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.
Each independent rounding error is uniform from -2.5 to 2.5, with mean zero and variance 25/12.
Ei∼Unif(−2.5,2.5)
E[Ei]=0,Var(Ei)=1225
Model
Model
For 48 errors, divide the individual variance by 48 to obtain a sample-mean standard deviation of 0.208333.
SD(Eˉ)=4825/12=0.208333
Compute
Compute
The symmetric bounds ±0.25 correspond to ±1.2 standard deviations. The central normal area is 0.769861.
P(∣Eˉ∣<0.25)≈P(∣Z∣<1.2)=2Φ(1.2)−1=0.769861
Answer
Answer
The approximate probability rounds to 0.77, corresponding to choice D.
0.77(D)
Test the method on unseen problems
Probability Univariate Distributions Practice Workbook
30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.
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
Answer and calculation
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.
