SOA Exam P Sample Question #679. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #679 — Verified Solution

Independent solution

How to solve this Normal Distribution question

Answer in brief

This is a normal-tail inversion after translating a payment threshold back to a loss threshold. The 86th loss percentile is 633.032, so subtracting the 500 payment threshold gives a deductible of 133.032 and choice E.

Setup

Let X denote the loss and d the deductible. A positive payment on one loss is the excess of X above d.

X\sim N(525,100^2),\qquad Y=(X-d)_+

Model

A payment greater than 500 occurs exactly when the underlying loss exceeds d+500. Convert the upper-tail probability to a cumulative probability.

\Pr(Y>500)=\Pr(X>d+500)=0.14

\Pr(X\le d+500)=0.86

Compute

Standardize the 86th percentile and solve the resulting linear equation for the deductible.

\frac{d+500-525}{100}=\Phi^{-1}(0.86)=1.0803193408\ldots

d=525+100(1.0803193408\ldots)-500=133.0319341\ldots

Answer

The deductible rounds to 133 in the units of the listed answers.

\boxed{d\approx 133\quad\text{(E)}}

Error diagnosis

Why the other letters tempt

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

AA value near 36 can result from reading 0.14 as an area from the mean, obtaining z near 0.36, and reporting only the standard-deviation displacement rather than the deductible.
BThe value 39 follows from treating the probability 0.14 itself as a z-score: 525+100(0.14)-500=39.
CUsing z near 0.36 inside the otherwise correct threshold equation produces about 61; the required upper-tail quantile is z=1.0803, not 0.36.
DUsing the lower-tail percentile with z=-1.0803 produces a negative deductible near -83; taking its magnitude loses the direction of the tail.

Original practice · fully worked

Original variant: equipment-service reimbursements

A service program models each equipment repair cost as normal with mean 850 dollars and standard deviation 120 dollars. A deductible d applies to each repair. Only 8% of repairs generate a reimbursement greater than 300 dollars. Calculate d.

A
169
B
550
C
681
D
719
E
1019

Variant answer in brief

A reimbursement above 300 requires a repair cost above d+300. The 92nd percentile of the repair-cost distribution is 1018.609, leaving a deductible of 718.609 and choice D.

Setup

Represent the repair cost by X and the reimbursement by the positive part after the deductible.

X\sim N(850,120^2),\qquad R=(X-d)_+

Model

Translate the reimbursement event into a repair-cost event and express the upper tail as a percentile.

\Pr(R>300)=\Pr(X>d+300)=0.08

\Pr(X\le d+300)=0.92

Compute

Use the 92nd standard-normal percentile and isolate d.

\Phi^{-1}(0.92)=1.4050715603\ldots

d=850+120(1.4050715603\ldots)-300=718.6085872\ldots

Answer

The deductible is approximately 719 dollars.

\boxed{d\approx 719\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #679 — Verified Solution.” Verified 2026-07-30.

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