Weibull's Law of Mortality : Practice Problems and Solutions

Weibull's law of mortality corresponds to the following survival distribution.

s(x) = exp[-uxⁿ⁺¹]

The force of mortality under Weibull's law is as follows.

µ_x = kxⁿ

Under Weibull's law, k and n are given constants, and u is defined as u = k/(n + 1).

The restrictions for using Weibull's law are as follows:

k > 0, n > 0, x ≥ 0.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. p. 78.

Original Problems and Solutions from The Actuary's Free Study Guide

Problem S3L25-1. The lives of rabbit-eared basilisks follow Weibull's law with k = 0.2 and n = 0.4. Find µ₁₈ for rabbit-eared basilisks.

Solution S3L25-1. We use the formula µ_x = kxⁿ. Thus, µ₁₈ = 0.2*18^0.4 = µ₁₈ = 0.6355343046.

Problem S3L25-2. The lives of rabbit-eared basilisks follow Weibull's law with k = 0.2 and n = 0.4. Find s(22) for rabbit-eared basilisks. The answer will be very small!

Solution S3L25-2. We use the formula s(x) = exp[-uxⁿ⁺¹]. We find u = k/(n+1) = 0.2/1.4 = 1/7.

Thus, s(22) = exp[-(1/7)22^1.4] = s(22) = about 0.00001996248653.

Problem S3L25-3. The lives of wild rainbow-spotted anchovies follow Weibull's law with k = 0.11. You know that µ₃ = 0.25 for wild rainbow-spotted anchovies. Find s(10) for wild rainbow-spotted anchovies.

Solution S3L25-3. First, we need to find n by using the formula µ_x = kxⁿ with x = 3.

Hence, µ₃ = 0.11*3ⁿ. Thus, 0.25 = 0.11*3ⁿ and 3ⁿ = 25/11, so n*ln(3) = ln(25/11) and n = ln(25/11)/ln(3) = n = 0.7472887028.

Now we find u = k/(n+1) = 0.11/1.7472887028 = u = 0.0629546793.

Now we use the formula s(x) = exp[-uxⁿ⁺¹] for x = 10.

s(10) = exp[-0.0629546793*10^1.7472887028] = s(10) = about 0.03072459

Problem S3L25-4. The lives of four-legged fish follow Weibull's law with n = 1 and s(20) = 0.14. Find k for four-legged fish.

Solution S3L25-4. We use the formula s(x) = exp[-uxⁿ⁺¹] to recognize that

s(20) = 0.14 = exp[-u*20²]

Thus, ln(0.14) = -u*20²

Thus, u = ln(0.14)/-400 = 0.0049152821

But we know that u = k/(n+1), so k = u*(n+1) = 0.0049152821*2 = k = about 0.0098305643.

Problem S3L25-5. The lives of rabbit-eared basilisks follow Weibull's law with k = 0.2 and n = 0.4. Find ₃p₃₄ for rabbit-eared basilisks.

Solution S3L25-5. We know that ₃p₃₄ = s(37)/s(34) and that s(x) = exp[-uxⁿ⁺¹]. We find u = k/(n+1) = 0.2/1.4 = u = 1/7. Thus, ₃p₃₄ = exp[-(1/7)37^1.4]/exp[-(1/7)34^1.4] = exp[-(1/7)37^1.4 +(1/7)34^1.4] =

exp[-2.501563687] = ₃p₃₄ = 0.0819567436.

See other sections of The Actuary's Free Study Guide for Exam 3L.