This Exam P sample reference tests Normal Distribution. The two upper-tail probabilities correspond to z-scores 0.449876 and 0.599859. Their 0.149984 difference spans 0.12 units, giving standard deviation 0.800088 and variance 0.640141, so choice B is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis results from an incorrect normal-table conversion that makes the z-score gap too large and therefore understates the variance.
CThis reports the standard deviation, approximately 0.80, rather than the requested variance.
DThis reports approximately the reciprocal of the standard deviation, 1/0.80=1.25.
EThis reports approximately the reciprocal of the variance, 1/0.64=1.56.
Original practice · fully worked
Original variant: demand mean from CV
A daily demand X is normally distributed with positive mean. Its coefficient of variation is 0.25, and its 90th percentile is 132.038789138615. Calculate the mean of X.
A 25.000
B 99.029
C 100.000
D 105.631
E 132.039
Variant answer in brief
CV 0.25 gives sigma=0.25mu. The 90th percentile is therefore mu[1+0.25Phi inverse 0.90]. Equating this to 132.03878914 gives mu=100, so choice C is correct.
Setup
Setup
Translate the coefficient of variation into a relation between standard deviation and mean.
μσ=0.25
σ=0.25μ
Model
Model
Write the normal 90th percentile using its standard-normal quantile.
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