GATE 2025 DA Q.19: solution and answer

GATE 2025 · DAQ.19 · 1 mark · MCQ

Let $X$ be a continuous random variable whose cumulative distribution function (CDF) $F_{X} (x)$ , for some $t$ , is given as follows:

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

If the median of $X$ is 3, then what is the value of $t$ ?

(A)
2
(B)
1
(C)
$- 1$
(D)
0

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Answer and solution

ANSWER

The value of $t$ is $2$ , option (A).

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

The median $m$ of a continuous random variable satisfies $F_{X} (m) = \frac{1}{2}$ . Here $m = 3$ , so we need $F_{X} (3) = \frac{1}{2}$ .
The CDF is $0$ for $x \leq t$ and $1$ for $x \geq 4$ . A value of $\frac{1}{2}$ is strictly between 0 and 1, so $3$ must lie in the middle piece, $t < 3 < 4$ . There $F_{X} (x) = \frac{x - t}{4 - t}$ .
Set $\frac{3 - t}{4 - t} = \frac{1}{2}$ .
Cross-multiply: $2 (3 - t) = 4 - t$ , so $6 - 2 t = 4 - t$ .
Rearranging gives $t = 2$ , which satisfies $t < 3$ , so the assumption in step 2 holds.
Check: $F_{X} (3) = \frac{3 - 2}{4 - 2} = \frac{1}{2}$ . Here $X$ is uniform on $[2, 4]$ , whose midpoint is $3$ , as expected. The other options give $F_{X} (3) = \frac{2}{3}$ for $t = 1$ , $\frac{3}{4}$ for $t = 0$ and $\frac{4}{5}$ for $t = - 1$ , none of which equals $\frac{1}{2}$ .

KEY IDEA

For a continuous random variable the median $m$ is the point where the CDF equals $\frac{1}{2}$ . Solve $F_{X} (3) = \frac{1}{2}$ using the piece of the CDF that contains 3.

WATCH OUT FOR

Setting the median to the midpoint of $0$ and $4$ , or using the wrong piece of the CDF. Another slip is cross-multiplying incorrectly and getting $t = 1$ from $3 - t = \frac{1}{2} (4 - t)$ treated as $3 - t = 2 - t$ . Always substitute the answer back and check that $F_{X} (3) = \frac{1}{2}$ .

From the GATE 2025 DA paper. Every question of this paper · Download the full paper · All 10 Continuous distributions questions, with notes

GATE 2025 DA Q.19: solution and answer

Answer and solution

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