You need to keep at least two pieces of information: the number of terms so far and the running mean (or something equivalent to it).
Let's suppose the $n$th component of the vector is $a_n$ and the running mean up to this is $m_n$ so $$m_n= \frac{1}{n}\sum_{i=1}^n a_i.$$
Starting with $m_0=0$, you can use the obvious single pass $$m_n = \frac{(n-1)m_{n-1}+a_n}{n}$$ but precision errors are likely to be smaller if you use the equivalent $$m_n = m_{n-1} + \frac{a_{n}-m_{n-1}}{n}.$$
