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The density operator is a representation of a state of a quantum system $\rho=|\psi\rangle\langle\psi|$, so it's just an alternative characterisation of a state (or more generally a statistical mixture of states). On the other hand, the density operator is a Hermitian operator on a Hilbert space and as such it can be thought of as representing some observable, i.e. some physical property. What is that physical property?

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Here's a possible way of thinking of the measurement defined by a general density operator $\rho$, not necessarily one representing a pure state $\lvert\psi\rangle$. If $\mathcal H\simeq \mathbb C^d$ is the Hilbert space under consideration and $\rho$ is a density operator on $\mathcal H$, i.e. a $d\times d$ positive semidefinite Hermitian matrix with trace $1$, then it can be decomposed into rank-one matrices via its singular value decomposition like this: $$\rho = \sum_{i=1}^d \sigma_i \lvert u_i\rangle\langle u_i\rvert$$ where the $\sigma_i=\sqrt{\lambda_i}$ are its singular values and the $\lvert u_i\rangle$ are an orthonormal basis of singular vectors. Using the interpretation you mentioned of a density operator as a "statistical mixture" of states, the above expresses $\rho$ as the density matrix of a statistical mixture of the orthonormal states $\lvert u_i\rangle$ with respective statistical frequencies $\sigma_i^2 = \lambda_i$.

Now, performing a measurement with respect to the operator $\rho$ on a qubit in the state $\lvert\psi\rangle$ is almost like performing a full measurement with respect to the basis $\{\lvert u_i\rangle\}$, except that there's a possibility of some of the probabilities $\lambda_i$ being equal, meaning that performing a measurement with respect to $\rho$ gives us strictly less information than a full measurement with respect to this basis of singular vectors. Because this measurement "groups together" the singular vectors with the same statistical frequencies $\lambda_i$, making them indistinguishable given the result of the measurement, we might think of a measurement by $\rho$ as telling us how unlikely of a result we would get if we performed a full measurement, without actually performing one. To be more precise, the probability of observing $\lambda$ when measuring $\lvert\psi\rangle$ with $\rho$ is the same as the probability of finding $\lvert\psi\rangle$ to be in one of the states with statistical frequency $\lambda$ after doing a measurement in the full basis.

To give an example: consider $\mathcal H \simeq \mathbb C^3$ and the following density matrix: $$\rho = \tfrac{1}{100}\lvert 0\rangle\langle 0\rvert + \tfrac{1}{100}\lvert 1\rangle\langle 1\rvert + \tfrac{49}{50}\lvert 2\rangle\langle 2\rvert$$ which represents a statistical mixture consisting of $1\%$ of qutrits in state $\lvert 0\rangle$, $1\%$ of qutrits in state $\lvert 1\rangle$, and $98\%$ of qutrits in state $\lvert 2\rangle$. Performing a measurement by $\rho$ on a qutrit in state $\lvert\psi\rangle$ does not necessarily tell us what state it will end up in - in particular, if we obtain a result of $\tfrac{1}{100}$ from this measurement, we still do not know whether, say, it is in the state $\lvert 0\rangle$ or $\lvert 1\rangle$ (it could be any of infinitely many combinations of these two basis states, i.e. any unit eigenvector with eigenvalue $\tfrac{1}{100}$). However, in this case we would know that, should we proceed to perform a full measurement in the basis $\{\lvert u_i\rangle\}$ on this qutrit, we would find it to be one of the vectors with statistical frequency $\tfrac{1}{100}$ in the mixture represented by $\rho$, i.e. either $\lvert 0\rangle$ or $\lvert 1\rangle$.

To summarize, a measurement by $\rho$ on the state $\lvert\psi\rangle$ can tell us intermediate information about the common-ness (in the statistical mixture represented by $\rho$) of the state we will find it to occupy after measuring completely with respect to $\{\lvert u_i\rangle\}$.

NOTE: An analogous situation in classical probability would be to consider the how the distribution of a random variable $X$ (say, with a finite sample space) changes when you are given the probability of its result. The probabilities of each possible outcome without knowing this information are given by $p(x) = \mathbb P(X = x)$, but the probabilities of each possible outcome knowing the probability of the specific outcome are given by $\mathbb P(X = x ~ \lvert ~ p(X))$.

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