Answer
(a) $k = \frac{2}{9}$
(b) $P(X \gt 1) = \frac{20}{27}$
(c) $\mu = \frac{3}{2}$
Work Step by Step
(a) We can find the value of $k$ such that $f$ is a probability density function:
$\int_{-\infty}^{\infty} f(x)~dx = 1$
$\int_{0}^{3}f(x)~dx = 1$
$\int_{0}^{3}k(3x-x^2)~dx = 1$
$k~\int_{0}^{3}(3x-x^2)~dx = 1$
$k~[(\frac{3}{2}x^2-\frac{1}{3}x^3)\vert_{0}^{3}~] = 1$
$k~[\frac{3}{2}(3)^2-\frac{1}{3}(3)^3-0] = 1$
$k~(\frac{27}{2}-9) = 1$
$k~(\frac{9}{2}) = 1$
$k = \frac{2}{9}$
(b) We can find $P(X \gt 1)$:
$P(X \gt 1) = \int_{1}^{3}f(x)~dx$
$P(X \gt 1) = \int_{1}^{3}(\frac{2}{9})(3x-x^2)~dx$
$P(X \gt 1) = (\frac{2}{9})[(\frac{3}{2}x^2-\frac{1}{3}x^3)\vert_{1}^{3}~]$
$P(X \gt 1) = (\frac{2}{9})[(\frac{3}{2}(3)^2-\frac{1}{3}(3)^3) -(\frac{3}{2}(1)^2-\frac{1}{3}(1)^3) ~]$
$P(X \gt 1) = (\frac{2}{9})(\frac{9}{2} -\frac{7}{6})$
$P(X \gt 1) = (\frac{2}{9})(\frac{27}{6} -\frac{7}{6})$
$P(X \gt 1) = (\frac{2}{9})(\frac{20}{6})$
$P(X \gt 1) = \frac{20}{27}$
(c) We can find the mean $\mu$:
$\mu = \int_{-\infty}^{\infty} xf(x)~dx$
$\mu = \int_{0}^{3} xf(x)~dx$
$\mu = \int_{0}^{3} \frac{2}{9}x~(3x-x^2)~dx$
$\mu = \frac{2}{9}~\int_{0}^{3} (3x^2-x^3)~dx$
$\mu = \frac{2}{9}~[(x^3-\frac{1}{4}x^4)\vert_{0}^{3}~]$
$\mu = \frac{2}{9}~[(3)^3-\frac{1}{4}(3)^4-0~]$
$\mu = \frac{2}{9}~(27-\frac{81}{4})$
$\mu = \frac{2}{9}~(\frac{108}{4}-\frac{81}{4})$
$\mu = \frac{2}{9}~\cdot \frac{27}{4}$
$\mu = \frac{3}{2}$
