9. It is known that at a bus terminal, three bus services $A, B, C$ depart simultaneously at 6:00 AM, and their departure intervals are 10 minutes, 12 minutes, and 15 minutes, respectively. If Xiao Ming arrives at the terminal to take one of these three bus services at some point between 8:00 AM and 12:00 PM, then the average waiting time for Xiao Ming is $\qquad$ minutes.
Given that $10=2 \times 5, 12=2^{2} \times 3, 15=3 \times 5$, then $[10,12,15]=2^{2} \times 3 \times 5=60$, so we only need to consider the situation within one hour.

As shown in the figure above, the average waiting time for the bus in the intervals $(0,10]$ and $(50,60]$ is 5 minutes; in the intervals $(10,12]$ and $(48,50]$ the average waiting time is 1 minute; in the intervals $(12,15]$ and $(45,48]$ the average waiting time is 1.5 minutes; in the intervals $(15,20]$ and $(40,45]$ the average waiting time is 2.5 minutes; in the intervals $(20,24]$ and $(36,40]$ the average waiting time is 2 minutes; in the intervals $(24,30]$ and $(30,36]$ the average waiting time is 3 minutes.
Therefore, the average waiting time for Xiao Ming is $\frac{20 \times 5 + 4 \times 1 + 6 \times 1.5 + 10 \times 2.5 + 8 \times 2 + 12 \times 3}{60} = \frac{19}{6}$ minutes.