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IAI Actuarial Core Principles · Economic Modelling · Role of insurance in reducing or removing risk

Policies in a pool each have a loss of ₹100,000 with probability 0.02 and zero otherwise, independently. The insurer charges a premium of ₹2,000 plus a loading, so that the standard deviation of the average loss per policy falls as the pool grows. What is the smallest number of policies n such that the standard deviation of the average loss per policy is at most ₹1,400 or less? (Individual standard deviation = 100,000 × √(0.02 × 0.98) = ₹14,000.)

The smallest pool is 100 policies. The individual standard deviation is ₹14,000, and the average's standard deviation is 14,000 divided by the square root of n. Setting this at most ₹1,400 requires the square root of n to be at least 10, so n is at least 100.

  1. A100Correct
  2. B10
  3. C1,000
  4. D14
  5. 10,000

Explanation

Need 14,000/√n ≤ 1,400, so √n ≥ 10 and n ≥ 100. With n = 10, the standard deviation is about 4,427, which is above 1,400. n=1,000 satisfies the bound but is not the smallest. n=14 gives about 3,742, still too high.

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