Question

Let \( X \) be a continuous random variable whose cumulative distribution function (CDF) \( F_X(x) \), for some \( t \), is given as follows:
\[ F_X(x) = \begin{cases} 0 & \text{if } x \leq t \\ \frac{x - t}{4 - t} & \text{if } t \leq x \leq 4 \\ 1 & \text{if } x \geq 4 \end{cases} \]
If the median of \( X \) is 3, then what is the value of \( t \)?

• A. 2
• B. 1
• C. -1
• D. 0

Hint:

To find the median from the CDF, set \( F_X(m) = 0.5 \) and solve for \( m \). Use the given piecewise expression for the CDF to substitute the median value and solve for \( t \).

Answer 1

The Correct Option is A

We are given the CDF of \( X \) as:
\[ F_X(x) = \begin{cases} 0 & \text{if } x \leq t \\ \frac{x - t}{4 - t} & \text{if } t \leq x \leq 4 \\ 1 & \text{if } x \geq 4 \end{cases} \]
The median of a random variable \( X \) is the value \( m \) such that \( F_X(m) = 0.5 \).

Given that the median of \( X \) is 3, we need to solve for \( t \) such that:
\[ F_X(3) = 0.5. \]
Using the second case of the CDF, for \( t \leq 3 \leq 4 \), we have:
\[ F_X(3) = \frac{3 - t}{4 - t}. \]
Equating this to 0.5:
\[ \frac{3 - t}{4 - t} = 0.5. \]
Multiplying both sides by \( 4 - t \):
\[ 3 - t = 0.5 \times (4 - t). \]
Expanding the right-hand side:
\[ 3 - t = 2 - 0.5t. \]
Rearranging:
\[ 3 + 2 = 0.5t + t \quad \Rightarrow \quad 5 = 1.5t. \]
Solving for \( t \):
\[ t = \frac{5}{1.5} = \frac{10}{3} \approx 2.5. \]
Thus, the value of \( t \) is approximately 2.5, which corresponds to Option (A).