$xp$ is not a hermitian operator and hence doesn't represent an observable. Then, how can we interpret the expression
$$ \langle x p \rangle \text{,} $$
the expectation value of position times momentum?
How to interpret the expectation value of a hermitian operator, say $p$, is clear: make several measurements on identically prepared systems and get the expectation value from the measured quantities.
But now we are talking about the expectation value of an operator that doesn't represent an observable. The discussion above hence isn't valid. However, computing $\langle x p \rangle$ knowing $\Psi$ according to the mathematical definition can be done, so $\langle x p \rangle$ can clearly be deduced if we know the wave function - but how can we interpret the quantity we get?
(The question arises from a line in my textbook - "in a stationary state, $\frac{d}{dt} \langle x p \rangle = 0$, ..." without further explanation - why is this obvious?)