This Exam P sample reference tests Uniform Distribution. This is a uniform-distribution probability centered at the mean. One standard deviation spans a total interval of length 30/√(3), so the requested probability is 1/√(3)=0.577350, which rounds to choice C.
For a uniform variable on an interval with endpoints a and b, write its mean and standard deviation.
μ=2a+b=80
σ=12b−a=1230=53
Model
Model
The interval one standard deviation from the mean remains completely inside the distribution's support. Uniform probability is interval length divided by total support length.
Pr(∣X−μ∣≤σ)=b−a(μ+σ)−(μ−σ)
Pr(∣X−μ∣≤σ)=302σ
Compute
Compute
Substitute the uniform standard deviation and simplify.
302(30/12)=122=31
31=0.5773502692…
Answer
Answer
The probability rounds to 0.58 at the precision of the listed answers.
Pr(∣X−μ∣≤σ)≈0.58(C)
Test the method on unseen problems
Probability Univariate Distributions Practice Workbook
30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 1/√(12)=0.288675 counts the interval from the mean to only one of the two standard-deviation boundaries.
BThis is close to the complement 1-1/√(3)=0.422650, so it reverses 'within' and 'outside' the centered interval.
DThe familiar 0.68 figure is the normal-distribution probability within one standard deviation; the distribution here is uniform.
EThe value 0.788675 is the one-sided probability above the lower boundary mu-sigma and fails to impose the upper boundary.
Original practice · fully worked
Original variant: calibration offsets
The calibration offset of a field sensor is uniformly distributed between -4 and 8 degrees. Calculate the probability that the offset is within 0.75 standard deviations of its mean.
A 0.217
B 0.375
C 0.547
D 0.433
E 0.750
Answer and calculation
For a uniform variable, the centered interval has length twice 0.75 standard deviations. Dividing that length by the 12-degree support gives 0.433013, so choice D is correct.
Setup
Setup
Compute the mean and standard deviation from the endpoints of the uniform support.
μ=2−4+8=2
σ=128−(−4)=12
Model
Model
A radius of 0.75 standard deviations stays inside the support, so probability is the centered interval length divided by 12.
Pr(∣X−2∣≤0.75σ)=122(0.75)σ
Compute
Compute
Insert the standard deviation and simplify the length ratio.
