Question-39

Consider the cumulative distribution function (CDF) of a random variable \(X\):

\[ F_{X}( x) =\begin{cases} 0, & x\leqslant -1\\ \frac{1}{4}( x+1)^{2} , & -1\leqslant x\leqslant 1\\ 1, & x\geqslant 1 \end{cases} \]

The value of \(P\left( X^{2} \leqslant 0.25\right)\) is:

• \(0.625\)
• \(0.25\)
• \(0.5\)
• \(0.5625\)

NoteAnswer

• \(0.625\)
• \(0.25\)
• \(0.5\)
• \(0.5625\)

NoteSolution

We have:

\[ \begin{aligned} P\left( X^{2} \leqslant 0.25\right) & =P( -0.5\leqslant X\leqslant 0.5)\\\\ & =F_{X}( 0.5) -F_{X}( -0.5)\\\\ & =\frac{1}{4}\left[( 1.5)^{2} -( 0.5)^{2}\right]\\\\ & =0.5 \end{aligned} \]