Discussion
If the failure rate of a random variable is \(h(t) = 3t + 2\), \(t \ge 0\), then its distribution function :
Answer: A
The distribution function is \(1 - \exp\left(-\frac{3t^{2}}{2} - 2t\right)\), \(t \ge 0\).
The failure rate determines the distribution through the cumulative hazard \(H(t) = \int_{0}^{t} h(u)\, du\):
\[ \begin{aligned} H(t) &= \int_{0}^{t} (3u + 2)\, du = \frac{3t^{2}}{2} + 2t \\ F(t) &= 1 - e^{-H(t)} = 1 - \exp\left(-\frac{3t^{2}}{2} - 2t\right) \end{aligned} \]
So the distribution can be obtained, which rules out the last option. The options with \((3t + 2)^{2}\) square the hazard instead of integrating it, and the one without '1 -' gives a survival-type function that is not even a valid distribution function, since it decreases.
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