Sums, the CLT and normal approximation

2 free worked examples

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1. Exam PSums, the CLT and normal approximationExam level

   A portfolio has 100 independent claims, each with mean 50 and standard deviation 12. Approximate the probability that total claims exceed 5,200.

   1. A$0.0478$
   2. B$0.0668$
   3. C$0.1587$
   4. D$0.4325$
   5. E$0.9522$

   Solution

   1. $E[S]=100(50)=5,000$ and $\mathrm{Var}(S)=100(144)=14,400$, so ${\sigma }_S=120$.
   2. Variances add; standard deviations do not - that is the whole content of the $\sqrt{n}$ scaling.
   3. $z=\frac{5200-5000}{120}=1.6667$.
   4. $P(S>5200)\approx 1-\Phi (1.6667)=1-0.952210=0.047790$.

   Trap. Using n·σ = 1,200 for the standard deviation of the sum instead of √n·σ = 120.

2. Exam PSums, the CLT and normal approximationExam level

   Claim counts across 50 independent policies are each Poisson with mean 2. Approximate the probability the total exceeds 115.

   1. A$0.0606$
   2. B$0.0668$
   3. C$0.1587$
   4. D$0.2266$
   5. E$0.9429$

   Solution

   1. The sum of independent Poissons is Poisson with mean $50\times 2=100$, and its variance is also 100.
   2. $\sigma =10$.
   3. $z=\frac{115-100}{10}=1.5$ (ignoring the continuity correction for a first pass).
   4. Applying the half-unit continuity correction at 115.5 gives $z=1.55$ and $P\approx 1-\Phi (1.55)=0.0606$.

   Trap. Using √(50) × 2 for the standard deviation rather than √(50 × 2).