 Solution

The Lorentz factor is $\gamma =E/(m_{\mu }{c}^2)=20000/106=188.6792$ and $\beta =\sqrt{1-{\gamma }^{-2}}$. Traversal of the height $H$ takes proper time $H/(\beta \gamma c)$. Using the exponential decay law with the given muon lifetime,The laboratory flight lasts about $83.39\mu \mathrm{s}$, while the elapsed proper time is about $0.4420\mu \mathrm{s}$. This is the probability of surviving decay to the surface, with atmospheric energy loss and detector efficiency neglected.