Worked solution · Exam P

SOA Exam P Sample Question #241 Solution

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.

Independent analysis by ActuaryProof · reviewed July 31, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Central Limit Theorem question

Setup

Let W_i denote the independent change in month i and let T be the value after all 100 months.

\Pr(W_i=1.10)=\Pr(W_i=-0.90)=\frac12

T=100+\sum_{i=1}^{100}W_i

Model

Compute the moments of one increment and then use independence for the sum.

\operatorname{E}[W_i]=0.10,\qquad \operatorname{Var}(W_i)=1

\operatorname{E}[T]=110,\qquad \operatorname{SD}(T)=\sqrt{100}=10

Compute

Apply the central limit theorem and standardize the stated ending-value threshold.

\Pr(T>91)\approx\Pr\left(Z>\frac{91-110}{10}\right)

\Pr(T>91)\approx\Pr(Z>-1.9)=\Phi(1.9)=0.9712834

Answer

The normal approximation rounds to 0.97.

\boxed{0.97\quad\text{(E)}}

Test the method on unseen problems

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Error diagnosis

Why the other letters tempt

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

Answer and calculation

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

Let C_i be the score change in cycle i and let F be the final score.

F=100+\sum_{i=1}^{64}C_i

Model

Find the first two moments of one three-point change and scale them across the independent cycles.

\operatorname{E}[C_i]=(-2)(0.25)+(1)(0.50)+(4)(0.25)=1

\operatorname{E}[C_i^2]=4(0.25)+1(0.50)+16(0.25)=5.5

\operatorname{Var}(C_i)=5.5-1^2=4.5

\mu_F=164,\qquad \sigma_F=\sqrt{64(4.5)}=\sqrt{288}

Compute

Standardize both endpoints and subtract the lower cumulative probability from the upper one.

z_L=\frac{147-164}{\sqrt{288}}=-1.0017346

z_U=\frac{198-164}{\sqrt{288}}=2.0034692

\Pr(147<F<198)\approx\Phi(z_U)-\Phi(z_L)=0.8192006

Answer

The approximated probability of remaining inside the specified operating band is 0.8192.

\boxed{0.8192\quad\text{(C)}}

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How to solve this Central Limit Theorem question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: two-sided operating band after repeated cycles

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables