This Exam P sample reference tests Linear Combinations of Independent Random Variables. 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.
How to solve this Linear Combinations of Independent Random Variables question
Setup
Setup
Let S_n be the sum for n independent employees. Add the means and variances before taking the standard deviation.
Sn∼N(2500n,7002n)
E[Sn]=2500n,SD(Sn)=700n
Model
Model
A normal 90th percentile is the mean plus z_{0.90} standard deviations.
167168=2500n+z0.90(700n),z0.90=1.2815516…
Compute
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.
2500x2+897.0861…x−167168=0
x=7.9997875…,n=x2=63.9966007…
Using z0.90=1.28:2500(64)+1.28(700)(8)=167168
Answer
Answer
The population count must be an integer, and the matching listed value is 64.
n=64(D)
Test the method on unseen problems
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30 original multivariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.
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 × 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
Answer and calculation
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
Setup
Let T_n be the combined energy use and add the independent normal means and variances.
Tn∼N(40n,36n)
SD(Tn)=6n
Model
Model
Use the standard normal 95th percentile to connect the total to n.
4098.69=40n+1.6448536…(6n)
Compute
Compute
With x equal to the square root of n, solve the positive quadratic root and then square it.
