Question 2.6: If T is a random variable representing the hours to failure ......
If T is a random variable representing the hours to failure for a device with the following pdf:
f(t)=t \exp \left(\frac{-t^2}{2}\right), \quad t \geq 0(a) Find the reliability function.
(b) Find the hazard function
(c) If 50 devices are placed in operation and 27 are still in operation 1 hour later, find approximately the expected number of failures in the time interval from 1 to 1.1 hours using the hazard function
(a) To develop the reliability function, R(t), we have
R(t)=\int_t^{\infty} f(\tau) d \tau=\int_t^{\infty} \tau \exp \left(-\tau^2 / 2\right) d \tauLet u=\tau^2 / 2, \mathrm{du}=\tau d \tau; then we have
R(t)=\int_{\frac{t^2}{2}}^{\infty} \exp (-u) d u=\exp \left(\frac{-t^2}{2}\right), \quad t \geq 0 .(b) To develop the hazard function h(t), we have
h(t)=f(t) / R(t)=t, \quad t \geq 0.
Thus the hazard rate is linearly increasing with a slope of 1.
(c) To answer this question, we can use the information in Section 2.2.1, and we have
N_S(0)=50, N_S(1 \text { hour })=27, \Delta N=N_S(1)-N_S(1.1)=?For small Δt, the expected number failing can be calculated using Equation 2.14:
\hat{h}(t) \approx \frac{N_1-N_2}{N_1 \Delta t} \text {. } (2.14)
\Delta N=N_S(t)-N_S(t+\Delta t) \approx h(t) \Delta t N_S(t) \\=1.0 \times 0.1 \times 27=2.7 \text {. }Note that by the using the concept of conditional reliability, we also get,
P[T>1.1 \mid T>1.0]=\frac{R(1.1)}{R(1.0)}=\frac{0.54607}{0.60653}=0.8904or
\Delta N=27 \times(1-0.9)=2.7