Suppose that computer failure data $x_1, x_2, \ldots, x_n$ are modelled as observations of i.i.d. Poisson random variables $X_1, X_2, \ldots, X_n$ with common mean $\mu$.
Recalling that, for this data, $n = 104$ and $\bar{x} = 3.75$, calculate an unbiased estimate for $\mu$, together with its standard error, under the Poisson model.
The best estimate of $\mu$ is $\hat \mu = \bar X$ and the standard error (standard deviation of $\bar X$) is $$SD(\bar X) = \sqrt{Var(\bar X)/n} = \sqrt{\mu/n},$$ because $E(X_i) = V(X_i) = \mu$ for $X_i \stackrel{iid}{\sim}\mathsf{Pois}(\mu).$
In testing and estimation one often needs the estimated standard error, which would be $\sqrt{\bar X/n}.$ (The word 'estimated' is often dropped.) I will leave the arithmetic to you.
This leads some authors to assume that $Z = \frac{\bar X = \mu}{\sqrt{\bar X/n}}$ is standard normal and hence to use $\bar X \pm 1.96\sqrt{\bar X/ n}$ as a 95% confidence interval for $\mu.$ However, unless $n$ is quite large this $Z$ is not reliably standard normal.
A better 95% CI for $\mu$ is to use $T + 2 \pm 1.96\sqrt{T + 1},$ where $T = n\bar X = \sum_{i=1}^n X_i,$ as 95% CI for $n\mu$ and to divide the endpoints by $n$ to get a 95% CI for $\mu.$ This confidence interval works well (has actual coverage of $\mu$ near 95%) for moderately large values of $T$ (say, $T > 15$).