This Exam P sample reference tests Linear Combinations of Independent Random Variables. The average of the two independent normals is normal with mean 75 and variance 149. Standardizing 80 gives z=0.4096 and an upper tail of 0.341044, so choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is approximately Pr(X>80)Pr(Y>80), which incorrectly replaces the event that the average exceeds 80 by the stricter event that both values exceed 80.
BThis is approximately Pr(X>80) for the first distribution alone and ignores the second value in the average.
DThis is close to the result of averaging the component standard deviations directly; independent variances must be combined with squared coefficients.
EThis is the lower-tail complement Pr(A less than or equal to 80), rather than the requested upper tail.
Original practice · fully worked
Original variant: blended laboratory score
Two independent laboratory instruments produce normally distributed quality scores. Instrument A has mean 10 and standard deviation 6; instrument B has mean 14 and standard deviation 8. A blended score is W=0.25A+0.75B. Calculate the probability that W exceeds 15.
A 0.091
B 0.373
C 0.395
D 0.450
E 0.627
Variant answer in brief
The weighted normal score has mean 13 and variance 38.25. Its threshold z-score is 0.32338, giving upper-tail probability 0.373203 and choice B.
Setup
Setup
Treat the reported score as a linear combination of the two independent readings.
W=0.25A+0.75B
Model
Model
Combine the normal means and independent variances using the linear weights.
E[W]=0.25(10)+0.75(14)=13
Var(W)=0.252(62)+0.752(82)=38.25
Compute
Compute
Standardize 15 under the resulting normal distribution.
z=38.2515−13=0.32338083…
Pr(W>15)=1−Φ(0.32338083)=0.37320342…
Answer
Answer
The blended score exceeds 15 with probability approximately 0.373.
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