This Exam P sample reference tests Normal Distribution. The 24 independent fills have total mean 288 and total standard deviation √(24) times the individual standard deviation. Matching the stated upper tail to z_0.80 gives sigma=0.4851, so choice C is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.01 results by first using 24sigma as the total standard deviation, obtaining about 0.10, and then unnecessarily squaring that value.
BThe value 0.10 comes from solving 2/[24(0.8416)], which makes standard deviations add directly across 24 independent fills instead of adding variances.
DThe value 2.38 is the standard deviation required for the entire 24-fill total, 2/0.8416, and omits the division by √(24) needed to recover one fill's standard deviation.
EA value near 10 comes from treating the tail probability 0.20 itself as a z-score and reporting the resulting case-level scale. A probability is not a standard-normal quantile.
Original practice · fully worked
Original variant: comparing two calibration teams
Two calibration teams each collect eight independent readings. Every reading is normally distributed with mean 50 and standard deviation 6, and the teams' readings are mutually independent. Calculate the probability that Team A's sample mean exceeds Team B's sample mean by more than 3.
A 0.0786
B 0.1587
C 0.2398
D 0.5000
E 0.8413
Variant answer in brief
The difference of the independent sample means is normal with mean zero and variance 36/8+36/8=9. The threshold is one standard deviation above zero, so the upper tail is 0.1587 and choice B is correct.
Setup
Setup
Let D be the difference between the two independent sample means.
D=XA−XB
E[D]=50−50=0
Model
Model
Independence makes the two sample-mean variances additive in the difference.
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