What is a 'statistical operator' in quantum mechanics? What is a 'statistical operator' in quantum mechanics? How is it different from just an operator? Are there any operator properties (e.g., normal, Hermitian, unitary, etc.) universally attributable to statistical operators? 
Or is it just an operator for which there's an expectation value with respect to some vector? 
 A: In mathematical sense, we say that operator $\rho$ is statistical if:


*

*It is hermitian  $\rho^\dagger = \rho$

*It is positive. This means that condition 1. is satisfied and also $\langle \psi| \rho |\psi \rangle \geq 0$

*It's trace is equal to unity  $\mathrm{Tr} \; \rho =1$
In quantum mechanics density matrix satisfies all three of these properties, so you will often hear that people refer to density matrix as a statistical operator (they basically use it as a synonym). 
A: Another way to see relation to statistical physics, is that the density matrix for system at temperature $T = 1 / \beta$ can be written in form:
$$
\rho = \sum_n e^{-\beta E_n} | n \rangle \langle n|
$$
Where the sum is over all eigenstates of Hamiltonian. So the partition function is:
$$
Z (\beta) = \text{tr} \rho  
$$
And expectation value of any observable:
$$
\langle A \rangle = \frac{\text{tr} \rho A}{\text{tr}  \rho} = \frac{1}{Z} \sum_n e^{-\beta E_n} \langle n| A | n \rangle
$$
A: A statistical operator (or, sometimes also a density matrix, especially when we talk about representations) is an operator that is used to compute the expectation values of observables via the trace, viz.
$$\operatorname E[O] = \operatorname{Tr}(\rho O)$$
whence the adjective "statistical". As pointed out in other answers, a statistical operator is self-adjoint, positive semi-definite and of trace 1. For, since the expectation of the identity operator is 1, we have
$$1=\operatorname E[I] = \operatorname{Tr}(\rho)$$
