Question

The p.d.f. of a continuous r.v. x is $f\left(x\right)\begin{array}{cc}=\frac{7}{9}{x}^2-2,& 0<x<3\\ =0,& otherwise\end{array}$. Then $P(X\le -1)$ is?

• A. $0$
• B. $1/2$
• C. $1/3$
• D. $1/4$

Answer 1

The correct option is A $0$

$\because f\left(x\right)\ne 0\forall x\in (0,3)$

$\therefore f\left(x\right)=0$ $\forall$ $x\in (-\infty ,0)\cup (3,\infty )$

$\therefore p(x\le -1)=0$

Hence, the answer is $0.$