This Exam P sample reference tests Normal Distribution. The normal percentile identifies an input variance of 25, after which the variance condition fixes the positive scale factor at one. The required output mean then gives an intercept of 30, matching choice D.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AWith the verified scale a=1, b=-34.72 would make E[Y]=20-34.72=-14.72 rather than the required 50.
BThe sign-error value b=-30 gives E[Y]=-10; solving 50=20+b requires adding 30, not subtracting it.
CSetting b=0 preserves the raw mean at 20 and ignores the condition that E[Y]=50.
EThe value b=34.72 would produce E[Y]=54.72, so it is incompatible with the mean constraint even though its sign is positive.
Original practice · fully worked
Original variant: acoustic calibration offset
An acoustic laboratory's uncalibrated noise index R is normally distributed with average 40, and 90% of readings do not exceed 52.8155. The lab publishes S=cR+d. Instrument records state that the variance of R is an integer, while calibration requires c to be positive, the standard deviation of S to be 15, and the average of S to be 180. Determine the programmed offset d.
A -120
B 60
C 100
D 120
E 180
Variant answer in brief
The 90th percentile identifies an input standard deviation of 10, so the positive output scale is 15/10=1.5. Matching the required output average then gives d=120, which is choice D.
Setup
Setup
Use the 90th-percentile z value to recover the spread of the uncalibrated index.
z0.90=1.2815516…
σR=z0.9052.8155−40=9.99999…
Model
Model
The integer variance is therefore 100, and a positive affine scale multiplies standard deviation by c.
Var(R)=100
15=SD(S)=cSD(R)=10c
c=1.5
Compute
Compute
Apply the output-average requirement to determine the remaining offset.
180=E[S]=cE[R]+d
180=1.5(40)+d
d=120
Answer
Answer
The laboratory must program the offset shown in choice D.
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