Estimate a population, then qualify a plug-in tail probability

$\bar x=33.6/60=0.56$ km and $\begin{aligned}&s^2\\ & =\frac{22.79-33.6^2/60}{59}\\ & =\frac{1987}{29500}\\ & =0.06735593220\ldots\text{ km}^2.\end{aligned}$ This is 0.067356 to 5 s.f. and 0.0674 to 3 s.f. The source prints 0.0675 at 3 s.f.; that rounding line is incorrect and the native source is unchanged.

The source substitutes $\hat\mu=0.56$ and $\hat\sigma^2\approx0.067356$ into $\bar X\approx N(0.56,0.067356/60)$, by CLT for 60 independent observations. With the exact estimated variance, $\begin{aligned}&P(\bar X<0.5)\approx\Phi\left(\frac{0.5-0.56}{\sqrt{(1987/29500)/60}}\right)\\ & =0.0366653904\approx0.0367.\end{aligned}$ This additionally treats estimated parameters as fixed and ignores estimation uncertainty.

The observed sample already has mean 0.56, so its realised mean is not random after observing the summaries and is not below 0.5. The source’s .0367 is a plug-in estimate for a hypothetical repeated independent size-60 sample from the same population, not a conditional probability that this observed mean changes. Normality of individual distances is not stated; the approximation relies on a suitable stable finite-variance sampling model.