Suppose Y has density function:
f(y) = ky(1-y) when 0<=y<=1
f(y)= 0 elsewhere
a)Find the value of k that makes f(y) a probability density function
for part a I found k to be 6 (by making the integral = 1 on the interval 0,1 and solving for k
for part b also used the integral between 0.4 and 1 and got 0.648
c) d) is where I have the problem. I suppose both must have the same answer.
Since this is conditional probability can we state that:
P(Y<=0.4|Y<=0.8) = [P(y<=0.4) AND P(y<=0.8)]/P(y<=0.8) = P(0.4<=Y<=0.8)/P(y<=0.8)
so I find using integral P(0.4<=Y<=0.8)=0.544
and P(y<=0.8) = P(0<=y<=0.8)=0.896
then we divide 0.544/0.896= 0.607
I feel that I am doing something wrong here, but can't see what.