This Exam P sample reference tests Normal Distribution. The sum of the independent normal totals is normal with mean 22 and variance 25. Standardizing 29 gives z=1.4, so the requested lower-tail probability is 0.919243 and choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.61 is approximately Φ(7/25)=Φ(0.28)=0.6103. It uses the aggregate variance 25 as though it were the standard deviation.
BThe value 0.69 is Φ(0.5)=0.6915. This results from adding the two standard deviations and then applying an extra factor of two, using 2(3+4)=14 as the scale.
CThe value 0.78 is approximately Φ(7/9)=0.7812. It uses the auto variance 3 squared as the aggregate standard deviation and omits the property uncertainty.
DThe value 0.84 is Φ(7/(3+4))=Φ(1)=0.8413. It adds standard deviations instead of adding independent variances.
Original practice · fully worked
Original variant: calibrating a gain from a normal tail target
Two independent calibration signals X and Y are normally distributed. Signal X has mean 8 and standard deviation 2, while signal Y has mean 3 and standard deviation 1. A controller records S=X+aY for a positive gain a. Determine a if the probability that S is at most 15 equals 0.841344746.
A 1.250
B 1.500
C 1.548
D 1.667
E 3.750
Variant answer in brief
The target probability is Φ(1). Standardizing S gives (7-3a)/√(4+a squared)=1. The admissible solution is a=1.5; the other squared-equation root has the wrong sign, so choice B.
Setup
Setup
Form the mean and variance of the gain-adjusted signal.
S=X+aY∼N(8+3a,22+a2(12))
Model
Model
Recognize the supplied probability as the standard-normal CDF at one and standardize the threshold.
0.841344746=Φ(1)
4+a215−(8+3a)=1
Compute
Compute
Square the equation, solve the resulting quadratic, and retain the root satisfying the original positive-sign equation.
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