= Solution
The <central limit theorem> states that for independent identically distributed variables with finite <mean> $m$ and positive finite <variance> $v$,
$$
\frac{\sum_{i=1}^nY_i-nm}{\sqrt{nv}}\xrightarrow{d}N(0,1).
$$
Applying it to the exponential gaps gives
$$
\frac{\lambda T_n-n}{\sqrt n}\xrightarrow{d}N(0,1),\qquad
\boxed{T_n\approx N\left(\frac n\lambda,\frac n{\lambda^2}\right).}
$$
Invert the approximate central probability $\mathbb P(-1.96\leq(\lambda T_n-n)/\sqrt n\leq1.96)\simeq0.95$. For an observed positive total $T_n$, an approximate 95% <confidence interval> is
$$
\boxed{\left[\frac{n-1.96\sqrt n}{T_n},\frac{n+1.96\sqrt n}{T_n}\right]\cap(0,\infty).}
$$
Equivalently its endpoints are $\widehat\lambda(1\pm1.96/\sqrt n)$, with a nonpositive lower endpoint replaced by the parameter boundary. This is a large-$n$ approximation; the exact gamma law of $T_n$ permits an exact interval if needed. It is a two-sided parameter interval, not a 95th-percentile annual loss estimate.
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