Worked 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 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

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)}}

Test the method on unseen problems

Probability Multivariate & Risk Models Practice Workbook

30 original multivariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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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

Answer and calculation

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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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: lower-tail reserve score across service visits

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables