Answer:

(a) The graph is \(y=kx^3\) on \(0\leq x\lt1\), then \(y=k(5-x)\) on \(1\leq x\leq5\), and \(y=0\) otherwise. It rises from \(0\) to \(k\) on \(0\leq x\lt1\), jumps to \(4k\) at \(x=1\), then decreases linearly to \(0\) at \(x=5\).

(b) \(k=\frac{4}{33}\).

(c) \(F(x)=\begin{cases}0,&x\lt0,\\\frac{x^4}{33},&0\leq x\lt1,\\\frac{4}{33}\left(5x-\frac{x^2}{2}-\frac{17}{4}\right),&1\leq x\leq5,\\1,&x\gt5.\end{cases}\)

(d) The median is \(\frac{10-\sqrt{33}}{2}\approx2.13\).

For \(f\) to be a probability density function, the total area under the graph must be \(1\).

For \(0\leq x\lt1\), the graph is the cubic curve \(y=kx^3\), starting at \(0\) and approaching \(k\) at \(x=1\). For \(1\leq x\leq5\), the graph is the straight line \(y=k(5-x)\), from \(4k\) at \(x=1\) to \(0\) at \(x=5\). The graph is zero outside this range.

Now

\(\int_0^1 kx^3\,dx=\frac{k}{4}\).

Also

\(\int_1^5 k(5-x)\,dx=k\left[5x-\frac{x^2}{2}\right]_1^5=8k\).

Therefore

\(\frac{k}{4}+8k=1\), so \(\frac{33k}{4}=1\), giving \(k=\frac{4}{33}\).

For the cumulative distribution function, first \(F(x)=0\) for \(x\lt0\).

For \(0\leq x\lt1\),

\(F(x)=\int_0^x \frac{4}{33}t^3\,dt=\frac{x^4}{33}\).

For \(1\leq x\leq5\),

\(F(x)=\frac{1}{33}+\int_1^x \frac{4}{33}(5-t)\,dt\).

So

\(F(x)=\frac{4}{33}\left(5x-\frac{x^2}{2}-\frac{17}{4}\right)\).

Finally, \(F(x)=1\) for \(x\gt5\).

Hence

\(F(x)=\begin{cases}0,&x\lt0,\\\frac{x^4}{33},&0\leq x\lt1,\\\frac{4}{33}\left(5x-\frac{x^2}{2}-\frac{17}{4}\right),&1\leq x\leq5,\\1,&x\gt5.\end{cases}\)

The median \(m\) lies in the interval \(1\leq m\leq5\), so set \(F(m)=\frac12\):

\(\frac{4}{33}\left(5m-\frac{m^2}{2}-\frac{17}{4}\right)=\frac12\).

This simplifies to

\(4m^2-40m+67=0\).

Therefore

\(m=\frac{10\pm\sqrt{33}}{2}\).

Only the value in the interval \(1\leq m\leq5\) is valid, so

\(m=\frac{10-\sqrt{33}}{2}\approx2.13\).