(a) hast du richtig gerechnet:
$$\int_{-\infty}^{\infty}f_X(x)\; dx = a\int_{0}^{1}x^2(1-x)\; dx\stackrel{!}{=}1\Rightarrow a=12$$
(b)
$$F_X(x) = \int_{-\infty}^{x}f_X(t)\; dx$$
X "lebt" nur auf [0,1]. Daher für \(x\in [0,1]\):
$$F_X(x) = 12\int_0^x t^2(1-t)\; dt = (4-3x)x^3$$ Also
$$ F_X(x) = = \left\{ \begin{array}{rl} 0 & x<0 \\ (4-3x)x^3 & 0\leq x\leq 1 \\ 1 & x> 1\end{array}\right.$$
(c)
\( E(X) =\int_{-\infty}^{\infty}xf_X(x)\; dx = 12\int_{0}^{1}x^3(1-x)\; dx = \frac 35 \)
\(D^2(X) = E(X^2) - E(X)^2\)
\( E(X^2) =\int_{-\infty}^{\infty}x^2f_X(x)\; dx = 12\int_{0}^{1}x^4(1-x)\; dx = \frac 25 \)
\(\Rightarrow D^2(X) = \frac 1{25}\)
(d)
\(P(X<0.5) = F_X(0.5) = \frac 5{16}\)
\(P(X<E(X)) = P(X<0.6) = F_X(0.6) = \frac {297}{625} = 0.4752\)
