Đáp án:
\(A.\ 0,0267\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad f(x) = \begin{cases}4x^3,\quad x \in [0;1]\\0,\qquad x\notin [0;1]\end{cases}\\
\text{Ta có:}\\
\bullet\quad E(X) = \displaystyle\int\limits_{-\infty}^{+\infty}xf(x)dx\\
\Leftrightarrow E(X) = \displaystyle\int\limits_0^14x^4dx\\
\Leftrightarrow E(X) = \dfrac45\\
\bullet\quad E(X^2) = \displaystyle\int\limits_{-\infty}^{+\infty}x^2f(x)dx\\
\Leftrightarrow E(X^2) = \displaystyle\int\limits_0^14x^5dx\\
\Leftrightarrow E(X^2) = \dfrac23\\
\text{Khi đó:}\\
\quad Var(X) = E(X^2) - [E(X)]^2\\
\Leftrightarrow Var(X)= \dfrac23 - \left(\dfrac45\right)^2\\
\Leftrightarrow Var(X) = \dfrac{2}{75}\approx 0,0267
\end{array}\)
