This Exam P sample reference tests Central Limit Theorem. Each monthly change has mean 0.10 and variance 1, so the ending value is approximately normal with mean 110 and standard deviation 10. Standardizing the threshold 91 gives -1.9, hence the required probability is Φ(1.9)=0.9713, choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AAn incorrect conversion to the underlying up-month count can impose K at least 49. With a continuity correction, that gives approximately P(Z>-0.30)=0.618, nearest 0.63; the correct cutoff is K at least 41.
BImposing the erroneous cutoff K at least 47 yields approximately P(Z>-0.70)=0.758, nearest 0.75. This loses eight favorable count outcomes through an algebra error in the ending value.
CIgnoring the positive mean change of 0.10 per month centers the ending value at 100. That makes the standardized gap 0.9 and gives Φ(0.9)=0.816, nearest 0.82.
DThe mistaken cutoff K at least 43 gives approximately P(Z>-1.50)=0.933, nearest 0.94. Solving the exact linear relation between K and the ending value gives K at least 41 instead.
Original practice · fully worked
Original variant: two-sided operating band after repeated cycles
A calibration score starts at 100. During each of 64 independent cycles it changes by -2, 1, or 4 points. The middle change occurs half the time, and each outer change occurs one quarter of the time. Using a normal approximation without a continuity correction, calculate the probability that the final score is greater than 147 but less than 198.
A 0.0226
B 0.1582
C 0.8192
D 0.8418
E 0.9774
Variant answer in brief
One cycle has mean 1 and variance 4.5, so the final score is approximately normal with mean 164 and variance 288. The two standardized bounds are -1.0017 and 2.0035, giving probability 0.8192, choice C.
Setup
Setup
Let C_i be the score change in cycle i and let F be the final score.
F=100+i=1∑64Ci
Model
Model
Find the first two moments of one three-point change and scale them across the independent cycles.
E[Ci]=(−2)(0.25)+(1)(0.50)+(4)(0.25)=1
E[Ci2]=4(0.25)+1(0.50)+16(0.25)=5.5
Var(Ci)=5.5−12=4.5
μF=164,σF=64(4.5)=288
Compute
Compute
Standardize both endpoints and subtract the lower cumulative probability from the upper one.
zL=288147−164=−1.0017346
zU=288198−164=2.0034692
Pr(147<F<198)≈Φ(zU)−Φ(zL)=0.8192006
Answer
Answer
The approximated probability of remaining inside the specified operating band is 0.8192.
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