The GATE Grind

GATE 2025 DA – Question 39

Probability and Statistics · Random variables, discrete and continuous distributions · 2 marks · Multiple choice

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

  1. 0.625
  2. 0.25
  3. 0.5
  4. 0.5625

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Correct answer: (C) 0.5

Explanation

$X^2 \le 0.25$ means $-0.5 \le X \le 0.5$. So $P = F_X(0.5) - F_X(-0.5) = \frac{1}{4}(1.5)^2 - \frac{1}{4}(0.5)^2 = \frac{2.25 - 0.25}{4} = 0.5$. (The distribution is continuous, so the end points do not matter.)