FRM Part I · FRM Exam Part I · Random Variables
A random variable X has the cumulative distribution function 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), which is 9/16 minus 1/16, or 0.5. The CDF gives the probability of an interval as the difference of its values at the endpoints; 0.5625 is only F(3) and omits the subtraction.
- A0.4375Correct
- B0.5000
- C0.5625
- D0.2500
Explanation
P(1 < X ≤ 3) = F(3) − F(1) = 9/16 − 1/16 = 8/16 = 0.5. Check: 9/16 = 0.5625 and 1/16 = 0.0625, so the difference is 0.5000. The option 0.4375 is wrong; 0.5625 is only F(3), which ignores subtracting F(1).
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