The time to failure, T, of a car battery is described by an exponential distribution...

Question

The time to failure, T, of a car battery is described by an exponential distribution: $$ F_{T}(t)=\left\{\begin{matrix} 1-e^{-\gamma t}, t\geq 0, \gamma >0\\ 0, t<0 \end{matrix}\right.$$ The expected time to failure is 4 years (i.e., the average battery fails after 4 years). Find the probability that a battery lasts at least 8 years.

I'm not exactly sure how to approach this problem. Any help would be appreciated. Thanks!

Answer 1

First of all, you should be aware that in the exponential distribution $E(T)=1/γ $, so γ=1/4 (rate of failures), not 4.

To solve the problem, you have two alternatives:

Alternative 1:
Due to the fact that $F_{T}(t)$ represents $P(T<t)$, the $P(T\geq 8)$ can be expressed as $1-(1-e^{-0.25 (8)})$, so $P(T\geq 8)=e^{-2}$.

Alternative 2:
1. Differentiate the CDF, $F_{T}(t)$, to obtain the pdf
2. Integrate the pdf from 8 to $+\infty$.

You should arrive to the same answer.

Answer 2

An outline of the solution:

1. Differentiate $F_T(t)$ to obtain the density function $f_T(t)$.
2. Use the density function to compute the expected value of $T$ as a function of $\gamma$.
3. Use the fact that you know the expectation to be $4$ to solve for $\gamma$.
4. Use your knowledge of $\gamma$ to answer the question.

Let me know if you need any more hints or a more thorough outline.