Worked solution · Exam P

SOA Exam P Sample Question #293 Solution

This Exam P sample reference tests Joint Distributions. Collapse the joint table by the possible values of the total cost. The resulting total has mean 2.04 and second moment 278.4, so its variance is 274.2384 and its standard deviation is 16.5601, selecting choice D.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Joint Distributions question

Setup

Let W be the sum of the two costs. Combine joint cells that produce the same total.

\Pr(W=0,40,80,200,240,400)=(0.9729,0.0200,0.0020,0.0040,0.0010,0.0001)

Model

Use the collapsed distribution to form the first two moments of the total.

E[W]=\sum_w w\Pr(W=w)=2.04

E[W^2]=\sum_w w^2\Pr(W=w)=278.4

Compute

Subtract the square of the mean from the second moment, then take a square root.

\operatorname{Var}(W)=278.4-(2.04)^2=274.2384

\operatorname{SD}(W)=\sqrt{274.2384}=16.56014493\ldots

Answer

The standard deviation rounds to 16.6.

\boxed{16.6\quad\text{(D)}}

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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.

ACounting only one cell from each symmetric off-diagonal pair gives the incomplete raw sums E[W]=1.12 and E[W squared]=153.6. Applying the variance formula anyway gives √(153.6-1.12 squared)=12.3428, which rounds to 12.3.
BThis retains only the one-disabled-person totals 40 and 200 and moves every both-disabled outcome to zero. The resulting standard deviation is 13.7637.
CThis treats the two costs as independent. Their marginal variances sum to 245.9192, whose square root is 15.6818, but the joint table shows positive covariance.
EThis value is impossible under the supplied table: 19.8 squared exceeds E[W squared]=278.4, whereas Var(W) cannot exceed its second moment.

Original practice · fully worked

Original variant: correlation from a shared demand component

Independent centered random variables A, B, and C have variances 4, 5, and 7. Two regional demand changes are modeled by X=A+B and Y=A+C, so the regions share only component A. Calculate the correlation between X and Y.

A
0.040
B
0.364
C
0.402
D
0.444
E
0.676

Answer and calculation

The shared component contributes covariance 4, while the two marginal variances are 9 and 11. Therefore the correlation is 4 divided by the square root of 99, or 0.402015, so choice C.

Setup

Add the independent variance components within each regional change.

\operatorname{Var}(X)=4+5=9

\operatorname{Var}(Y)=4+7=11

Model

Only the shared centered component contributes to the covariance; all cross-covariances between distinct components vanish.

\operatorname{Cov}(X,Y)=\operatorname{Cov}(A+B,A+C)=\operatorname{Var}(A)=4

Compute

Normalize the covariance by the product of the marginal standard deviations.

\rho_{X,Y}=\frac{4}{\sqrt{9}\sqrt{11}}=\frac4{\sqrt{99}}

=0.4020151261\ldots

Answer

The correlation rounds to 0.402.

\boxed{0.402\quad\text{(C)}}

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How to solve this Joint Distributions question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: correlation from a shared demand component

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables