Let \(X\) be a continuous random variable whose cumulative distribution function (CDF) \(F_{X}(x)\), for some \(t\), is given as follows:
\[F_{X}(x) = \begin{cases} 0 & x \leq t \\ \dfrac{x - t}{4 - t} & t \leq x \leq 4 \\ 1 & x \geq 4 \end{cases}\]
If the median of \(X\) is 3, then what is the value of \(t\)?
Answer
(A)
Explanation
The median \(m\) is the point where the CDF reaches one half, \(F_{X}(m) = \frac{1}{2}\). Since 3 lies in the middle piece (\(t \leq 3 \leq 4\)),
\[\frac{3 - t}{4 - t} = \frac{1}{2} \;\Rightarrow\; 6 - 2t = 4 - t \;\Rightarrow\; t = 2.\]
Check: with \(t = 2\), \(X\) is uniform on \([2, 4]\), whose midpoint is 3. The other options fail; for example, \(t = 0\) gives \(F_{X}(3) = \frac{3}{4}\), so the median would be 2 instead.
Answer: A