This Exam P sample reference tests Normal Distribution. Full-precision normal quantiles give a shared standard deviation of 105.1017 and a second mean of 79.3068. The official table-rounded calculation gives 79.25; both select 79 and choice D.
Write the common reported value as a normal quantile for the first distribution.
214=30+σz0.96
z0.96=Φ−1(0.96)=1.750686071…
Model
Model
Solve for the shared standard deviation and write the second percentile equation.
σ=z0.96214−30=105.1016530…
214=μB+σz0.90
Compute
Compute
Insert the 90th-percentile standard-normal value and isolate the second mean. A standard table rounded at the 96th percentile reproduces the slightly different official intermediate value.
z0.90=1.281551566…
μB=214−(105.1016530)(1.281551566)
μB=79.30681204…
μBtable=214−1.75184(1.2816)=79.24891429…
Answer
Answer
The second mean rounds to 79.
79(D)
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.
AA mean of 33 would place 214 at z=(214-33)/105.1017=1.722, about the 95.75th percentile, not the 90th percentile.
BReplacing normal quantiles by the raw percentile levels gives 214-184(0.90/0.96)=41.5, which rounds toward 42. Percentile probabilities cannot substitute for z-scores.
CA mean of 54 would make the standardized value 1.522, about the 93.60th percentile rather than the required 90th percentile.
EThe value 105 is the recovered common standard deviation, not the second distribution's mean.
Original practice · fully worked
Original variant: percentile of a squared normal deviation
A calibration score X is normally distributed with mean 50 and standard deviation 8. Define the squared deviation L=(X-50)². Calculate the 90th percentile of L.
A 10.25
B 64.0
C 105.1
D 173.2
E 245.9
Answer and calculation
The event L at most ell is a central normal interval. Central probability 0.90 leaves 0.05 in each tail, so √(ell)/8=z_0.95 and ell=(8z_0.95)²=173.1548, choice D.
Setup
Setup
Standardize the score and translate the squared-deviation event.
Z=8X−50∼N(0,1)
{L≤ℓ}={∣Z∣≤8ℓ}
Model
Model
A central probability of 0.90 leaves probability 0.05 in each normal tail.
8ℓ=z0.95=1.644853627…
Compute
Compute
Return to squared score units.
ℓ=(8z0.95)2
ℓ=173.1547811…
Answer
Answer
The 90th percentile of the squared deviation is approximately 173.2.
