Answer:

(a) The cumulative distribution function is

\(F(x)=\begin{cases}0, & x\lt 0 \\ \frac{x^2}{10}, & 0\le x\lt 2 \\ \frac{10x-x^2}{15}-\frac{2}{3}, & 2\le x\le 5 \\ 1, & x\gt 5\end{cases}\)

(b) The median is \(m=5-\frac{\sqrt{30}}{2}\approx 2.26\).

(c) \(\mathrm{E}(X^2)=\frac{13}{2}=6.5\).

(d) \(\mathrm{P}(1\le X\le 3)=\frac{19}{30}\).

(a) The cumulative distribution function is \(F(x)=\mathrm{P}(X\le x)=\int_{-\infty}^x f(t)\,dt\).

For \(x\lt 0\), the density is zero, so \(F(x)=0\).

For \(0\le x\lt 2\),

\(F(x)=\int_0^x \frac{1}{5}t\,dt=\frac{1}{5}\left[\frac{t^2}{2}\right]_0^x=\frac{x^2}{10}.\)

For \(2\le x\le 5\), split the integral at \(2\):

\(F(x)=\int_0^2 \frac{1}{5}t\,dt+\int_2^x \frac{2}{15}(5-t)\,dt.\)

The first part is

\(\int_0^2 \frac{1}{5}t\,dt=\frac{1}{5}\left[\frac{t^2}{2}\right]_0^2=\frac{2}{5}.\)

The second part is

\(\int_2^x \frac{2}{15}(5-t)\,dt=\frac{2}{15}\left[5t-\frac{t^2}{2}\right]_2^x\)

\(=\frac{2}{15}\left(5x-\frac{x^2}{2}-(10-2)\right)=\frac{2}{15}\left(5x-\frac{x^2}{2}-8\right).\)

So

\(F(x)=\frac{2}{5}+\frac{2}{15}\left(5x-\frac{x^2}{2}-8\right)=\frac{10x-x^2}{15}-\frac{2}{3}.\)

For \(x\gt 5\), all the probability has been included, so \(F(x)=1\).

Hence

\(F(x)=\begin{cases}0, & x\lt 0 \\ \frac{x^2}{10}, & 0\le x\lt 2 \\ \frac{10x-x^2}{15}-\frac{2}{3}, & 2\le x\le 5 \\ 1, & x\gt 5\end{cases}\)

(b) The median \(m\) satisfies \(F(m)=\frac12\).

Since \(F(2)=\frac{2^2}{10}=\frac{2}{5}\lt \frac12\), the median lies in the interval \(2\le m\le 5\).

So use the second part of \(F(x)\):

\(\frac{10m-m^2}{15}-\frac{2}{3}=\frac12.\)

Multiply by \(30\):

\(2(10m-m^2)-20=15\)

\(20m-2m^2-20=15\)

\(2m^2-20m+35=0.\)

Using the quadratic formula,

\(m=\frac{20\pm\sqrt{400-280}}{4}=\frac{20\pm\sqrt{120}}{4}=5\pm\frac{\sqrt{30}}{2}.\)

Only the value in the interval \([2,5]\) is valid, so

\(m=5-\frac{\sqrt{30}}{2}\approx 2.26.\)

(c) To find \(\mathrm{E}(X^2)\), use

\(\mathrm{E}(X^2)=\int_{-\infty}^{\infty} x^2 f(x)\,dx.\)

So

\(\mathrm{E}(X^2)=\int_0^2 x^2\cdot \frac{x}{5}\,dx+\int_2^5 x^2\cdot \frac{2}{15}(5-x)\,dx.\)

First integral:

\(\int_0^2 \frac{x^3}{5}\,dx=\frac{1}{5}\left[\frac{x^4}{4}\right]_0^2=\frac{1}{5}\cdot 4=\frac{4}{5}.\)

Second integral:

\(\int_2^5 \frac{2}{15}(5x^2-x^3)\,dx=\frac{2}{15}\left[\frac{5x^3}{3}-\frac{x^4}{4}\right]_2^5.\)

At \(x=5\):

\(\frac{5(125)}{3}-\frac{625}{4}=\frac{625}{3}-\frac{625}{4}=\frac{625}{12}.\)

At \(x=2\):

\(\frac{5(8)}{3}-\frac{16}{4}=\frac{40}{3}-4=\frac{28}{3}.\)

Difference:

\(\frac{625}{12}-\frac{28}{3}=\frac{625}{12}-\frac{112}{12}=\frac{513}{12}=\frac{171}{4}.\)

So the second integral is

\(\frac{2}{15}\cdot \frac{171}{4}=\frac{57}{10}.\)

Therefore

\(\mathrm{E}(X^2)=\frac{4}{5}+\frac{57}{10}=\frac{8}{10}+\frac{57}{10}=\frac{65}{10}=\frac{13}{2}=6.5.\)

(d) We need

\(\mathrm{P}(1\le X\le 3)=F(3)-F(1).\)

Now

\(F(1)=\frac{1^2}{10}=\frac{1}{10}.\)

Also, since \(3\ge 2\), use the second part:

\(F(3)=\frac{10(3)-3^2}{15}-\frac{2}{3}=\frac{30-9}{15}-\frac{2}{3}=\frac{21}{15}-\frac{10}{15}=\frac{11}{15}.\)

Hence

\(\mathrm{P}(1\le X\le 3)=\frac{11}{15}-\frac{1}{10}=\frac{22}{30}-\frac{3}{30}=\frac{19}{30}.\)