FRM Part I · FRM Exam Part I · Random Variables
A continuous random variable X has CDF F(x) = 0 for x < 0, F(x) = x^2/16 for 0 ≤ x ≤ 4, and F(x) = 1 for x > 4. What is P(1 < X ≤ 3)?
The probability equals F(3) minus F(1). F(3) is 9/16 and F(1) is 1/16, so the difference is 8/16, or 0.5. Subtracting the CDF at the lower bound removes the mass below 1.
- A0.2500
- B0.5000Correct
- C0.5625
- D0.7500
Explanation
P(1 < X ≤ 3) = F(3) − F(1) = 9/16 − 1/16 = 8/16 = 0.5. Using F(3) alone gives 0.5625, a common mistake of ignoring the lower bound. Using 3/4 reflects a linear misreading of the CDF.
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