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

The mean of this distribution is m, where

m= ó õ 3 0  x r(x) dx.

Thus

m = 2 27 ó õ 3 0  x æ è 6+x-x2 ö ø  dx = 2 27 ó õ 3 0  (6x+x2-x3) dx,

hence

m= 2 27 é ë 3x2 + x3/3 - x4/4 ù û 3 0  = 2 27 [27 +9 -81/4] = 7 6 .

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

Prob{0 £ X £ t} = Prob {t £ X £ 3}.

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

ó õ t 0  r(x) dx = 1 2

which gives:

2 27 ó õ t 0  (6+x-x2)dx = 1 2 .

I obtain the following equation

4t3-6t2-72t+81=0

which has the solutions:

t1 = 9 2 , t2 = 3 2 (-1-Ö3),  3 2 (-1 + Ö3).

The only solution in the interval [0,3] is t₃ 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 r¢(x) and then r¢¢(x):

r¢(x)= 2 27 (1 - 2x)

r¢¢(x) = -4 27 .

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