Question

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

\[ F_X(x) = \begin{cases} 0 & \text{if } x \leq -1 \\ \frac{1}{4}(x + 1)^2 & \text{if } -1 \leq x \leq 1 \\ 1 & \text{if } x \geq 1 \end{cases} \]

The value of \( P(X^2 \leq 0.25) \) is:

• A. 0.625
• B. 0.25
• C. 0.5
• D. 0.5625

Hint:

When solving for probabilities involving a square, first determine the range of the variable that satisfies the inequality. Then, use the CDF to calculate the desired probability.

Answer 1

The Correct Option is C

To find \( P(X^2 \leq 0.25) \), we solve for the range \( -0.5 \leq X \leq 0.5 \). Using the CDF, we calculate the probabilities for the corresponding values of \( X \). The probability is: \[ P(X^2 \leq 0.25) = F_X(0.5) - F_X(-0.5) = 0.5625 - 0.0625 = 0.5 \]