This Exam P sample reference tests Normal Distribution. The two positive-profit probabilities imply standardized means about 1.05 and 1.40. With a common mean, the two standard deviations are therefore proportional to their reciprocals; adding the independent variances gives a two-year standardized mean about 1.68 and positive probability 0.9535, choice D.
Let the independent annual outcomes be X and Y, each with mean mu and standard deviations s1 and s2. Convert the two positive-tail probabilities into standard normal quantiles.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis multiplies the two annual positive probabilities, 0.8531 × 0.9192 = 0.7842, which is the probability both years are individually positive. A positive total can also occur when one year offsets a negative year.
BThis correctly aggregates the variance but uses only one copy of the common mean. The resulting z-score is about 1 / 1.1905 = 0.84, and Φ(0.84) = 0.7995; the two-year sum has mean 2μ.
CThis adds standard deviations rather than independent variances. It gives z-score 2 / (1 / 1.05 + 1 / 1.40) = 1.20 and Φ(1.20) = 0.8849.
EThis adds the two annual z-scores, 1.05 + 1.40 = 2.45, and reports Φ(2.45) = 0.9929. Standardized scores cannot be added without accounting for the combined variance.
Original practice · fully worked
Original variant: sign of a corrected laboratory score
Independent laboratory readings A and B are normally distributed with means 3 and 1 and standard deviations 2 and 3, respectively. A corrected score is R=2A-B. Calculate the probability that R is positive.
A 0.1587
B 0.5793
C 0.7625
D 0.8413
E 0.9172
Variant answer in brief
The linear score is normal with mean 5 and variance 4(2²)+3²=25. Its standardized mean is 1, so P(R>0)=Φ(1)=0.8413, selecting choice D.
Setup
Setup
Apply the linear-transformation rules for independent normal variables.
R=2A−B
Model
Model
Transform the means linearly and the independent variances with squared coefficients.
E[R]=2(3)−1=5
Var(R)=22(22)+(−1)2(32)=16+9=25
Compute
Compute
Standardize the zero threshold using the score's standard deviation of five.
Pr(R>0)=Φ(55)=Φ(1)
Pr(R>0)=0.8413447461
Answer
Answer
The corrected score is positive with probability 0.8413.
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