Question 2.9: The failure rate per year of a component is given by: h(t) =......

(a) R(t)=\exp \left(-\int_0^t h(\tau) d \tau\right)=\exp \left(-\int_0^t 0.003 \tau^2 d \tau\right)

= exp(-0.001t³)

and for the probability density function, we have

(b) We have

B_{20}=\left(\frac{\ln 0.80}{-0.001}\right)^{1 / 3}=6.065 years.

(c) E[T]=\int_0^{\infty} R(t) d t=\int_0^{\infty} t \cdot f(t) d t=\int_0^{\infty} 0.003 t^3 \exp \left(-0.001 t^3\right) d t .

Let u = 0.00t³ , du = 0.003t² dt

E[T]=\frac{1}{0.001^{ 1/ 3}} \int_0^{\infty} u^{(1 / 3+1)-1} e^{-u} d u=\frac{1}{0.001^{1 / 3}} \Gamma(1.333)=10 \times 0.89302=8.9302 years

where the value of the gamma function is found from the table in Appendix B.