Worked solution · Exam P

SOA Exam P Sample Question #289 Solution

This Exam P sample reference tests Discrete Distributions. This is the standard deviation of a capped step benefit derived from a discrete snowfall model. The benefit has mean 92 and second moment 35,200, giving variance 26,736 and standard deviation 163.5, so choice A.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Discrete Distributions question

Setup

Collapse snowfall categories that lead to the same policy benefit. Let Y denote the payment amount.

\Pr(Y=0)=0.72,\quad \Pr(Y=200)=0.14

\Pr(Y=400)=0.06,\quad \Pr(Y=500)=0.08

Model

Compute the first two raw moments of the discrete payment.

E[Y]=\sum_y y\Pr(Y=y)

E[Y^2]=\sum_y y^2\Pr(Y=y)

Compute

Evaluate the moments and convert the variance to a standard deviation.

E[Y]=200(0.14)+400(0.06)+500(0.08)=92

E[Y^2]=200^2(0.14)+400^2(0.06)+500^2(0.08)=35200

\operatorname{Var}(Y)=35200-92^2=26736

\operatorname{SD}(Y)=\sqrt{26736}=163.5114675

Answer

The policy payment has standard deviation about 163.5.

\boxed{163.5\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.

BThe value 187.6 is √(35,200), the root mean square of the payment. It omits subtraction of the squared mean before taking the square root.
CThe value 208.7 results from ignoring the policy cap, using payments 0, 200, 400, 600, and 800. That model gives mean 108, second moment 55,200, and standard deviation 208.65.
DThe value 234.9 is √(55,200), combining the uncapped-payment error with the further error of reporting a raw root second moment instead of a standard deviation.
EThe value 336.6 results from assigning uncapped 200-unit steps across all eight snowfall bands, with payments from 0 through 1400. That removes both the threshold and cap, producing variance 113,296 and standard deviation 336.59.

Original practice · fully worked

Original variant: variability of a cumulative lab award

A laboratory can finish at most three calibrations in one shift. The tail probabilities for the completed count K are P(K at least 1)=0.60, P(K at least 2)=0.30, and P(K at least 3)=0.10. A team earns 80 credits for the first calibration, 120 additional credits for the second, and 200 additional credits for the third. Calculate the standard deviation of its total credits.

A
64.62 credits
B
99.68 credits
C
104.00 credits
D
122.90 credits
E
161.00 credits

Answer and calculation

Differencing the three tail probabilities gives count masses 0.40, 0.30, 0.20, and 0.10. The cumulative awards 0, 80, 200, and 400 then have variance 15,104 and standard deviation 122.90 credits, so choice D.

Setup

Convert the nested tail probabilities into disjoint count probabilities.

\Pr(K=0)=0.40,\quad \Pr(K=1)=0.60-0.30=0.30

\Pr(K=2)=0.30-0.10=0.20,\quad \Pr(K=3)=0.10

Model

Accumulate the incremental credits. The total C takes values 0, 80, 200, and 400 for counts zero through three.

C(0)=0,\quad C(1)=80,\quad C(2)=200,\quad C(3)=400

Compute

Calculate the first two moments and center the second moment.

E[C]=0.30(80)+0.20(200)+0.10(400)=104

E[C^2]=0.30(80^2)+0.20(200^2)+0.10(400^2)=25920

\operatorname{Var}(C)=25920-104^2=15104

\operatorname{SD}(C)=\sqrt{15104}=122.898332

Answer

The total-credit standard deviation rounds to 122.90 credits.

\boxed{122.90\text{ credits}\quad\text{(D)}}

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How to solve this Discrete Distributions question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: variability of a cumulative lab award

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables