This Exam P sample reference tests Continuous Random Variables. This is a percentile calculation under a symmetric triangular density. Normalization gives c=1/9, and solving (x-5)²⁄¹⁸=0.30 on the rising half gives x=5+√(5.4)=7.32379, so choice E is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis places the percentile almost at the lower endpoint, where the triangular density is near zero and too little probability has accumulated.
BThis also understates the quantile by treating the rising density too much like a constant height instead of integrating its triangular area.
CThis is the 30%-of-range location 5+0.30(11-5)=6.80, which would apply to a uniform density rather than the stated triangular density.
DThis results from underestimating the area needed on the rising branch; the correct CDF is quadratic in the distance x-5, not linear.
Original practice · fully worked
Original variant: maximum strength-ratio median
Three independently manufactured fibers have strength ratios X1, X2, and X3, each with density f(x)=2x for 0<x<1. A batch is summarized by its strongest fiber, M=max(X1,X2,X3). Calculate the median of M.
A 0.5000
B 0.7071
C 0.7937
D 0.8909
E 0.9439
Variant answer in brief
Each fiber has CDF x², so the maximum has CDF m⁶. Setting that CDF equal to one-half gives median 2⁻¹⁄⁶=0.890899, which selects choice D.
Setup
Setup
Integrate the common fiber density to obtain its distribution function.
FX(x)=∫0x2tdt=x2,0<x<1
Model
Model
The maximum is at most m exactly when all three independent fibers are at most m.
FM(m)=Pr(X1≤m,X2≤m,X3≤m)
FM(m)=FX(m)3=m6
Compute
Compute
Set the maximum's CDF equal to one-half and solve for its median.
m6=21
m=2−1/6=0.8908987181…
Answer
Answer
The median strongest-fiber ratio is approximately 0.8909.
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