# Variance of mean of two independent normally distributed values

Apologies for a silly question - may not make sense. If I have two measurements (assumed to come from normal distribution) X1 = 250 and X2 = 260 where variances are V1 = 5 and V2 = 6 (variance could be thought as the uncertainty in the measurements). Now the average of the two is m = 255 but what is the resulting variance of this average? Is it possible to calculate this?

It's not a silly question, but the concepts should be clear. If we denote your random variables as $$X_1$$ and $$X_2$$, the average will be $$\bar{X}=\frac{X_1+X_2}{2}$$, and we can find the variance of it via a simple calculation noting that the two RVs are independent: $$\operatorname{var}\left(\frac{X_1+X_2}{2}\right)=\frac{\operatorname{var}(X_1)+\operatorname{var}(X_2)}{4}=\frac{5+6}{4}=\frac{11}{4}$$ This calculation has nothing to do with the means of $$X_1,X_2$$, as well as the measurement values.

Informally, variance of an unknown $$X_1$$ can be non-zero. However, variance of a known $$X_1$$ is $$0$$ because it is known and therefore constant. So, variance of $$X_1=250$$ will be $$0$$, and it actually means something like $$\operatorname{var}(X_1|X_1=250)$$, mathematically, which is $$0$$. Measurement random variables have non-zero variance, but the measurement itself is fixed and have no variance (e.g. it is 250 always since you measured it).