in your set (in your example $a_1=32.5$, $a_2=32.0$, $a_3=32.3$ so $n=3$)

Using the numbers from your question I got $\sigma \approx 0.20548$

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The standard deviation of the data is not the same as the uncertainty of the mean...with enough data you can find the mean to high precision even when the data have a broad distribution. Do a little more reading. – dmckee Jun 9 '12 at 20:39
Thank you for your help. I was looking for the mean, but I think standard deviation is more suited to what I'm doing. – Nyx Jun 9 '12 at 22:08
It is also incorrect if you were trying to estimate the standard deviation. In that case you should use the "sample standard deviation" as a reasonable estimate.… – Rob Jeffries Oct 28 '14 at 15:29
This is not the uncertainty in the average, it is the uncorrected sample standard deviation. It is only valid to use this if the three things measured are the entire set of possible measurements. – Rob Jeffries Oct 28 '14 at 15:32

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