Perhaps using an iterative method like Newton Raphson,or you could rearrange it into f(x) = 0 and try and guess factors.
I ended up with x =~ 1.33, is that right?
The Continuous random variable Y has pdf f(y) defined by:
f(y)=12y^2(1-y) for 0<y<1, 0 otherwise
Find the median:
so you want when p<y=0.5
So i did this by integrating 12y^2-12y^3
which is 4y^3-3y^4 = 0.5.
(then someone's going to tell me i've integrated it wrong!)
Due to the nature of question I would have thought there would be an exact answer using a method, other than Newton Raphson, etc etc...
I get 0.6143 and 1.2475 as my solutions to 4DP.
How do I solve: 4x^3-3x^4=0.5?
There are no nice factors of 8x^3 - 6x^4 - 1 that I can see, so an approximations might be the best you can get from me...
We just started that and it had a few questions like that and we were told not to bother doing them by our teacher because a) they are unsoluable using methods they expect for s3 and b) we'd never get asked such a thing in an exam. In fact our teacher said it probably was an error in printing.
Yeah i understand your methods...i think the book must have done a printing error...or some sort...as it thinks the answer is 2/3