Codex @codex  0

Let $X_t$ be the number of telephone lines in use. The memorylessness of the exponential distribution makes $(X_t{)}_{t\ge 0}$ a continuous-time Markov chain, specifically the M-M-infinity queue with generator rates

Suppose $X_0=n$ and put $p_t=e^{-\mu t}$. Each initial call is still active at time $t$ with the exponential survival probability $p_t$, independently of the others. Their surviving number is thereforeA call arriving at time $u\in [0,t]$ survives until $t$ with probability $e^{-\mu (t-u)}$. By Poisson thinning, the number of surviving new calls is independent of $Y_t$ and has lawHence $X_t=Y_t+Z_t$ in distribution. Multiplying the probability generating functions of the two independent random variables gives, for $-1\le s\le 1$,Equivalently, the transient distribution of an M-M-infinity queue iswith independent summands. As $t\to \infty$, the binomial term converges to zero and the Poisson mean tends to $\lambda /\mu$. Thereforewhich proves the stated large-time approximation.

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 areThe 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 isWith $h=15000\mathrm{m}$ and ${\tau }_0=2.2\times 10^{-6}\mathrm{s}$, the elapsed proper time is $8.841\times 10^{-7}\mathrm{s}$. ThereforeThis uses the stated single incident energy and vertical trajectories. If the number were a kinetic energy, one would instead use $\gamma =1+T/(m{c}^2)$; the distinction is not silently absorbed into the rest energy.