Question

The lifetime of a bird can be modeled as a Weibull distribution with a=0.9.

a) If the probability that a bird lives longer than 10 years is 0.4, find the value of the parameter λ?

b) What is the time to which 80% of the birds live?

Answer 1

a)

Let X be the lifetime of a bird.

Probability that a bird lives longer than 10 years is 0.4.

=> P(X > 10) = 0.4

=> exp(-(x / )^a) = 0.4

=> exp(-(10 / )^0.9) = 0.4

=> -(10 / )^0.9 = log(0.4)

=> -(10 / )^0.9 = -0.9163

=> 10 / = 0.9163^1/0.9

=> 10 / = 0.9074

=> = 10 / 0.9074 = 11.02

b)

Let t be time to which 80% of the birds live. Then,

P(X t) = 0.8

=> P(X >  t) = 1 - 0.8 = 0.2

=>  exp(-(x / )^a) = 0.2

=> exp(-(x / 11.02)^0.9) = 0.2

=> -(x / 11.02)^0.9   = log(0.2)

=> -(x / 11.02)^0.9   = -1.6094

=>  x / 11.02 = 1.6094^1/0.9

=> x / 11.02 = 1.6968

=> x = 11.02 * 1.6968 = 18.6987