Let \(h\) be the probability density function for a continuous random variable \(X\), where
\(h(x)=\left\{
\begin{array} {c}
\rule{0pt}{2.5ex} \ \ \ \ \ \dfrac{x}{6}+k \rule[-1ex]{0pt}{0pt} & -3 \leq x<0 \\
\rule{0pt}{2.5ex} \ \ -\dfrac{x}{2}+k \rule[-1ex]{0pt}{0pt} & 0 \leq x \leq 1 \\
\rule{0pt}{2.5ex} 0 \rule[-1ex]{0pt}{0pt} & \text { elsewhere } \\
\end{array}\right.\)
and \(k\) is a positive real number.
The value of \(\text{Pr}(X<0.5)\) is
- \(\dfrac{1}{2}\)
- \(\dfrac{15}{16}\)
- \(\dfrac{3}{16}\)
- \(\dfrac{49}{48}\)
Show Worked Solution
\(\text{Using CAS}:\)
\(\text{Alternatively, given that }k=\dfrac{1}{2}\ \text{by CAS:}\)
| \(\text{Pr}\left(X<\dfrac{1}{2}\right)\) | \(=\displaystyle \int_{-3}^0 \dfrac{x}{6}+\dfrac{1}{2}\,dx +\int_0^{0.5} \dfrac{1}{2}+\dfrac{x}{2}\,dx\) |
| \(=\dfrac{1}{2}{\left[\dfrac{x^2}{6}+x\right] _{-3}^0} +\dfrac{1}{2}{\left[x-\dfrac{x^2}{2}\right] _0^{0.5}}\) | |
| \(=\dfrac{1}{2}\left[0-\left(\dfrac{9}{6}-3\right)\right]+\dfrac{1}{2}\left[\left(0.5-\dfrac{0.5^2}{2}\right)-0\right]\) | |
| \(=\dfrac{15}{16}\) |
\(\Rightarrow B\)
