91Ó°ÊÓ

The probability density function of a random variable X is given as,

\(f\left( x \right) = \left\{ \begin{array}{l}\frac{1}{8}x\;\;\;for\;0 \le x \le 4,\\0\;\;\;\;\;\;\;\;\;\;otherwise\end{array} \right.\)

The probability in general for a continuous variable is the value of integral over the given interval.

a.

The required value of t is computed as,

\(\begin{aligned}{c}\Pr \left( {X \le t} \right) &= \frac{1}{4}\\\int\limits_0^t {\frac{1}{8}xdx} &= \frac{1}{4}\\\frac{1}{8}\left( {\frac{{{x^2}}}{2}} \right)_0^t &= \frac{1}{4}\\\frac{1}{{16}}{t^2} &= \frac{1}{4}\\t &= \pm 2\end{aligned}\)

Therefore, the value of t is 2.

b.

The required value of t is computed as,

\(\begin{aligned}{c}\Pr \left( {X \ge t} \right) &= \frac{1}{2}\\\int\limits_t^4 {\frac{1}{8}xdx} &= \frac{1}{2}\\\frac{1}{8}\left( {\frac{{{x^2}}}{2}} \right)_t^4 &= \frac{1}{2}\\\frac{1}{{16}}\left( {16 - {t^2}} \right) &= \frac{1}{2}\\{t^2} &= 8\\t &= \pm \sqrt 8 \end{aligned}\)

Therefore, the value of t is \(\sqrt 8 \).