This Exam P sample reference tests Conditional Normal Probability. Standardizing profit 60 and 0 gives z-scores -2 and -5. The conditional probability is [Φ(5)-Φ(2)]/Φ(5), choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AChoice A uses the unconditional upper-tail probability beyond two standard deviations and omits conditioning on the value being positive.
BChoice B forms a ratio of cumulative probabilities rather than the probability of the interval between the two standardized boundaries.
CChoice C replaces the lower-bound cumulative probability by one. That approximation is extremely close numerically but is not the exact conditional expression.
DChoice D uses ratios of the original monetary boundaries instead of standardizing by subtracting the mean and dividing by the standard deviation.
Original practice · fully worked
Original variant: normal output within a conditioned range
Daily output X is normal with mean 50 and standard deviation 10. Let Phi denote the standard normal CDF. Determine P(X≤40 | X>20).
A 1-Φ(1)
B [Φ(-1)-Φ(-3)]/Φ(3)
C Φ(-1)/Φ(-3)
D [Φ(3)-Φ(1)]/Φ(1)
E Φ(1)-Φ(-3)
Variant answer in brief
The standardized interval is -3<Z≤-1 and the conditioning event has probability Φ(3). Their ratio is [Φ(-1)-Φ(-3)]/Φ(3), choice B.
Setup
Setup
Standardizing 20 and 40 gives boundaries negative three and negative one.
X=20↦Z=−3,X=40↦Z=−1
Model
Model
The numerator is the probability between those boundaries, while the denominator is the probability above negative three.
Pr(X≤40∣X>20)=Pr(Z>−3)Pr(−3<Z≤−1)
Compute
Compute
Writing both probabilities with the standard normal cumulative distribution gives the expression in choice B, numerically about 0.157518.
Pr=Φ(3)Φ(−1)−Φ(−3)=0.157518
Answer
Answer
The conditional probability is the expression in choice B.
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