Answer to Question #207028 in Statistics and Probability for Sajid

Question

A component has a Weibull failure time distribution with parameters α = 0.25 and β = 3.0 (a) Find the probability that the component is still operating at time 5. (b) Find the probability that the component fails before time 3. (c) find the failure rate function. (d) How much more likely to suddenly fail is a component operating at time 5 compared with a component operating at time 3?

Answer 1

Given,

$\alpha=0.25,\beta=3$

(a) Probability that the component is still operating at time 5-

$P(X>5) =r(5)=e^{(-\alpha\times 5)^\beta}=e^{-(0.25\times 5)^3}=0.1418$

(b)  Probability that the component fails before time 3.

$P(X\le 3)=F(3)=1-e^{(-\alpha\times 3)^\beta}=1-e^{(-0.25\times 3)^3}=0.3441$

(c) The failure rate function is-

$h(t)=\dfrac{f(t)}{r(t)}=\dfrac{\beta \alpha ^{\beta-1}e^{-(\alpha .t)^\beta}}{e^{-(\alpha.t)^\beta}}=\beta \alpha^{\beta}t^{\beta-1}=3\times (0.25)^3\times t^{3-1}=0.04687t^2$

(d) The chance that component that has not failed by time 5 suddenly fails is h(5). Similarly, The chance that a component that has not failed by time 3 suddenly fails is h(3).

$\dfrac{h(12)}{h(8)}=\dfrac{0.04687 (5)^2}{0.04687(3)^2}=\dfrac{25}{9}=2.777$