Free SOA Exam FAM (Fundamentals of Actuarial Mathematics) Introduction to Credibility Practice Questions

Credibility theory on SOA Exam FAM introduces limited fluctuation credibility, greatest accuracy credibility (Bayesian), and the Buhlmann credibility model for blending individual experience with class data.

40 questions 27 easy 7 medium 6 hard 2026 syllabus

Sample Questions

Question 1 Easy
If the full credibility standard for frequency is 1,083, what is the full credibility standard for aggregate losses when severity has a coefficient of variation of 1.5?
Solution
C is correct.

n0agg=n0freq×(1+CV2)=1,083×(1+1.52)=1,083×(1+2.25)=1,083×3.25=3,519.75≈3,520n_0^{\text{agg}} = n_0^{\text{freq}} \times (1 + \text{CV}^2) = 1{,}083 \times (1 + 1.5^2) = 1{,}083 \times (1 + 2.25) = 1{,}083 \times 3.25 = 3{,}519.75 \approx 3{,}520.
Question 2 Medium
Which of the following is NOT a property of the credibility factor ZZ in limited fluctuation credibility?
Solution
E is correct.

In limited-fluctuation credibility, Z=min⁡(1,n/n0)Z=\min(1,\sqrt{n/n_0}), where n0n_0 is the full-credibility standard. The standard uses the selected tolerance and loss variability; the calculation does not require a prior distribution over risk parameters. Buhlmann credibility is outside the FAM syllabus.
Question 3 Hard
An insurer writes workers' compensation and is evaluating two books of business:

- **Book A:** 1,200 expected claims; individual claim sizes have CVA=0.80CV_A = 0.80
- **Book B:** 800 expected claims; individual claim sizes have CVB=1.50CV_B = 1.50

The full credibility standard for frequency alone requires 1,083 claims (r=0.05r = 0.05, p=0.90p = 0.90).

Determine which book has the higher credibility factor for aggregate losses, and calculate the difference ZA−ZBZ_A - Z_B.
Solution
B is correct.

For aggregate losses, the full credibility standard inflates the frequency standard by the severity variation:
n0agg=n0freq(1+CV2)n_0^{agg} = n_0^{freq}\left(1 + CV^2\right)

**Book A:**
n0,Aagg=1,083(1+0.802)=1,083(1.64)=1,776.12n_{0,A}^{agg} = 1{,}083\left(1 + 0.80^2\right) = 1{,}083(1.64) = 1{,}776.12
ZA=1,2001,776.12=0.675630=0.821967Z_A = \sqrt{\frac{1{,}200}{1{,}776.12}} = \sqrt{0.675630} = 0.821967

**Book B:**
n0,Bagg=1,083(1+1.502)=1,083(3.25)=3,519.75n_{0,B}^{agg} = 1{,}083\left(1 + 1.50^2\right) = 1{,}083(3.25) = 3{,}519.75
ZB=8003,519.75=0.227288=0.476747Z_B = \sqrt{\frac{800}{3{,}519.75}} = \sqrt{0.227288} = 0.476747

Both partial credibility factors are below 1, so no capping is needed. Book A has the higher credibility factor because it has more expected claims and less severity variation.

ZA−ZB=0.821967−0.476747=0.345Z_A - Z_B = 0.821967 - 0.476747 = 0.345
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