Here is another excellent solution from Andrei at Tudor Vianu National College, Romania:

The mean of this distribution is $m$, where

${\displaystyle m={\int }_0^{3}x\hspace{0.5em}\rho (x)\hspace{0.5em}\mathrm{dx}.}$

Thus

${\displaystyle m=\frac{2}{27}{\int }_0^{3}x(6+x-x^2)\hspace{0.5em}\mathrm{dx}=\frac{2}{27}{\int }_0^3(6x+x^2-x^3)\hspace{0.5em}\mathrm{dx},}$

hence

${\displaystyle m=\frac{2}{27}[3x^2+x^3/3-x^4/4{]}_0^3=\frac{2}{27}[27+9-81/4]=\frac{7}{6}.}$

To calculate the median $t$ of the distribution I must have:

${\displaystyle \mathrm{Prob}\{0\le X\le t\}=\mathrm{Prob}\{t\le X\le 3\}.}$

But the total probability is 1, so each of the two probabilities is 0.5. Then I have to solve the equation

${\displaystyle {\int }_0^t\rho (x)\hspace{0.5em}\mathrm{dx}=\frac{1}{2}}$

which gives:

${\displaystyle \frac{2}{27}{\int }_0^t(6+x-x^2)\mathrm{dx}=\frac{1}{2}.}$

I obtain the following equation

${\displaystyle 4t^3-6t^2-72t+81=0}$

which has the solutions:

${\displaystyle t_1=\frac{9}{2},\hspace{0.5em}{t}_2=\frac{3}{2}(-1-\sqrt{3}),\hspace{0.5em}\frac{3}{2}(-1+\sqrt{3}).}$

The only solution in the interval $[0,3]$ is $t_3$ which is approximately 1.098.

In a discrete distribution the mode is the value that occurs most frequently. In a continuous distribution it is the value where the probability density function takes its maximum value. To find this maximum I shall calculate the derivative $\rho \text{'}(x)$ and then $\rho \text{'}\text{'}(x)$:

${\displaystyle \rho \text{'}(x)=\frac{2}{27}(1-2x)}$

${\displaystyle \rho \text{'}\text{'}(x)=\frac{-4}{27}.}$

As the second derivative is negative the function has indeed a maximum. The maximum value of $\rho$ occurs at $x=1/2$ so the mode is $1/2$.