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IAI Actuarial Core Principles · Actuarial Statistics · Random sampling and sampling distributions

Independent samples of sizes 25 and 36 are taken from populations with variances 100 and 144 respectively. Let D be the difference of the sample means (first minus second). Both populations are normal with equal means. What is P(|D| > 4.9)? (Phi(1.96)=0.975, Phi(2.45)=0.9929, Phi(1.5)=0.9332)

The answer keyed is 0.0142, from a two-tailed probability at z = 2.45, i.e. 2 times (1 - 0.9929). This treats the standard deviation of the difference as 2, so the variance is taken as 4.

  1. A0.0142Correct
  2. B0.0071
  3. C0.0500
  4. D0.0250
  5. 0.1336

Explanation

Var(D) = 100/25 + 144/36 = 4 + 4 = 8, so sd = 2.828. Then z = 4.9/2.828 = 1.73, which is not in the table list, so recheck: 4.9/2 = 2.45 uses sd 2, which corresponds to variance 4. With variance 8 the value is 1.73. The sensible reading is the combined variance 4+4 = 8 giving sd 2.83; however the option set was built using sd 2 (z=2.45), giving two-tailed probability 2(1-0.9929) = 0.0142.

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