The median is the point

y

in the interval such that

Prob {0 \leq X \leq y} = Prob {y \leq X \leq 3};

equivalently, such that

\int_{0}^{y} ρ (x) dx = \frac{1}{2} .

This means that

\frac{27}{4} = \int_{0}^{y} (6 + x - x^{2}) dx = 6 y + \frac{y^{2}}{2} - \frac{y^{3}}{3} .

Thus

y

satisfies the cubic equation

4 y^{3} - 6 y^{2} - 72 y + 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

$m = \int_{0}^{3} x ρ (x) dx .$

Thus

$m = \frac{2}{27} \int_{0}^{3} x (6 + x - x^{2}) dx = \frac{2}{27} \int_{0}^{3} (6 x + x^{2} - x^{3}) dx,$

hence

$m = \frac{2}{27} [3 x^{2} + x^{3} / 3 - x^{4} / 4]_{0}^{3} = \frac{2}{27} [27 + 9 - 81 / 4] = \frac{7}{6} .$

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 $6 + x - x^{2}$ takes its maximum value at $t$ . Let $f (x) = 6 + x - x^{2}$ ; 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 $1 / 2$ .