8. A station has exactly one bus arriving every $8: 00 \sim 9: 00, 9: 00 \sim 10: 00$, but the arrival time is random, and the arrival times of the two buses are independent, with the following pattern:

A passenger arrives at the station at $8: 20$, then the expected waiting time for the bus is $\qquad$ (to the nearest minute).
The arrival time of each bus is unique, and the arrival times of two buses are independent of each other. Therefore, the probabilities that the passenger boards the bus at 8:30 and 8:50 are $\frac{1}{2}$ and $\frac{1}{3}$, respectively, and the probability of missing the first bus is $\frac{1}{6}$. The expected waiting time for the passenger is $E(X)=10 \times \frac{1}{2}+30 \times \frac{1}{3}+50 \times \frac{1}{36}+70 \times \frac{1}{12}+90 \times \frac{1}{18}$ $=15+\frac{110}{9}=27$ (minutes).