Taking equipment precision into standard deviation I have a set of $n$ measurements, $x_i$, representing distance to a point, and want to find the mean and the standard deviation of the distance, the problem is, I don't know how to take the precision of the measuring equipment into the equation. 
The precision of the measuring equipment is given by the producer, say $\sigma_p$, so each measurement has a uncertainty: 
$x_i \pm \sigma_p$
When calculating the mean, I do this: 
$\mu = \frac{1}{n} \sum_{i=1}^n (x_i \pm \sigma_p) = \frac{x_i}{n} \pm \sigma_p$ 
but I'm not sure if $\sigma_p$ should be in the equation at all or how to interpret it. Furthermore, I need to calculate the standard deviation, which usually is 
$\sigma_x = \sqrt{\frac{1}{n-1} \sum_{i = 1}^n (x_i - \bar{x})^2}$, 
but if I substitute my calculations I obtain
$\sigma_x = \sqrt{\frac{1}{n-1} \sum_{i = 1}^n (x_i \pm\sigma_p - (\bar{x}\pm \sigma_p))^2}$,
which I do not not how to manipulate further. Do the $+\pm\sigma_p$ and $-\pm\sigma_p$ cancel out or add up? 
Should I rather just add the two together? That is
$\sigma_x^2 = \sigma_p^2 + \frac{1}{n-1}\sum_{i = 1}^n (x_i - \bar{x})^2$,
where the mean is calculated as normally, 
$\bar{x} = \frac{1}{n} \sum x_i$
Any help is much appreciated!
 A: The primary assumption (which is often very reasonable) is that the measurement errors are independent, identically distributed, and conditionally independent on the true value of the variable of interest. With electronic devices, this is often the case. 
So denote the 'true' distance variable $X_d$, the measurement error random variable $X_p$, and you observe a joint variable, $X_x=X_d+X_p$. You are asking for an estimate of $X_d$ which is corrupted with measurement error. So $\mu_x=\mu_d+\mu_p$ whose expected value is $E[X_x]=E[X_d + X_p] = E[X_d]+E[X_p]$ under independence. If your measurement error is zero-centered (that is, the measurement is unbiased) than $E[X_p]=0$, therefore $E[X_x]=E[X_d]$ and your best estimate is $\hat{X_d}=\hat{X_x}= \frac{1}{n}\sum{x_x^i}$. So basically, if you can assume zero centered, independent measurement errors, the best estimate of $\mu_d$ is the mean of your measurements.
Also, the variance of two independent random variables is the sum of the two individual variances. You can think of the variance of your data as representing $\hat{\sigma_x^2}=\frac{1}{n-1}\sum{(x_x^i-\bar{x})^2}$. Now we know $\sigma_x^2 = \sigma_p^2+\sigma_d^2 \to \sigma_d^2=\sigma_x^2-\sigma_p^2$. We have the estimator $\hat{\sigma_x^2}$ and we know $\sigma_p^2$, therefore our estimator $\hat{\sigma_d^2}=\hat{\sigma_x^2}-\sigma_p^2$. 
