I am trying to solve the following question. I have come up a solution to this but need to know that is my approach correct in solving the question. The question is as follows
A Test is 1 hour long, and has two questions. The time you give to Q1 is represented by the random variable , which has the following PDF:
$f_Z(z)= \begin{cases} 12z(1-z)^2,& \text{if } 0 \leq z\leq 1\\ 0, & \text{otherwise} \end{cases} $
The rest of the time is given to Q2. The value of Z is chosen based on the relative lengths of the questions. Let be the ratio between the time you give to the longer question and the time you give the shorter question:
= longer question / shorter question.
Note that ≥ 1.
a) Find ( ≤ 2)
b) Find the probability distribution of i.e. ( ≤ )
$\\$
Well i am doing it in the following way
First of all changing hour unit to minutes the PDF of z will now become
$f_Z(z)= \begin{cases} 720z(1-60z)^2,& \text{if } 0 \leq z\leq 60\\ 0, & \text{otherwise} \end{cases}$
and the PDF of Q2 $= 1 - f_Z(z)$
so the of $X$ will become
$f_X(x)= \begin{cases} \frac{720z(1-60z)^2}{1 - 720z(1-60z)^2},& \text{if } 30 \leq z\leq 60\\ \frac{1 - 720z(1-60z)^2}{720z(1-60z)^2},& \text{if } 0 \leq z\leq 30\\ 0, & \text{otherwise} \end{cases} $
Am I going in the right direction?
First note that PDF of $1-Z$ is not $1-f_Z(z)$ and also PDF of $U/V$ is not the ratio of PDF's of $U$ and $V$. Note also that your PDF of $Z$ in minutes is not PDF: it integrates not to 1 but to 8391860496000.
If I understand the question correctly, $X=\frac{\max(Z,1-Z)}{\min(Z,1-Z)}$.
First of all, you can solve inequality $X\leq 2$ with respect to $Z$: $$\tag{1}\label{1} \frac{\max(Z,1-Z)}{\min(Z,1-Z)} \leq 2. $$ Consider separately the cases $Z\leq \frac12$ and $Z>\frac12$.
In the first case $\max(Z, 1-Z)=1-Z$, $\min(Z,1-Z)=Z$ and the equality (\ref{1}) leads to $1-Z\leq 2Z$ or $\frac13\leq Z\leq \frac12$.
Solve the equality for the second case the same way and get $\frac12 < Z\leq \frac23$.
Then find probability in (a) as $$ \mathbb P(X\leq 2)= \mathbb P\left(\frac{\max(Z,1-Z)}{\min(Z,1-Z)} \leq 2\right) = \mathbb P\left(\frac13\leq Z \leq \frac23\right) = \int\limits_{1/3}^{2/3} f_Z(z)\,dz.$$
In (b) replace $2$ by $x$ and repeat.