$F(y) = \left\{\begin{array}{ll} 0 & : y \lt 0\\ cy^3 & : 0 \le y \le 2\\ 1 - \left(\frac{1}{y^2}\right) &: y \ge 2 \end{array} \right.$
1) Find c
2) Find the density function (pdf) of Y
3) Find the interquartile range, that is, $\phi_{0.75} - \phi_{0.25}$
What I have done so far:
a) For this, I took the limits from both sides around 2. So: $$\lim_{y\to2^{-}} F(y) =\lim_{y\to2^{-}} cy^3 = 8c$$ and $$\lim_{y\to2^{+}} F(y) =\lim_{y\to2^{+}} 1- \left(\frac{1}{y^2}\right) = \left(\frac{3}{4}\right)$$
Then setting those two equal, I got $c = \left(\frac{3}{32}\right)$
b) For this part, I believe I was supposed to just take the derivative at each place. So I got:
$f(y) = F'(y)= \left\{\begin{array}{ll} 0 & : y \lt 0\\ \left(\frac{9}{32}\right)y^2 & : 0 \le y \le 2\\ \left(\frac{2}{y^3}\right) &: y \ge 2 \end{array} \right.$
c) For this part, I'm not really sure how to go about this yet. I am struggling to understand how to set these equations up and how to solve them. I know that I do need to solve $\phi_{0.75}$ and $\phi_{0.25}$. As far as I understand, I can start with the general formula $P(Y \le \phi_{p}) = F(\phi_{p}) \ge p$. I'm not sure how to pick the correct piece of the CDF to plug into for both $p=0.75$ and $p=0.25$ to solve that equation though.
The median $M$ is defined by a CDF equal to $1/2$. So you get
$$\frac{3}{32} y^3=\frac{1}{2} $$ $$M=2 \sqrt[3]{\frac{2}{3}}$$
Similarly, you get the $q_1$ and $q_3$ values of the IQR by setting
$$\frac{3}{32} y^3=\frac{1}{4} $$ $$q_1=\frac{2}{ \sqrt[3]{3}}$$
and
$$\frac{3}{32} y^3=\frac{3}{4} $$ $$q_3=2$$
Note that this last value was already known from your calculations.
So the IQR $\displaystyle \{ q_1,q_3\}= \left\{ \frac{2}{ \sqrt[3]{3} },2 \right\}$