This Exam P sample reference tests Variance and Standard Deviation. Independence makes the variance of the annual total equal to 52 times the common weekly variance. Equating this to the annual variance 9 gives weekly variance 9/52 and weekly standard deviation 3 divided by the square root of 52, so choice D is correct.
How to solve this Variance and Standard Deviation question
Setup
Setup
Let W_i denote the amount in week i and let sigma be the common weekly standard deviation.
S=i=1∑52Wi
SD(Wi)=σ
Model
Model
Because the weekly amounts are independent, their variances add in the annual total.
Var(S)=i=1∑52Var(Wi)=52σ2
Compute
Compute
Square the stated annual standard deviation, solve for the weekly variance, and then take its positive square root.
32=52σ2
σ2=529
σ=523=0.4160251472…
Answer
Answer
Each weekly amount has standard deviation 3 divided by the square root of 52.
523(D)
Test the method on unseen problems
Probability Multivariate & Risk Models Practice Workbook
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.
AThis divides a standard deviation by 52 squared. Variances add across independent weeks, so only one factor of 52 belongs in the variance equation.
BThis divides the annual standard deviation directly by 52, as though standard deviations rather than variances added for independent amounts.
CThis treats the annual standard deviation 3 as though it were the annual variance, producing the square root of 3/52 instead of 9/52.
EThis scales the annual standard deviation upward by the square root of 52; the weekly component must be smaller than the annual total.
Original practice · fully worked
Original variant: weighted navigation errors
Three independent navigation errors A, B, and C have standard deviations 2, 3, and 6 meters, respectively. A corrected position error is defined as T = 2A − B + 0.5C. Calculate the standard deviation of T.
A 5.000
B 5.831
C 5.916
D 7.000
E 10.000
Answer and calculation
The three transformed errors have standard deviations 4, 3, and 3. Independence makes their variances add, so Var(T)=16+9+9=34 and SD(T)=√(34)=5.831, which is choice B.
Setup
Setup
Record the standard deviations and the coefficients in the corrected error.
SD(A)=2,SD(B)=3,SD(C)=6
T=2A−B+0.5C
Model
Model
For independent inputs, each variance is multiplied by the square of its coefficient before the contributions are added.
Var(T)=22Var(A)+(−1)2Var(B)+(0.5)2Var(C)
Compute
Compute
Insert the three component variances and take the square root of the resulting total variance.
Var(T)=4(22)+1(32)+0.25(62)=34
SD(T)=34=5.830951895…
Answer
Answer
The corrected position error has standard deviation approximately 5.831 meters.
