This Exam P sample reference tests Normal Distribution. The independent profit components sum to a normal variable with mean 600 and standard deviation √(400²+300²)=500. Standardizing zero gives -1.20, so the requested probability is P(Z>-1.20) and choice B is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis uses an aggregate spread that is too small, producing an excessively large standardized mean magnitude.
CThis adds the component standard deviations to get 700, yielding 600/700 about 0.86 rather than using the root-sum-of-squares spread.
DThe displayed z-score 0.83 implies an aggregate standard deviation near 600/0.83=723, which is even larger than the unsupported arithmetic sum 400+300=700.
EThis standardizes the first profit component alone, using 200/400=0.50 and omitting the second component.
Original practice · fully worked
Original variant: inferred reading correlation
Two jointly normal calibration readings A and B have means 10 and 6 and standard deviations 3 and 4, respectively. The probability that A-B is positive is 0.8413447461. Calculate the correlation between A and B.
A -0.375
B 0.000
C 0.375
D 0.750
E 1.000
Variant answer in brief
The difference has mean 4, and the stated probability is Φ(1), so its standard deviation must be 4. Equating Var(A-B)=25-24rho to 16 gives rho=0.375, so choice C is correct.
Setup
Setup
Let D be the difference and express its mean and variance in terms of the unknown correlation.
D=A−B
E[D]=10−6=4
Var(D)=32+42−2ρ(3)(4)=25−24ρ
Model
Model
The stated positive probability corresponds to a standard-normal score of one.
Pr(D>0)=Φ(SD(D)4)=Φ(1)
SD(D)=4
Compute
Compute
Set the variance expression equal to 16 and solve for the correlation.
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