Đáp án:

\(\begin{array}{l}
a)\quad  c = \dfrac34\\
b)\quad E(X) = \dfrac65\\
\qquad Var(X) = \dfrac{4}{25}\\
c)\quad P(X > 1) = \dfrac{11}{16}\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
\quad f(x) = \begin{cases}cx^2(2-x)\quad khi\quad x\in [0,2]\\0\kern50pt khi\quad x\notin [0,2]\end{cases}\\
a)\quad \text{$(f(x)$ là hàm mật độ xác suất khi và chỉ khi}\\
\quad \displaystyle\int\limits_{-\infty}^{+\infty}f(x)dx = 1\\
\Leftrightarrow \displaystyle\int\limits_0^2cx^2(2-x)dx = 1\\
\Leftrightarrow c\left(\dfrac{2x^3}{3} - \dfrac{x^4}{4}\right)\Bigg|_0^2 = 1\\
\Leftrightarrow \dfrac43c = 1\\
\Leftrightarrow c = \dfrac34\\
b)\quad \text{Kỳ vọng toán:}\\
\quad E(X) = \displaystyle\int\limits_{-\infty}^{+\infty}xf(x)dx \\
\Leftrightarrow E(X) = \displaystyle\int\limits_0^2\dfrac34x^3(3-x)dx\\
\Leftrightarrow E(X) = \dfrac65\\
\quad E(X^2) = \displaystyle\int\limits_{-\infty}^{+\infty}x^2f(x)dx \\
\Leftrightarrow E(X^2) = \displaystyle\int\limits_0^2\dfrac34x^4(3-x)dx\\
\Leftrightarrow E(X^2) = \dfrac85\\
\text{Phương sai:}\\
\quad Var(X) = E(X^2) - [E(X)]^2\\
\Leftrightarrow Var(X) = \dfrac85 - \left(\dfrac65\right)^2\\
\Leftrightarrow Var(X) = \dfrac{4}{25}\\
c)\quad P(X > 1) = \displaystyle\int\limits_1^{+\infty}f(x)dx\\
\Leftrightarrow P(X > 1) = \displaystyle\int\limits_1^2\dfrac34x^2(2-x)dx\\
\Leftrightarrow P(X > 1) = \dfrac{11}{16}
\end{array}\)