Question

Let X be a random variable with distribution function \( F_X(x) = \begin{cases} 0 & \text{for } x < 0 \\ \frac{x}{8} & \text{for } 0 \le x < 1 \\ \frac{1}{4} + \frac{x}{8} & \text{for } 1 \le x < 2 \\ \frac{3}{4} + \frac{x}{12} & \text{for } 2 \le x < 3 \\ 1 & \text{for } x \ge 3 \end{cases} \) then \( P(1 \le X \le 2) \) is

• A. $\frac{3}{8}$
• B. $\frac{7}{16}$
• C. $\frac{13}{24}$
• D. $\frac{19}{24}$

Hint:

In CDF problems, always check if the function is continuous at the boundaries. If $F(a) \neq \lim F(x \to a^-)$, there is a discrete mass $P(X=a)$. For this variable, $P(X=1) = 5/8 - 1/4 = 3/8$ and $P(X=2) = 3/4 - 6/8 = 0$. Knowing these jumps is key.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:

Using the distribution function printed in the paper, we need \( P(1 \le X \le 2) \), which includes both end points.

Step 2: Detailed Explanation:

• For a distribution function, \( P(a \le X \le b) = F(b) - F(a^-) \).

• \( F(2) = \frac{3}{4} + \frac{2}{12} = \frac{11}{12} \) (the branch for \( 2 \le x < 3 \)).

• \( F(1^-) = \frac{1}{8} \) (the branch for \( 0 \le x < 1 \) at \( x = 1 \)).

• \( P(1 \le X \le 2) = \frac{11}{12} - \frac{1}{8} = \frac{22 - 3}{24} = \frac{19}{24} \).

Step 3: Final Answer:
The correct option is (D) \( \frac{19}{24} \).