Let $f_X(x)=k\sin x,0≤x≤π$.
- (a) Find $k$ so $f_X$ is a probability density function.
I got $k = -1/2$ - (b) Find $P(π/3≤X ≤π/2)$.
I got $1/4$ - (c) Find $P(X ≥ π/3)$.
I stuck here, this function is divergent right? - (d) Sketch the graph of $f_X$ and shade the area represented by the probability in
- (e) Find $F_X$ and sketch its graph.
(a) Try again; since you want a density, then $$\int_0^{\pi}k\sin x\,dx = 1.$$ You got the wrong $k$.
(b) Since you got (a) wrong, I assume this is wrong too. Should be easy to fix.
(c) No, they told you that $0\leq x\leq \pi$. Hence $$P(X\geq \pi/3) = \int_{\pi/3}^\pi f_X(x)\,dx.$$
(d),(e) This should not be too bad.