Bus arrival density function

Question

Fred lives in Blissville, where buses always arrive exactly on time, with the time
between successive buses fixed at 10 minutes. Having lost his watch, he arrives at
the bus stop at a random time (assume that buses run 24 hours a day, and that the
time that Fred arrives is uniformly random on a particular day).

 **What is the distribution of how long Fred has to wait for the next bus? What is
the average time that Fred has to wait, and what is the density function (PDF) or his waiting time till the next bus?

So the answer is: $E[X] = 5$ and the PDF is $\frac{1}{10}$.

Can you explain me why? Why do I need to get the average between 0 and 10 for the $E[X]$ and why the pdf is $\frac{1}{10}$?

I mean I want to understand the inside of the calculation.. the lambda here is 1/10 but other than that.. I know that PDF is not just the lambda so how can be it 1/10?

Thanks.

Answer 1

The buses arrive always on time every 10' thus the distribution is uniform in $[0;10]$ The density of a continuous uniform is

$$f(x)=\frac{1}{b-a}=\frac{1}{10-0}=\frac{1}{10}$$

Its expectation ( the average time he has to wait) is

$$E(X)=\frac{a+b}{2}=\frac{0+10}{2}=5$$