SOA Exam P Sample Question #582. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #582 — Verified Solution

Independent solution

How to solve this Variance and Standard Deviation question

Answer in brief

Independence makes the variance of the annual total equal to 52 times the common weekly variance. Equating this to the annual variance 9 gives weekly variance 9/52 and weekly standard deviation 3 divided by the square root of 52, so choice D is correct.

Setup

Let W_i denote the amount in week i and let sigma be the common weekly standard deviation.

S=\sum_{i=1}^{52}W_i

\operatorname{SD}(W_i)=\sigma

Model

Because the weekly amounts are independent, their variances add in the annual total.

\operatorname{Var}(S)=\sum_{i=1}^{52}\operatorname{Var}(W_i)=52\sigma^2

Compute

Square the stated annual standard deviation, solve for the weekly variance, and then take its positive square root.

3^2=52\sigma^2

\sigma^2=\frac{9}{52}

\sigma=\frac{3}{\sqrt{52}}=0.4160251472\ldots

Answer

Each weekly amount has standard deviation 3 divided by the square root of 52.

\boxed{\frac{3}{\sqrt{52}}\quad\text{(D)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis divides a standard deviation by 52 squared. Variances add across independent weeks, so only one factor of 52 belongs in the variance equation.
BThis divides the annual standard deviation directly by 52, as though standard deviations rather than variances added for independent amounts.
CThis treats the annual standard deviation 3 as though it were the annual variance, producing the square root of 3/52 instead of 9/52.
EThis scales the annual standard deviation upward by the square root of 52; the weekly component must be smaller than the annual total.

Original practice · fully worked

Original variant: weighted navigation errors

Three independent navigation errors A, B, and C have standard deviations 2, 3, and 6 meters, respectively. A corrected position error is defined as T = 2A - B + 0.5C. Calculate the standard deviation of T.

A
5.000
B
5.831
C
5.916
D
7.000
E
10.000

Variant answer in brief

The three transformed errors have standard deviations 4, 3, and 3. Independence makes their variances add, so Var(T)=16+9+9=34 and SD(T)=sqrt(34)=5.831, which is choice B.

Setup

Record the standard deviations and the coefficients in the corrected error.

\operatorname{SD}(A)=2,\quad \operatorname{SD}(B)=3,\quad \operatorname{SD}(C)=6

T=2A-B+0.5C

Model

For independent inputs, each variance is multiplied by the square of its coefficient before the contributions are added.

\operatorname{Var}(T)=2^2\operatorname{Var}(A)+(-1)^2\operatorname{Var}(B)+(0.5)^2\operatorname{Var}(C)

Compute

Insert the three component variances and take the square root of the resulting total variance.

\operatorname{Var}(T)=4(2^2)+1(3^2)+0.25(6^2)=34

\operatorname{SD}(T)=\sqrt{34}=5.830951895\ldots

Answer

The corrected position error has standard deviation approximately 5.831 meters.

\boxed{5.831\text{ meters}\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #582 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.