distribution function of time T an ambulance station is located 30 miles from one end of a 100-mile road. the station services accidents along the entire road. suppose that an accident occurs. suppose that Suppose accidents occur with uniform distribution along the road and suppose that the ambulance travels 60 miles per hour. Let T be the amount of time it takes the ambulance to arrive at the sence with the assumption that the ambulance leaves the station right after the accident occurs.  find the distribution function FT(t) and the density function Ft(t)
 A: Let random variable $X$ be the distance the ambulance has to travel. The time $T$ is then $\frac{X}{60}$. We will find the distribution of $X$. From that it is not hard to find the distribution of $T$. We do $X$ instead of $T$ for two reasons: (i) It is closer to the intuition and (ii) There might as well be something left for you to do.
We will find the cumulative distribution function $F_X(x)$ of the random variable $X$. Recall that $F_X(x)=\Pr(X\le x)$.
First let's knock off the easy stuff. If $x\lt 0$, then $F_X(x)=0$. (The probability that the ambulance has to travel $\le -17$ miles is $0$. Also, if $x\gt 70$, then $F_X(x)=1$ (for sure the ambulance will have to travel fewer than $88$ miles).
So from now on we confine attention to $0\le x\le 70$. Because of the non-central location of the ambulance, we need to break up this interval.
Suppose $0\le x\le 30$. The ambulance has to travel a distance $\le x$ if te accident takes place within $\le x$ miles from the normal position of the ambulance. This is an interval of length $2x$, so the probability the accident takes place in this interval is $\frac{2x}{100}$.
In symbols, $F_X(x)=\frac{2x}{100}$ if $0\le x\le 30$.
Now suppose that $30\lt x\le 70$. Then we have to travel $\le x$ if (i) the accident takes place within $30$ miles of the ambulance (probability $\frac{60}{100}$) or the accident takes place at a distance between $30$ and $x$ from the ambulance (probability $\frac{x-30}{100}$. Add the two probabilities. 
Thus if $30\lt x\le 0$, then $F_X(x)=\frac{x+30}{100}$.
Now we have the cdf. Differentiate to get the density. there are points of non-differentiability at $0$, $30$, and $70$, but we won't worry about them We get $f_X(x)=0$ if $x\lt 0$ and also if $x\gt 70$. For $0\le x\le 30$, $f_X(x)=\frac{2}{100}$. For $30\lt x\le 70$, $f_X(x)=\frac{1}{100}$.
Remark: For your homework problem, you can either imitate the analysis, using $T$ or (easier) directly write down the cdf and density of $T$ (they are close relatives of the cdf and density of $X$.)
