Question

For a given product, suppose the probability of an initial failure (a failure prior to time \(t=\alpha\) ) is \(\theta_{1}\), the probability of a wear-out failure (a failure beyond time \(t=\beta\) ) is \(\theta_{2}\), and that for the interval \(\alpha \leq t \leq \beta\) the failure-time density is given by
\[
f(t)=\frac{1-\theta_{1}-\theta_{2}}{\beta-\alpha}
\]
(a) Find an expression for \(F(t)\) for the interval \(\alpha \leq t \leq \beta\)
(b) Show that for the interval \(\alpha \leq t \leq \beta\), the failure rate is given by
\[
Z(t)=\frac{1-\theta_{1}-\theta_{2}}{(\beta-\alpha)\left(1-\theta_{1}\right)-\left(1-\theta_{1}-\theta_{2}\right)(t-\alpha)}
\]
(c) Suppose that the failure of a digital television set is considered to be an initial failure if it occurs during the first 100 hours of usage and a wear-out failure if it occurs after 15,000 hours. Assuming that the model given in this exercise holds and that \(\theta_{1}\) and \(\theta_{2}\) equal \(0.05\) and \(0.75\), respectively, sketch the graph of the failure-rate function from \(t=100\) to \(t=15,000\) hours.

Answer 1

(a) The cumulative distribution function (CDF) of a random variable is defined as [F(t) = P(X \leq t) = \int_{-\infty}^t f(u) , du,]where $f(t)$ is the probability density function (PDF) of the random variable $X$.

In this case, the PDF of the failure time for the interval $\alpha \leq t \leq \beta$ is given by [f(t) = \frac{1 - \theta_1 - \theta_2}{\beta - \alpha}.]Therefore, the CDF for the interval $\alpha \leq t \leq \beta$ is given by [F(t) = \int_{\alpha}^t \frac{1 - \theta_1 - \theta_2}{\beta - \alpha} , du = \frac{1 - \theta_1 - \theta_2}{\beta - \alpha} (t - \alpha).]

(b) The failure rate of a product is the derivative of the CDF with respect to time, i.e., [Z(t) = \frac{dF(t)}{dt} = \frac{d}{dt} \left( \frac{1 - \theta_1 - \theta_2}{\beta - \alpha} (t - \alpha) \right) = \frac{1 - \theta_1 - \theta_2}{\beta - \alpha}.]

(c) If the failure of a digital television set is considered to be an initial failure if it occurs during the first 100 hours of usage and a wear-out failure if it occurs after 15,000 hours, then we have $\alpha = 100$ hours and $\beta = 15,000$ hours. Assuming that $\theta_1 = 0.05$ and $\theta_2 = 0.75$, the failure rate is given by [Z(t) = \frac{1 - \theta_1 - \theta_2}{\beta - \alpha} = \frac{1 - 0.05 - 0.75}{15,000 - 100} = \frac{1}{2,999}.]The graph of this function is shown below for the interval $100 \leq t \leq 15,000$ hours:

[asy] unitsize(1 cm);

real Z(real t) { return 1/2999; }

draw(graph(Z,-1.5,16.5)); draw((-1.5,0)--(16.5,0),Arrows);