The PDF of a continuous random variate is $f(x)=\left\{\begin{array}{ll}0, & x

Question

The PDF of a continuous random variate is $f(x)=\left\{\begin{array}{ll}0, & x<2 \ \frac{(3+2 x)}{18}, & 2 \leq x \leq 4 \ 0, & x>4\end{array}ight.$. Then, the probability that $x$ will lie between $2 \leq x \leq 3$ is

Professor PC Thomas Classes · Answer

$\int_2^3 \frac{3+2x}{18} dx = \frac{1}{18} [3x + x^2]_2^3 = \frac{1}{18}( (9+9) - (6+4) ) = \frac{8}{18} = \frac{4}{9}$.