Gate DA 2025 | Question - 19 | Probability and Statistics

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

$F_{X} (x) = \{\begin{cases} \begin{array}{ll} 0 & x \leq t \\ \frac{x - t}{4 - t} & t \leq x \leq 4 \\ 1 & x \geq 4 \end{array} \end{cases}$

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

-1

Solution:

We are given the CDF $F_X(x)$ of a continuous random variable $X$ as:

$F_{X} (x) = \{\begin{cases} \begin{array}{ll} 0 & x \leq t \\ \frac{x - t}{4 - t} & t \leq x \leq 4 \\ 1 & x \geq 4 \end{array} \end{cases}$

We are told the median of X is 3.

The median is the value $m$ such that:

$F_X(m) = 0.5$

Given: $m = 3$

So,

$F_X(3) = 0.5$

From the middle case of the CDF:

$F_X(x) = \frac{x - t}{4 - t}, \quad \text{for } t \leq x \leq 4$

Substitute $x = 3$ and set equal to 0.5:

$\frac{3 - t}{4 - t} = 0.5$

$\frac{3 - t}{4 - t} = \frac{1}{2}$

Cross-multiply:

$2(3 - t) = 4 - t$ $6 - 2t = 4 - t$ $6 - 4 = 2t - t \Rightarrow 2 = t$

(A) 2

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