GATE 2025 DA Q.39: solution and answer

GATE 2025 · DAQ.39 · 2 marks · MCQ

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

F_{X} (x) = ⎩ ⎨ ⎧ 0 \frac{1}{4} (x + 1)^{2} 1 x \leq - 1 - 1 \leq x \leq 1 x \geq 1

The value of $P (X^{2} \leq 0.25)$ is

(A)
0.625
(B)
0.25
(C)
0.5
(D)
0.5625

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Answer and solution

ANSWER

$P (X^{2} \leq 0.25) = 0.5$ , option (C).

Official answer: CSource: Official answer key, GATE 2025 DA (IIT Roorkee)

Rewrite the event: $X^{2} \leq 0.25 ⟺ - 0.5 \leq X \leq 0.5$ .
The CDF is continuous on $[- 1, 1]$ (it equals $0$ at $x = - 1$ and $1$ at $x = 1$ , and $\frac{1}{4} (x + 1)^{2}$ is continuous), so $X$ has no point masses and $P (X = - 0.5) = 0$ .
Hence $P (- 0.5 \leq X \leq 0.5) = F_{X} (0.5) - F_{X} (- 0.5)$ .
$F_{X} (0.5) = \frac{1}{4} (1.5)^{2} = \frac{2.25}{4} = 0.5625$ .
$F_{X} (- 0.5) = \frac{1}{4} (0.5)^{2} = \frac{0.25}{4} = 0.0625$ .
$P (X^{2} \leq 0.25) = 0.5625 - 0.0625 = 0.5$ , which is option (C).

KEY IDEA

Turn the event on $X^{2}$ into an interval for $X$ , then subtract CDF values at the two ends: $P (a \leq X \leq b) = F (b) - F (a)$ when $X$ has no point mass at $a$ .

WATCH OUT FOR

Using only $F_{X} (0.5) = 0.5625$ (option D) and forgetting that $X^{2} \leq 0.25$ also needs $X \geq - 0.5$ , so the lower tail $F_{X} (- 0.5) = 0.0625$ must be subtracted.

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GATE 2025 DA Q.39: solution and answer

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