Consider the cumulative distribution function (CDF) of a random variable \(X\):
\[F_X(x) = \begin{cases} 0 & x \le -1 \\ \frac{1}{4}(x+1)^{2} & -1 \le x \le 1 \\ 1 & x \ge 1 \end{cases}\]
The value of \(P(X^{2} \le 0.25)\) is
Answer
(C)
Explanation
\(X^{2} \le 0.25\) means \(-0.5 \le X \le 0.5\). The CDF has no jumps (it rises continuously from 0 at \(x = -1\) to 1 at \(x = 1\)), so
\[P(-0.5 \le X \le 0.5) = F_X(0.5) - F_X(-0.5) = \frac{(1.5)^{2} - (0.5)^{2}}{4} = \frac{2}{4} = 0.5.\]
Two slips lead to the distractors: \(F_X(0.5) = 0.5625\) alone forgets the lower limit (D), and reading the event as \(|X| \le 0.25\) gives \(F_X(0.25) - F_X(-0.25) = 0.25\) (B).
Answer: C