= Solution
Interpret the quoted muon <energy> as total <relativistic energy>, and neglect energy loss in the atmosphere. The <Lorentz factor>, <speed> and Earth-frame <travel time> are
$$
\gamma=\frac{6000\,\mathrm{MeV}}{106\,\mathrm{MeV}}=56.60,\qquad
\beta=\sqrt{1-\gamma^{-2}}=0.999844,\qquad
t=\frac{h}{\beta c}.
$$
The <muon decay> law is exponential in <proper time>. Equivalently, <time dilation> makes the Earth-frame mean lifetime $\gamma\tau_0$. The <exponential survival probability> over the flight is
$$
P=\exp\!\left(-\frac{t}{\gamma\tau_0}\right)
=\exp\!\left(-\frac{h}{c\tau_0\sqrt{\gamma^2-1}}\right).
$$
With $h=15000\,\mathrm m$ and $\tau_0=2.2\times10^{-6}\,\mathrm s$, the elapsed <proper time> is $8.841\times10^{-7}\,\mathrm s$. Therefore
$$
\boxed{P\simeq e^{-0.40186}\simeq0.669,\quad\text{about }67\%.}
$$
This uses the stated single incident <energy> and vertical trajectories. If the number were a <kinetic energy>, one would instead use $\gamma=1+T/(mc^2)$; the distinction is not silently absorbed into the <rest energy>.
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