This Exam P sample reference tests Uniform Distribution. For a uniform interval, the mean is its midpoint and the standard deviation is its width divided by √(12). The width is 2√(3), so the lower endpoint is 3-√(3)≈1.27, which is choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis reports 1/√3 = 0.577…, the reciprocal of the correct half-width factor. The half-width is √3 standard deviations, and it must be subtracted from the mean.
CThis is the half-width √3 = 1.732… by itself. The requested endpoint is the midpoint minus that half-width.
DThis value is 4/√3 = 2.309…, obtained by dividing an unsupported four-hour span by the uniform half-width factor. The supplied equations instead determine both the sum 6 and width 2√3.
EThis uses a reciprocal half-width and computes 3 − 1/√3 = 2.423…. From Var(X) = (b − a)²/12, the half-width is √3, not 1/√3.
Original practice · fully worked
Original variant: recover spread from a tail fraction
A laboratory reading X follows a continuous uniform law over an unknown interval. The reading averages 10, while only 20% of readings are above 14. Determine Var(X).
A 3.85
B 10.00
C 13.33
D 14.81
E 177.78
Variant answer in brief
The threshold 14 is four units above the midpoint and is the 80th percentile. That four-unit shift equals 0.30 of the support width, giving width 40/3 and variance 400/27≈14.81, choice D.
Setup
Setup
Write the support in terms of its known midpoint 10 and unknown width w.
X∼Uniform(10−2w,10+2w)
Model
Model
A 0.20 upper tail makes 14 the 80th percentile, located 0.30w above the midpoint.
14=10−2w+0.80w=10+0.30w
Compute
Compute
Solve for the support width and insert it into the continuous-uniform variance formula.
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