I am stuck with finding the pdf for Z = X+Y, I have the pdf for X and Y.
The problem:
$f_Y(y) = 1/(b-a) ~~\big[25 \leqslant y \leqslant 35\big]$
$f_X(x) = 0.1 - 0.00667x+1.1\times 10^{-4} x^2$
I was trying to find the $\Pr(Z\geqslant z)$
I tried to solve it as following
$\Pr(Z \geqslant z) = 1 - \Pr(Z < z) = 1- Pr(X+Y<z) = 1- Pr(X< z - Y)$
as the problem is with sea level (X) and wave level (Y) I think they are independent
$\int f_Y(y)\times\int f_X(x) $
Is this true? or should I use double integral and if yes how?
Thanks
The formula for the density of $X+Y$ is $\int f_X(x-y)f_Y(y)\, dy$ provided $X$ and $Y$ are independent. Without independence it is not possible to find the distribution of $X+Y$.
For $25 <x<35$ integrate w.r.t. $y$ from $25$ to $x$; $35 <x<55$ integrate w.r.t. $y$ from $25$ to $55$; $55 <x<65$ integrate w.r.t. $y$ from $x-30$ to $x$. For $x \notin (25, 65)$ the density of $X+Y$ is $0$. Once you find $f_{x+y}$ you can use $P(X+Y >50)=\int_{50}^{65} f_{x+y} (t)\, dt$.