Verified solution · Exam P

SOA Exam P Sample Question #298 Solution

This Exam P sample reference tests Central Limit Theorem. The claim has mean 0.8 and variance 0.76, making the per-policy premium 1. For 76 independent policies, total claims have approximate mean 60.8 and standard deviation 7.6; the premium boundary is z=2, so the upper tail is 0.02275 and choice A.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Central Limit Theorem question

Setup

Compute the first two moments of one policy's claim by collecting the positive-value contributions.

E[X]=0.2+0.6=0.8

E[X^2]=0.2+1.2=1.4

\operatorname{Var}(X)=1.4-(0.8)^2=0.76

Model

The loaded premium is one per policy. Aggregate the independent claim moments over 76 policies.

1.25E[X]=1,\qquad P_{\mathrm{total}}=76

E[S]=76(0.8)=60.8,\qquad \operatorname{Var}(S)=76(0.76)=57.76

Compute

Apply the central limit approximation and standardize the premium boundary.

\operatorname{SD}(S)=\sqrt{57.76}=7.6

\Pr(S>76)\approx1-\Phi\!\left(\frac{76-60.8}{7.6}\right)=1-\Phi(2)

=0.02275013195\ldots

Answer

The approximate probability rounds to 0.02.

\boxed{0.02\quad\text{(A)}}

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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.

BThis uses the raw second moment 1.4 as if it were the individual variance. Then z=15.2/√(76·1.4)=1.474 and the upper tail is about 0.070.
CUsing a 12.5% loading instead of the stated 25% puts the boundary one aggregate standard deviation above the mean, giving an upper tail near 0.16, which rounds toward this choice.
DThis corresponds to a standard-normal boundary near z=0.67, requiring an aggregate standard deviation about three times the verified 7.6. The independent-variance calculation supplies no such inflation.
EThis multiplies the one-policy standard deviation by 76 instead of √(76). The resulting z is about 0.229 and the upper tail about 0.409.

Original practice · fully worked

Original variant: lower-tail reserve score across service visits

A maintenance reserve score changes independently on each of 81 service visits. The change is -2 points with probability 4/21, 1 point with probability 2/3, and 5 points with probability 1/7. Use the central limit theorem to approximate the probability that the total change over all 81 visits is below 54 points.

A
0.000
B
0.067
C
0.090
D
0.434
E
0.933

Variant answer in brief

One visit has mean 1 and variance 4. The 81-visit total therefore has approximate mean 81 and standard deviation 18. The lower boundary standardizes to -1.5, giving probability 0.066807 and choice B.

Setup

Compute the first two moments of the score change Y from one visit.

E[Y]=-2\left(\frac4{21}\right)+1\left(\frac23\right)+5\left(\frac17\right)=1

E[Y^2]=4\left(\frac4{21}\right)+1\left(\frac23\right)+25\left(\frac17\right)=5

\operatorname{Var}(Y)=5-1^2=4

Model

Aggregate the moments over 81 independent visits and apply the central limit theorem.

E[T]=81(1)=81

\operatorname{SD}(T)=\sqrt{81(4)}=18

T\ \dot\sim\ N(81,18^2)

Compute

Standardize the lower-tail boundary and evaluate the normal CDF.

\Pr(T<54)\approx\Phi\!\left(\frac{54-81}{18}\right)

=\Phi(-1.5)=0.06680720127\ldots

Answer

The approximate lower-tail probability rounds to 0.067.

\boxed{0.067\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #298 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Central Limit Theorem question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: lower-tail reserve score across service visits

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution