# Standard deviation of quantities with errors [duplicate]

Suppose I have taken 5 measurements, each with an associated error, let's call these values $\eta_i\pm\sigma_i$, where $i$ is an integer from 1 to 5.

I would like to find the mean and standard deviation of these quantities, $\bar{\eta}\pm\bar{\sigma}$.

The mean can be calculated simply: $\bar{\eta}=\sum_{i=1}^{5}\eta_i$.

How is the standard deviation $\bar{\sigma}$ calculated, including the contribution from the deviation of $\eta_i$ about $\bar{\eta}$, and also the contribution from the errors $\sigma_i$?

EDIT To give some more context, the $\eta_i$ are a function of the number of counts $N$ of a Poisson distributed variable. Due to the high number of counts, $\sigma_i$ is a function of $\sqrt{N}$, which is the standard deviation of the count number.

Thus, I definitely know the errors $\sigma_i$. However, I would like to combine my repeated measurements to find a mean and deviation.

EDIT#2 My reading into the topic thus far has led me to this page on propagating errors (https://en.wikipedia.org/wiki/Propagation_of_uncertainty).

The example formulas tab has an example for the function $f=aA+bB$, which should be applicable in my case where I would like to find the mean. For my purposes (finding the mean) $a=1, b=1$. Using this entry in the table, $\sigma_f=\sqrt{\sigma_A^2+\sigma_B^2+2\sigma_{AB}}$, where $\sigma_{AB}$ is the covariance.

If I were to apply this reasoning to my problem, I should get:

$\bar{\sigma}=\sqrt{\sum_{i=1}^5\sigma_i^2 + 2cov}$, where $cov$ is the covariance.

I am rather unfamiliar with the covariance, but I think that in my example, it is equal to the variance of my 5 different $\eta_i$.

In other words, $cov=\frac{\sum_{i=1}^5(\eta_i-\bar{\eta})^2}{5}$

My question, is this interpretation right for the propagation of my errors, and if not, where has my reasoning gone wrong?

## marked as duplicate by whuber♦Apr 16 '18 at 15:18

• Assuming I do know the errors, it seems like I shouldn't just ignore them. To use your analogy of the size of the Earth, 3 measurements of $6000\pm500$km, when averaged, doesn't have a standard deviation of 0, which is what happens when they are eliminated as you suggest. – ocelto Apr 15 '18 at 21:46
• But you don't know the errors. If you knew the errors, you would incorporate them into the estimate and would have error free estimates. You never get a measurement of $6000 \pm 500$ or anything like that. You estimate the errors from the set of measurements. – Peter Flom Apr 15 '18 at 21:54