This Exam P sample reference tests Central Limit Theorem. The rounding error has variance 25/12. The average of 48 errors has standard deviation 0.20833, so the two-sided probability within 0.25 is 0.7699, choice D.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
CNo standard one-sided portion of the central interval was identified that produces 0.57.
EThe value 0.88 is the rounded one-sided cumulative probability at z=1.2. The requested interval needs the area between -1.2 and 1.2.
Original practice · fully worked
Original variant: average sensor rounding error
Independent sensor readings are rounded to the nearest 2 units, making each rounding error uniform from -1 to 1. For 75 readings, approximate the probability that the mean rounding error is between -0.10 and 0.10.
A 0.6827
B 0.7500
D 0.8203
C 0.8664
E 0.9332
Variant answer in brief
The mean error has standard deviation √((1/3)/75)=1/15. The bounds correspond to plus or minus 1.5 standard deviations, giving 0.8664.
Setup
Setup
Each rounding error is uniform from -1 to 1, giving mean zero and variance 1/3.
Ei∼Unif(−1,1)
Var(Ei)=31
Model
Model
For 75 readings, the sample-mean standard deviation is the square root of (1/3)/75, which equals 1/15.
SD(Eˉ)=751/3=151
Compute
Compute
The bounds ±0.10 correspond to ±1.5 standard deviations, whose central normal area is 0.8663856.
1/150.10=1.5
P(∣Eˉ∣<0.10)≈2Φ(1.5)−1=0.8663856
Answer
Answer
The approximate probability is 0.8664, corresponding to choice C.
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