For a Poisson process, $15$ occurrences were observed in $10$ minutes. The mean rate of occurrence was estimated as $\hat{\lambda}=1.5$ and the uncertainty of the estimate as $\sigma_\hat{\lambda}=0.387$. Find the probability of exactly one occurrence being observed in the next minute and find the uncertainty in the estimate.
The probability is easy, by the PMF it is $P(X=1) = .335$
The uncertainty is where I am confused. I understand I am meant to use propagation of uncertainty to find this, with
$\sigma=|\frac{dU}{d\lambda}|\sigma_X$
I find $U$ by substituting $1$ into the PMF and I get
$U(\lambda)=\lambda e^{\lambda}$
So the uncertainty is
$\sigma=|\frac{d}{d\lambda}(\lambda e^{\lambda})|\sigma_X = (e^{\lambda}+\lambda e^{\lambda})\sigma_X = (e^{1.5}+1.5e^{1.5}).387 = 4.336$
However, this answer doesn't make sense to me since I am finding the uncertainty in an estimate of probability. Shouldn't I expect the uncertainty to be a value between $0$ and $1$?
The next jump time $\tau_i$ of a Poisson process is exponentially distributed with parameter $\lambda\,,$ that is, $\mathbb P(\tau_i\in[t,t+dt))=\lambda e^{-\lambda}$ is the probability density of $\tau_i$. The probability that $\tau_i$ is within the next one minute is $$\tag{1} \mathbb P(\tau_i\in[0,1])=1-e^{-\lambda}\,. $$ You estimated this $\lambda$ by $\hat\lambda=1.5$ (jumps per minute) with an accuracy of $\sigma_{\hat\lambda}=0.387\,.$ Plugging the values $$ \lambda\in\{\hat\lambda-\sigma_{\hat\lambda},\hat\lambda,\hat\lambda+\sigma_{\hat\lambda}\} $$ into (1) gives probabilities approximately as $0.67,0.78,0.85$. Thus, the probability is around $0.76$ within the range $0.76\pm 0.09$.