Solving the two normal-quantile equations at z_0.05 = -1.644854 and z_0.99 = 2.326348 gives mean 19.733110 and standard deviation 27.195800. Their requested absolute ratio is 1.378181, which rounds to 1.38 and matches choice E.
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Solving the two normal-quantile equations at z_0.05 = -1.644854 and z_0.99 = 2.326348 gives mean 19.733110 and standard deviation 27.195800. Their requested absolute ratio is 1.378181, which rounds to 1.38 and matches choice E.
Setup
Setup
Write each supplied percentile as the mean plus its standard-normal quantile times the standard deviation.
qp=μ+zpσ
z0.05=−1.644853627,z0.99=2.326347874
Model
Model
Substitute the two given percentile values to obtain a linear system for the mean and standard deviation.
−25=μ−1.644853627σ
83=μ+2.326347874σ
Compute
Compute
Subtract the equations to find the standard deviation, then back-substitute for the mean and form the requested ratio.
σ=2.326347874−(−1.644853627)83−(−25)=27.19579955
μ=−25+1.644853627(27.19579955)=19.73310953
μσ=19.7331095327.19579955=1.378181148
Answer
Answer
The ratio rounds to 1.38, the value listed under choice E.
∣σ/μ∣≈1.38(E)
Error diagnosis
Why the other letters tempt
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
ADropping the minus sign from the lower percentile and solving with 25 instead of -25 gives an absolute ratio about 0.298, which rounds to 0.30.
BPairing the lower observation with +2.326 and the upper observation with +1.645, then ignoring the resulting negative standard deviation, gives an absolute ratio about 0.461.
CUsing positive signs for both standard scores while keeping their order gives an absolute ratio about 0.555, which rounds to 0.56.
DTaking the ratio of the two standard-score magnitudes, 1.644854 divided by 2.326348, gives about 0.707 or 0.71; that is not sigma divided by mu.
Original practice · fully worked
Original variant: resin-strength percentiles
A composite resin's strength index R is normally distributed with a positive mean and coefficient of variation 0.20. Its percentile corresponding to standard-normal score z = 1 is 52. Calculate the value at the symmetric percentile corresponding to z = -1.
A 34.7
B 41.6
C 43.3
D 52.0
E 62.4
Variant answer in brief
The named percentiles correspond to z = 1 and z = -1, while sigma = 0.20 mu. Solving mu + sigma = 52 gives mu = 43.3333 and sigma = 8.6667, so the lower percentile is 34.6667 and choice A.
Setup
Setup
Translate the coefficient of variation into a relationship between the positive mean and standard deviation.
μσ=0.20⟹σ=0.20μ
Model
Model
The 84.1345th percentile has standard score 1, so use its value to solve for the distribution parameters.
52=μ+σ=μ+0.20μ=1.20μ
μ=43.3333333,σ=8.6666667
Compute
Compute
The 15.8655th percentile has standard score -1, so evaluate the symmetric lower point.
q0.158655=μ−σ
q0.158655=43.3333333−8.6666667=34.6666667
Answer
Answer
The lower resin-strength percentile rounds to 34.7.
q0.158655≈34.7(A)
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