= Solution
Use the parameters from (e), and interpret overflow as the number needing simultaneous care exceeding capacity at a randomly chosen time in <stationary distribution>. The occupancy of the <immigration--death process> is not fixed at its mean. <Detailed balance> gives
$$
\pi_n\alpha=\pi_{n+1}(n+1)\delta,\qquad\pi_n=\pi_0\frac{(\alpha/\delta)^n}{n!}.
$$
Normalizing yields $\pi_0=e^{-25}$, so \b[stationary occupancy has a <Poisson distribution> with mean 25]. Thus integer capacity $K$ must satisfy
$$
P(X>K)=1-e^{-25}\sum_{n=0}^{K}\frac{25^n}{n!}<0.1.
$$
The adjacent tails are
$$
P(X>31)\approx0.1000679170,\qquad P(X>32)\approx0.0714560312.
$$
Since the tail decreases with $K$, the \b[minimum capacity is $\boxed{32\ \text{animals}}$]. The small excess above $0.1$ at 31 matters; a <normal approximation> rounded too early can choose the wrong integer.
This is a stationary occupancy criterion. If the event is instead an offered arrival finding at least $K$ animals already present in this unrestricted process, <Poisson arrivals see time averages> gives $P(X\geq K)$, whose minimum qualifying capacity is 33. That uses a different event definition. From an initially empty centre, occupancy at a finite time is <Poisson distributed> with mean $N(t)\leq25$, so the stationary criterion also bounds the fixed-time overflow probability during initial filling. A probability of ever exceeding capacity over an unlimited future is a different question: this irreducible recurrent process will eventually exceed every finite capacity with probability one. If instead arrivals are refused when all places are occupied, that changes the model to an <loss network>, whose blocking probability is not the unrestricted occupancy tail calculated here.
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