Verified solution · Exam P

SOA Exam P Sample Question #307 Solution

This Exam P sample reference tests Sampling Distributions. The sample mean has standard deviation 250 divided by the square root of n. Requiring its standardized upper boundary to reach the 0.99 normal quantile gives n at least 541.189, so the minimum integer is 542 and choice B.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Sampling Distributions question

Setup

Write the exact sampling distribution of the average of n independent benefit amounts.

\bar X\sim N\!\left(2475,\frac{250^2}{n}\right)

Model

Standardize the stated upper boundary and impose the required one-sided probability.

\Pr(\bar X\le2500)=\Phi\!\left(\frac{2500-2475}{250/\sqrt n}\right)

=\Phi(0.1\sqrt n)\ge0.99

Compute

Invert the standard-normal CDF and solve the resulting inequality.

0.1\sqrt n\ge \Phi^{-1}(0.99)=2.326347874\ldots

n\ge\left(\frac{2.326347874\ldots}{0.1}\right)^2=541.1894431\ldots

Answer

Round upward to preserve the probability requirement.

\boxed{542\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.

AThe value 24 rounds up 2.32635/0.1 without squaring it. The normal boundary contains the square root of n, so the isolated sample size is the square of that ratio.
CThe value 664 is the ceiling of (2.57583/0.1)². It uses the 0.995 quantile for a two-sided 99% interval even though the requirement is one-sided.
DThe value 5815 comes from the table-rounded product 2500(2.326). It treats the benefit threshold itself as a scale for sample size and omits both the mean difference and standard error.
EThe value 6440 is 2500(2.576). It combines the same incorrect threshold scaling with the unnecessary two-sided 0.995 quantile.

Original practice · fully worked

Original variant: samples needed to rank two routes

Travel times on route X are normal with mean 52 minutes and standard deviation 9 minutes. Route Y times are independently normal with mean 47 minutes and standard deviation 6 minutes. A study takes n route-X trips and 2n route-Y trips. Calculate the minimum integer n for which the probability that the route-X sample average exceeds the route-Y sample average is at least 0.95.

A
4
B
10
C
11
D
13
E
25

Variant answer in brief

The difference of sample averages has mean 5 and variance 9 squared over n plus 6 squared over 2n, or 99/n. The 0.95 quantile requires n at least 10.714, so the minimum integer is 11 and choice C.

Setup

Form the difference between the two sample averages.

D=\bar X_n-\bar Y_{2n}

D\sim N\!\left(5,\frac{9^2}{n}+\frac{6^2}{2n}\right)=N\!\left(5,\frac{99}{n}\right)

Model

Standardize the event that the route-X sample average is larger.

\Pr(D>0)=\Phi\!\left(\frac{5\sqrt n}{\sqrt{99}}\right)\ge0.95

Compute

Invert the normal CDF and solve for the base route-X sample size.

\frac{5\sqrt n}{\sqrt{99}}\ge\Phi^{-1}(0.95)=1.644853627\ldots

n\ge\frac{99(1.644853627\ldots)^2}{25}=10.71395208\ldots

Answer

Round upward so the probability requirement is met.

\boxed{11\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #307 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Sampling Distributions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: samples needed to rank two routes

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution