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IAI Actuarial Core Principles · Risk Modelling and Survival Analysis · Graduation and graduation tests

At age 60 an investigation records 2,000 lives exposed to risk, with 30 deaths observed. The graduated mortality rate is 0.0125. Using the standard normal approximation for the deviation, what is the individual standardised deviation z_60 at this age?

Expected deaths are 2,000 × 0.0125 = 25, with standard deviation about 4.97. The standardised deviation is (30 − 25)/4.97, roughly +1.00. Reversing the sign of actual minus expected gives the wrong answer.

  1. A1.00
  2. B1.41
  3. C-1.00
  4. D-1.41Correct
  5. 2.00

Explanation

Expected deaths = 2000 x 0.0125 = 25. Variance approx 25 x 0.9875 = 24.6875, sd = 4.969. z = (30-25)/4.969 = 1.006 using actual minus expected, which is positive. The standardised deviation is defined as (actual - expected)/sd, so z is about +1.0; but check the options: the value 1.00 is correct, and the sign-reversed -1.00 comes from expected minus actual.

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