This Exam P sample reference tests Normal Distribution. The averaged error is normal with standard deviation 0.0035609h. The central probability inside ±0.005h is 0.83972, choice D.
Let Y be the average of the two independent normal measurement errors.
Y=2X1+X2
Model
Model
A linear combination of independent normal variables is normal. Divide the sum of their variances by four to obtain the variance of the average.
Var(Y)=4(0.0056h)2+(0.0044h)2
σY=0.0035609h
Compute
Compute
The standard deviation of Y is 0.0035609h. Standardizing the symmetric limits ±0.005h and subtracting the two tails gives a central probability of 0.839723.
Pr(∣Y∣≤0.005h)=2Φ(0.005/0.0035609)−1=0.839723
Answer
Answer
The probability that the averaged error lies within the stated tolerance rounds to 0.84, which is choice D.
0.84(D)
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CUsing 0.005h as the standard deviation gives the central standard-normal probability Pr(|Z|≤1)=0.6827, approximately 0.68. This ignores the variance calculation for the average.
Original practice · fully worked
Original variant: upper percentile of a Gaussian composite score
Independent measurements U and V are normal with means 2 and 5 and variances 1 and 4. A composite score is T=2U+V. Determine the 90th percentile of T.
A 9.000
B 10.812
C 11.326
D 12.625
E 14.544
Variant answer in brief
The composite is normal with mean 9 and variance 8. Adding 1.28155 standard deviations gives the 90th percentile 12.625, choice D.
Setup
Setup
Compute the mean of T=2U+V by linearity of expectation.
E[T]=2(2)+5=9
Model
Model
Independence makes the variance of T the sum of the scaled component variances, so T is normal with mean 9 and variance 8.
Var(T)=22(1)+4=8
Compute
Compute
Add the 90th-percentile standard-normal quantile 1.28155 times the standard deviation √8 to the mean, obtaining 12.624775.
t0.90=9+Φ−1(0.90)8=12.624775
Answer
Answer
The 90th percentile of the composite score rounds to 12.625, corresponding to choice D.
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