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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.
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(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);
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