This Exam P sample reference tests Central Limit Theorem. The aggregate has approximate mean 6250 and standard deviation 200. The threshold is 2.5 standard deviations below the mean, so the upper-tail probability is 0.993790 and choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is the opposite tail, P(S ≤ 5750) ≈ Φ(−2.5) = 0.0062, rather than the probability of exceeding the threshold.
BA probability below one-half contradicts the fact that 5750 is below the aggregate mean of 6250; it reflects a tail-direction or centering error.
CThis value is only slightly above one-half and corresponds to using a standardized distance near 0.10 instead of the correct distance 2.5.
DThis value corresponds to an upper tail at approximately z = −0.40, understating the concentration created by summing 625 independent payments.
Original practice · fully worked
Original variant: bike-share battery reimbursements
A city bike-share program will process 400 independent battery-replacement reimbursements this quarter. Each reimbursement has mean 15 dollars and standard deviation 6 dollars. Using a normal approximation, calculate the probability that the total reimbursement amount exceeds 6240 dollars.
A 0.0228
B 0.1587
C 0.5000
D 0.9772
E 0.9987
Variant answer in brief
The total has approximate mean 6000 and standard deviation 120. The threshold has standardized value 2, giving an upper tail of 0.022750 and choice A.
Setup
Setup
Write T for the total of the 400 independent reimbursements and aggregate the first two moments.
E[T]=400(15)=6000
Var(T)=400(62)=14400
Model
Model
The normal approximation uses standard deviation 120 for the quarterly total.
T∼˙N(6000,1202)
Compute
Compute
Standardize 6240 and evaluate the standard normal upper tail.
z=1206240−6000=2
Pr(T>6240)≈1−Φ(2)=0.0227501319…
Answer
Answer
The probability that the quarter exceeds the stated reimbursement budget is about 2.28 percent.
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