Hazard rate function

Question

The cumulative distribution function of the lifetime $T$ of an electrical component of a device is give by

$$F(t) = \begin{cases} 0, & t\le 0 \\ 1 - e^{-(t/4 ~+ ~2t^2)}, & 0 < t < \infty \end{cases}$$

$a)$ Derive the hazard rate function, $h(t)$ for this lifetime variable.

$b)$ Calculate the expected time to failure of the electrical component.

I'm not quite sure how to do $a)$ wouldn't the hazard rate function just be the derivative of $\left(\dfrac{t}{4} + 2t^2\right)$

Answer 1

Yes that's correct. To see this, recall (e.g. see Wikipedia) that $$h(t) = -\frac{d}{dt}\big{(}\ln (S(t))\big{)},$$ where $S(t)=1-F(t)$ is the survival function.