SOA Exam P Sample Question #715. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #715

Weight each office's relative frequencies by its share of combined sales. The combined cumulative probability is 0.41 through 160 and 0.51 through 165, so the median is 165 and choice B is correct.

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 Discrete Random Variables question

Answer in brief

Weight each office's relative frequencies by its share of combined sales. The combined cumulative probability is 0.41 through 160 and 0.51 through 165, so the median is 165 and choice B is correct.

Setup

Treat the combined sales distribution as a mixture of the two office distributions.

p(x)=0.70p_1(x)+0.30p_2(x)

Model

Compute the combined masses at prices no greater than 165.

p(150)=0.70(0.20)+0.30(0.25)=0.215

p(155)=0.70(0.10)+0.30(0.15)=0.115

p(160)=0.70(0.05)+0.30(0.15)=0.080

p(165)=0.70(0.10)+0.30(0.10)=0.100

Compute

Accumulate the mixture masses and locate the first support value whose CDF reaches one-half.

F(160)=0.215+0.115+0.080=0.410

F(165)=0.410+0.100=0.510

F(160)<0.5\le F(165)

Answer

The smallest combined selling price with cumulative probability at least one-half is 165 thousand.

\boxed{165.0\quad\text{(B)}}

Error diagnosis

Why the other letters tempt

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

AThe combined CDF is only 0.41 at 160, so this stops before half of the sales have accumulated.
CThis interpolates between 165 and 170. A discrete distribution has a probability jump at 165, so interpolation is unnecessary.
DThis skips the mass at 165 even though that mass is exactly what moves the cumulative probability above one-half.
EThis is the median of Office 1 considered alone; it ignores the 30% contribution from Office 2.

Original practice · fully worked

Original variant: repair-desk wait time

A repair center combines tickets from two service desks. Desk A handles 60% of all tickets and has wait-time probabilities 0.10, 0.20, 0.30, 0.25, and 0.15 at 4, 8, 12, 16, and 20 minutes, respectively. Desk B handles the remaining 40% and has corresponding probabilities 0.20, 0.30, 0.25, 0.15, and 0.10. Calculate the median wait time for a randomly selected ticket.

A
8 minutes
B
10 minutes
C
12 minutes
D
14 minutes
E
16 minutes

Variant answer in brief

The mixed CDF is 0.38 through 8 minutes and 0.66 through 12 minutes. Its first crossing of one-half is therefore 12 minutes, choice C.

Setup

Combine the desk distributions using their ticket shares.

p(t)=0.60p_A(t)+0.40p_B(t)

Model

Calculate the first three mixture masses.

p(4)=0.60(0.10)+0.40(0.20)=0.14

p(8)=0.60(0.20)+0.40(0.30)=0.24

p(12)=0.60(0.30)+0.40(0.25)=0.28

Compute

Compare the cumulative probabilities immediately below and at 12 minutes.

F(8)=0.14+0.24=0.38

F(12)=0.38+0.28=0.66

F(8)<0.5\le F(12)

Answer

The median wait time is the first support value at which the CDF reaches one-half.

\boxed{12\text{ minutes}\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #715 — Verified Solution.” Verified 2026-07-30.

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