I was trying to solve the following renewal process problem:
A policeman spends his entire day on the lookout for speeders. The policeman cruises on average approximately 10 minutes before stopping a car for some offense. Of the cars he stops, 90% of the drivers are given speeding tickets with an \$80 fine. It takes the policeman an average of 5 minutes to write such a ticket. The other 10% of the stops are for more serious offenses, leading to an average fine of \$300. These more serious charges take an average of 30 minutes to process. Let's suppose we start to observe the policeman when he finishes writing a ticket.
Suppose that the time the policeman takes is exactly 10 minutes, from the time he finishes fining one car until he stops the next car and the time that it takes to process the fine is uniform, with the lower bound being 0 (both minor and serious offenses)
1 - What would be the the distribution of the time that elapses since the policeman ends with a car until the next fine is processed?
2 - What is the distribution of the number of fines processed in 1 hour would (from the beginning)?
But I found some problems to give the distribution in an explicit manner.
Until the moment I have:
$$f(z) = \begin{cases} U(10, 15) & \text{for 90% of the cases} \\ U(10, 40) & \text{for 10% of the cases} \\ \end{cases}$$
And from there, I was looking here to calculate the sum of the two distributions. But
How can I deal with the percentages?
In case this can not be given as a explicit formula, how can I calculate the last part of the exercise (2)?