The median is the point y in the interval such that
Prob{0 £ X £ y} = Prob{y £ X £ 3};
equivalently, such that
ó
õ 		y

0 
 r(x) dx = 1
2
 .
This means that
27
4
 = ó
õ 		y

0 
 (6+x-x2) dx = 6y+ y2
2
 - y3
3
 .
Thus y satisfies the cubic equation
4y3-6y2-72y+81=0.
This has one root y=9/2 so, by factorizing the cubic (or by using any other numerical method) we find that the only solution for y (which most satisfy 0 < y < 3) is (approximately) 1.1.

The mean of the 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
 .
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. Thus the mode is t, where 6+x-x2 takes its maximum value at t. Let f(x) = 6+x-x2; then f¢(x)=0 when x=1/2. Thus the maximum value of f is taken an one of the points 0, 1/2, 3. In fact the maximum value is at 1/2 so the mode is 1/2.