SOA Exam P Sample Question #724. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #724 — Verified Solution

Independent solution

How to solve this Linear Combinations of Independent Random Variables question

Answer in brief

Model the company total as normal with mean proportional to n and standard deviation proportional to the square root of n, then impose the 90th-percentile equation. Its positive root gives n approximately 64, and the integer answer is choice D.

Setup

Let S_n be the sum for n independent employees. Add the means and variances before taking the standard deviation.

S_n\sim N\!\left(2500n,\,700^2n\right)

\mathbb{E}[S_n]=2500n,\qquad \operatorname{SD}(S_n)=700\sqrt{n}

Model

A normal 90th percentile is the mean plus z_{0.90} standard deviations.

167168=2500n+z_{0.90}(700\sqrt{n}),\qquad z_{0.90}=1.2815516\ldots

Compute

Set x equal to the square root of n and solve the resulting quadratic. The tiny deviation from 64 comes from using the full normal quantile while the target value reflects table rounding.

2500x^2+897.0861\ldots x-167168=0

x=7.9997875\ldots,\qquad n=x^2=63.9966007\ldots

\text{Using }z_{0.90}=1.28:\quad 2500(64)+1.28(700)(8)=167168

Answer

The population count must be an integer, and the matching listed value is 64.

\boxed{n=64\quad\text{(D)}}

Error diagnosis

Why the other letters tempt

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

AAdding individual standard deviations linearly gives 167168 = n(2500 + 1.28155 times 700), or n=49.21, which leads to A instead of using a square-root scaling.
BFor n=54 the reconstructed 90th percentile is only about 141,592, so B fails a direct substitution check by more than 25,000.
CFor n=59 the reconstructed percentile is about 154,391, still roughly 12,777 below the supplied total; stopping at this nearby trial value is premature.
EUsing a minus sign for the upper percentile gives n=69.87, which points to E. The 90th percentile lies above, not below, the mean.

Original practice · fully worked

Original variant: incubator energy total

The daily energy use of each of n laboratory incubators is normally distributed with mean 40 kWh and standard deviation 6 kWh. Uses by different incubators are independent. The 95th percentile of their combined daily energy use is 4098.69 kWh. Calculate n.

A
82
B
99
C
100
D
102
E
105

Variant answer in brief

The combined use has mean 40n and standard deviation 6 times the square root of n. Solving its 95th-percentile equation gives n essentially equal to 100, so choice C is correct.

Setup

Let T_n be the combined energy use and add the independent normal means and variances.

T_n\sim N\!\left(40n,\,36n\right)

\operatorname{SD}(T_n)=6\sqrt{n}

Model

Use the standard normal 95th percentile to connect the total to n.

4098.69=40n+1.6448536\ldots(6\sqrt{n})

Compute

With x equal to the square root of n, solve the positive quadratic root and then square it.

40x^2+9.8691218\ldots x-4098.69=0

x=9.9999985\ldots,\qquad n=x^2=99.9999699\ldots

Answer

The integer number of incubators is 100.

\boxed{n=100\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #724 — Verified Solution.” Verified 2026-07-30.

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