Worked solution · Exam P

SOA Exam P Sample Question #315 Solution

This Exam P sample reference tests Normal Distribution. Because the exponential function is increasing, each loss exceeds the stated threshold exactly when its underlying normal variable exceeds 16. Independence then gives 0.5 times the standard-normal upper tail at 0.5, or 0.154269 and choice B.

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 Normal Distribution question

Setup

Use monotonicity of the exponential transformation to translate the two loss inequalities to the normal scale.

\min(Y_1,Y_2)>e^{16}\iff X_1>16\ \text{and}\ X_2>16

Model

Standardize the two normal variables and use their independence.

\Pr(X_1>16)=1-\Phi(0)=0.5

\Pr(X_2>16)=1-\Phi\!\left(\frac{16-15}{2}\right)=1-\Phi(0.5)

Compute

Multiply the two upper-tail probabilities.

\Pr(X_1>16,X_2>16)=0.5[1-\Phi(0.5)]

=0.1542687694\ldots

Answer

The joint exceedance probability rounds to 0.154.

\boxed{0.154\quad\text{(B)}}

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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.

AUsing 1.50 instead of 2.00 as the standard deviation of the second normal variable gives 0.5[1-Φ(2/3)]=0.126246, approximately this choice.
CThis treats the threshold as the mean of both normal variables, assigning upper-tail probability 0.5 to each and multiplying to obtain 0.25.
DThis is 1-Φ(0.5)=0.308538, the probability that the second loss alone exceeds the threshold. It omits the simultaneous first-loss requirement.
EThis multiplies 0.5 by Φ(0.5), giving 0.345731. It uses the lower tail for the second normal variable instead of the required upper tail.

Original practice · fully worked

Original variant: ratio of two exponential readings

Independent normal variables X and Y have respective means 5 and 3 and respective standard deviations 1 and 1.5. Two positive instrument readings are defined by A=exp(X) and B=exp(Y). Calculate the probability that A is greater than exp(3) times B.

A
0.159
B
0.252
C
0.290
D
0.345
E
0.710

Answer and calculation

Taking logarithms converts the ratio event to X-Y>3. The difference is normal with mean 2 and variance 3.25, producing an upper-tail probability of 0.289550 and choice C.

Setup

Take logarithms to turn the multiplicative comparison into a difference.

A>e^3B\iff e^X>e^{3+Y}\iff X-Y>3

Model

The difference of independent normal variables is normal; its variances add.

X-Y\sim N\!\left(5-3,\,1^2+1.5^2\right)=N(2,3.25)

Compute

Standardize the threshold and take the upper tail.

z=\frac{3-2}{\sqrt{3.25}}=0.5547001962\ldots

\Pr(X-Y>3)=1-\Phi(z)=0.2895498710\ldots

Answer

The probability rounds to 0.290.

\boxed{0.290\quad\text{(C)}}

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How to solve this Normal Distribution question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: ratio of two exponential readings

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables