I am a probability noob and I was solving the following problem, but my answer doesn't match the book's.
The length and width of panels used for interior doors(in inches) are denoted as Xand Y, respectively. Suppose that Xand Yare independent, continuous uniform random variables for 17.75<x<18.25 and 4.75<y<5.25, respectively.determine the probability that thearea of a panel exceeds 90 squared inches.
I thought the area random variable will have a minimum value of 17.75*4.75=84.3125 Maximum value of area will be 18.25*5.25=95.8125
Also if X and Y are uniformly distributed and independent then area should be uniformly distributed too. Every single value that area can take has equal chance. Please correct me here.
Now with this when I integrate area(A) from 90 to 95.8125 for a constant PDF f(A)=1/(95.8125-84.3125), my answer comes out to be 0.5054. But the answer according to the solutions given at the back of book is 0.499. I am not able to understand where I am doing a mistake. Can anyone please help me understand where I am going wrong?
A Start: Let random variable $W$ represent the area. The distribution of $W$ is not uniform, though it is not terribly far from being uniform.
If we assume independence (which is not entirely plausible), then the joint density of $X$ and $Y$ is $4$ on the rectangle given by $17.75\le x\le 18.25$ and $4.75\le y\le 5.25$.
Draw the rectangle, and draw carefully the hyperbola $xy=90$. The probability that the area is $\gt 90$ is the integral of $4$ over the part of the rectangle which is "above" the hyperbola. So it is $4$ times the area of a certain region.