Discussion
The distribution of a random variable X is given by :
\[ F(x) = P[X \le x] = \begin{cases} 0 & \text{if } x \lt 0 \\ \frac{1}{4} & \text{if } 0 \le x \lt \frac{1}{4} \\ \frac{1}{2} & \text{if } \frac{1}{4} \le x \lt \frac{1}{2} \\ \frac{3}{4} & \text{if } \frac{1}{2} \le x \lt \frac{3}{4} \\ \frac{x + 3}{5} & \text{if } \frac{3}{4} \le x \lt 2 \\ 1 & \text{if } x \ge 2 \end{cases} \]
Then \(P\left[\frac{1}{2} \le X \le \frac{3}{4}\right]\) is :
Answer: C
The probability is \(\frac{1}{4}\).
For a distribution function with jumps, \(P[a \le X \le b] = F(b) - F(a^{-})\), where \(F(a^{-})\) is the left limit at \(a\). Here
\[ \begin{aligned} F\left(\tfrac{3}{4}\right) &= \frac{3/4 + 3}{5} = \frac{3.75}{5} = \frac{3}{4} \\ F\left(\tfrac{1}{2}^{-}\right) &= \frac{1}{2} \\ P &= \frac{3}{4} - \frac{1}{2} = \frac{1}{4} \end{aligned} \]
This 1/4 is the jump at \(x = 1/2\) (from 1/2 to 3/4); F is continuous at 3/4, since the formula \((x + 3)/5\) gives 3/4 there, so there is no extra mass at the right endpoint.
The value 3/4 is \(F(3/4)\) alone, 1/2 forgets to subtract the right amount, and 1/8 does not arise from the distribution function.
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