= Solution
For independent daily counts $X_1,\ldots,X_N$ modeled by a <Poisson distribution>, the <log-likelihood> is
$$
\ell(\mu)=-N\mu+\left(\sum_jX_j\right)\log\mu-\sum_j\log(X_j!).
$$
Differentiating gives the <maximum-likelihood estimator> $\widehat\mu=\overline X$. Here the observed frequencies sum to $N=1096$, and the expected frequencies are consistent with $\widehat\mu\approx2.157$ (their ratio $E_1/E_0$ estimates the same parameter). The final class is open-ended: its single observation is only reported as at least 9, so its exact contribution to the sample total cannot be recovered from this grouped table alone. The underlying ungrouped daily observations determine the exact <sample mean>.
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