= Solution
One complete cycle consists of ten stops of mean one minute and ten journeys of mean four minutes, so its mean duration is fifty minutes. Over the eight-hour shift, the elementary renewal approximation gives \b[$480/50=9.6$ completed circuits on average].
For the end-of-shift calculation use the long-run occupation <probabilities> of the twenty successive phases. Each stop phase has <probability> $1/50$, and each travel phase <probability> $4/50$. Alighting stop $j$ is therefore selected with <probability> $1/50+4/50=1/10$, from either stopping there or travelling toward it. If that stop is the station, no further stops are passed. If it is stop $j\ge2$, the intermediate stops are $j+1,\ldots,10$, numbering $10-j$. Thus the requested approximate mean is
$$
\boxed{\frac1{10}\left(0+8+7+\cdots+1+0\right)=3.6.}
$$
The approximation treats eight hours as long compared with a cycle and ignores the initial-phase transient. Exponential <holding times> justify the twenty-phase Markov description; independent cycle times justify the renewal rate.
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