 Solution

The central limit theorem states that for independent identically distributed variables with finite mean $m$ and positive finite variance $v$,Applying it to the exponential gaps givesInvert the approximate central probability $\mathbb{P}(-1.96\le (\lambda T_n-n)/\sqrt{n}\le 1.96)\simeq 0.95$. For an observed positive total $T_n$, an approximate 95% confidence interval isEquivalently its endpoints are $\hat{\lambda }(1\pm 1.96/\sqrt{n})$, 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.