Original practice · fully worked
Original variant: conditional interval under an increasing density
A quality score X has density f(x)=x/8 for 0<x<4. Given that X exceeds 1, determine the probability that X is less than 2.
- A
0.125 - B
0.1875 - C
0.250 - D
0.200 - E
0.750
Answer and calculation
The CDF is x²/16. The mass from one to two is 3/16 and the mass above one is 15/16, giving 1/5, choice D.
Setup
Setup
Integrate the density from zero to x to obtain the cumulative distribution F(x)=x²/16.
F(x)=∫0xt/8dt=x2/16 Model
Model
Within the condition X>1, the target event is the interval from 1 to 2; divide that interval's mass by the mass above 1.
Pr(X<2∣X>1)=1−F(1)F(2)−F(1) Compute
Compute
The numerator is 3/16 and the denominator is 15/16, so the conditional probability is 3/15=0.20.
1−1/164/16−1/16=153=0.20 Answer
Answer
Thus Pr(X<2 given X>1)=0.20, which is choice D.
0.20(D)