This Exam P sample reference tests Uniform Distribution. This is a uniform-moment inversion followed by a percentile calculation. The stated second moment forces the first lower endpoint to be 60, making the second interval [75,100] and its 80th percentile 95, so choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 80 is 75+0.20(25), the 20th percentile of Y. It uses the upper-tail complement 1-0.80 as the lower-tail percentile.
BThe value 85 is 75+0.40(25), only 40% of the way across Y's support. It results from halving the requested percentile position.
CThe value 90 is 75+0.60(25), the 60th percentile of Y. It stops before the required 80% position.
DThe value 92 is 60+0.80(40), the 80th percentile of X. It does not replace X's lower endpoint by the specified endpoint for Y.
Original practice · fully worked
Original variant: expected correction above a threshold
A calibration drift U is uniformly distributed on [a,16], where a is less than 16, and its second raw moment is 112. A correction amount is C=max(U-10,0). Calculate E[C].
A 0.50 units
B 1.50 units
C 3.00 units
D 6.00 units
E 10.00 units
Variant answer in brief
The uniform second-moment equation gives the valid lower endpoint a=4. Averaging the positive excess U-10 over the resulting interval [4,16] gives E[C]=1.5 units, so choice B.
Setup
Setup
Use the raw second moment of a uniform variable to recover the unknown lower endpoint.
E[U2]=3a2+16a+162=112
a2+16a−80=0
Model
Model
Retain the endpoint that produces a valid ordered support.
(a−4)(a+20)=0
a=4,U∼Uniform(4,16)
Compute
Compute
Integrate the positive correction over the portion of the support above the threshold.
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