= Solution
Write $z=e^{\beta(\mu-\epsilon)}$. For one nondegenerate fermionic <quantum state>, the allowed occupations 0 and 1 give $\Xi_F=1+z$. For a bosonic state, summing all nonnegative occupations gives $\Xi_B=(1-z)^{-1}$, requiring $z<1$ or $\mu<\epsilon$. Differentiating the logarithm,
$$
\bar n=z\frac{\partial\ln\Xi}{\partial z},\qquad
\boxed{\bar n_F=\frac1{e^{\beta(\epsilon-\mu)}+1},\qquad
\bar n_B=\frac1{e^{\beta(\epsilon-\mu)}-1}.}
$$
The <interacting point defect> has two distinct <spin> states, not one fermionic state. Its four <microstates> have grand-canonical weights $1,z,z,qz^2$, with $q=e^{-\beta U}$. The first two single-electron weights are separate up/down <spin> possibilities; the double state has one <electron> of each <spin>. Thus
$$
\Xi=1+2z+qz^2,\qquad
\boxed{\bar n=\frac{2z+2qz^2}{1+2z+qz^2}.}
$$
For $U=0$, $\Xi=(1+z)^2$ and $\bar n=2z/(1+z)$. The sketch is a single smooth step from 0 to 2, centred on $\mu=\epsilon$, where $\bar n=1$. As $T\to0$ its width tends to zero.
For positive U with $\beta U\gg1$, the lowest grand energies are successively $0$, $\epsilon-\mu$ and $2\epsilon+U-2\mu$. Away from the two crossings,
$$
\boxed{\bar n\simeq\begin{cases}
0,&\mu<\epsilon,\\
1,&\epsilon<\mu<\epsilon+U,\\
2,&\mu>\epsilon+U.
\end{cases}}
$$
Finite <temperature> smooths each step over an <energy> range of order $k_BT$. Near the lower step double occupation is exponentially suppressed, so $\bar n\simeq2z/(1+2z)$. Near the upper step put $y=e^{\beta(\mu-\epsilon-U)}$ and neglect the empty state, giving $\bar n\simeq1+y/(2+y)$. <Spin degeneracy> shifts the half-step positions to $\epsilon-k_BT\ln2$ and $\epsilon+U+k_BT\ln2$.
For a grand-canonical probability $p_s\propto e^{-\beta(E_s-\mu n_s)}$, differentiation gives $\partial_\mu p_s=\beta(n_s-\bar n)p_s$. Multiplying by $n_s$ and summing proves
$$
\boxed{\sigma^2=k_BT\left(\frac{\partial\bar n}{\partial\mu}\right)_T.}
$$
For this defect, direct state counting gives
$$
\langle n^2\rangle=\frac{2z+4qz^2}{\Xi},\qquad
\sigma^2=\frac{2z(1+2qz+qz^2)}{(1+2z+qz^2)^2}.
$$
At $\mu=\epsilon$, $z=1$ and therefore
$$
\boxed{\sigma^2=\frac{2+6e^{-\beta U}}{(3+e^{-\beta U})^2}\longrightarrow\frac29.}
$$
The same limit gives $\bar n=2/3$: empty, up and down states have equal weights, while the double state is suppressed. Replacing this degeneracy by a two-outcome equal-weight model would incorrectly give <variance> $1/4$ at this <chemical potential>.
Across the full chemical-potential range the <variance> has two well-separated narrow peaks and tends to zero on the 0, 1 and 2 plateaux. The limiting shapes are $2z/(1+2z)^2$ near the lower step and $2y/(2+y)^2$ near the upper one. Each peak has height $1/4$, at the shifted half-step position; the <variance> at the unshifted crossings is instead $2/9$.
\Image[/past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2013/paper-1-tsp2013-defect-statistics.png]
{title=Original grand-canonical defect occupations and number variance, showing the interaction-induced one-electron plateau and spin-degeneracy shifts of the variance peaks}
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