Verified solution · Exam P

SOA Exam P Sample Question #251 Solution

This Exam P sample reference tests Normal Distribution. The percentage comparison is equivalent to a positive tail for W=X-1.2Y. This normal contrast has mean -2b and variance 2.44b squared, giving upper-tail probability 0.100208 and choice B.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Normal Distribution question

Setup

Translate the percentage comparison into a linear normal contrast.

\{X\ge1.2Y\}=\{W\ge0\},\qquad W=X-1.2Y

Model

Apply linearity to the means and independence to the variance.

E[W]=10b-1.2(10b)=-2b

\operatorname{Var}(W)=b^2+1.2^2b^2=2.44b^2

Compute

Standardize the zero threshold; the positive scale parameter cancels.

z=\frac{0-(-2b)}{\sqrt{2.44}\,b}=\frac2{\sqrt{2.44}}=1.2803687993\ldots

\Pr(W\ge0)=1-\Phi(z)=0.1002077308\ldots

Answer

The probability rounds to 0.100.

\boxed{0.100\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 only the first estimate's standard deviation b gives z=2 and upper tail 0.02275, producing 0.023. It omits the second estimate's variability.
CTreating the multiplier 1.2 itself as the standardized boundary gives 1-Φ(1.2)=0.11507. The z-score must use the contrast mean and variance.
DUsing 1.2 as the numerator but retaining the correct denominator gives z=1.2/√(2.44)=0.7682 and upper tail 0.22118. The actual mean gap is 2b, not 1.2b.
ETreating the 20 percent gap as 0.2 and combining two unit standard deviations gives z=0.2/√(2)=0.1414 and upper tail 0.44377. This discards the given mean-to-standard-deviation scale.

Original practice · fully worked

Original variant: a normal comparison under two calibration modes

Independent normal readings A and B have means 12 and 10 and standard deviation 2 each. A calibration mode, selected independently of the readings, uses multiplier 1.0 with probability 0.40 and multiplier 1.2 with probability 0.60. A check passes when A exceeds the selected multiplier times B. Calculate the pass probability.

A
0.500
B
0.604
C
0.630
D
0.656
E
0.760

Variant answer in brief

The multiplier-1.0 branch has pass probability 0.760250, while symmetry makes the multiplier-1.2 branch probability 0.5. Weighting them by 0.40 and 0.60 gives 0.604099976 and choice B.

Setup

Compute the normal contrast in the multiplier-1.0 mode.

A-B\sim N(2,8)

p_1=\Pr(A-B>0)=\Phi\left(\frac2{\sqrt8}\right)=0.7602499389\ldots

Model

Compute the contrast in the multiplier-1.2 mode.

A-1.2B\sim N\!\left(0,\,2^2+1.2^2(2^2)\right)

p_{1.2}=\Pr(A-1.2B>0)=\frac12

Compute

Average the two branch probabilities using the mode probabilities.

\Pr(\text{pass})=0.40p_1+0.60p_{1.2}

=0.40(0.7602499389)+0.60(0.5)=0.6040999756\ldots

Answer

The unconditional pass probability is approximately 0.604.

\boxed{0.604\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #251 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Normal Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: a normal comparison under two calibration modes

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution