Suppose $F_Y (y) =\begin{cases} y^3 & 0 ≤ y < 1/2\\ 1 − y^3& 1/2 ≤ y ≤ 1.\\\end{cases}$
Compute each of the following.
(a) $P(1/3 < Y < 3/4)$
Attempt:
$$P(1/3 < Y < 3/4) = P(Y < 3/4) - P(Y \leq 1/3) = (1-(3/4)^3)-(1/3)^3 = \frac{935}{1728}$$
Why is this wrong?
Looking at the formula for $F_Y$, which I assumed it was the (cumulative) distribution function of $Y$, I see that this is not a CDF by any means. It is not only undefined outside $[0,1]$ (which could be easily arranged), but it is decreasing between $y=\frac12$ and $y=1$, which is impossible. So, either it is a a probability density function (usually noted $f_Y$) and the correct answer would be $$\int_{1/3}^{3/4} f_Y(y) dy$$ (although you should check before that $$\int_{0}^{1} f_Y(y) dy=1$$ to be able to assume this is a PDF) or something is wrong somewhere.
UPDATE: I've checked the numbers and it turns out that $$\int_{0}^{1} f_Y(y) dy=\frac9{32},$$ so this isn't a PDF either (nevertheless, $\frac{32}9 F_Y$ would).