FRM Part I · FRM Exam Part I · Fundamentals of Probability
A continuous random variable X has CDF F(x) = x²/16 for 0 ≤ x ≤ 4 (F = 0 below 0 and 1 above 4). What is the median of X?
The median solves F(m) = 0.5. Setting m squared over 16 equal to 0.5 gives m squared = 8, so m is about 2.83. The support midpoint of 2 is wrong because the density rises with x.
- A2.00
- B2.83Correct
- C3.00
- D8.00
Explanation
The median m satisfies F(m) = 0.5, so m²/16 = 0.5, m² = 8, and m = 2.83. Check: 2.83² = 8.0 and 8/16 = 0.5. The value 2.00 is the midpoint of the support, which is not the median because the density is not uniform.
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