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IAI Actuarial Core Principles · Actuarial Statistics · Central limit theorem

The number of claims in a year at an Indian insurer is Poisson with mean 64. Using a normal approximation with continuity correction, which expression approximates P(X ≥ 72)?

The Poisson has mean and variance 64, so sd is 8. P(X ≥ 72) becomes the area above 71.5, giving 1−Φ((71.5−64)/8). Dividing by 64 would wrongly use the variance rather than the standard deviation.

  1. A1−Φ((72−64)/8)
  2. B1−Φ((71.5−64)/8)Correct
  3. C1−Φ((72.5−64)/8)
  4. D1−Φ((71.5−64)/64)
  5. Φ((71.5−64)/8)

Explanation

Poisson mean and variance both equal 64, so sd = 8. P(X ≥ 72) corresponds to the continuous region above 71.5. Hence 1−Φ((71.5−64)/8) = 1−Φ(0.9375). Dividing by 64 uses the variance instead of the sd.

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