Minimum average lifespan using an exponential law distribution? It is assumed that the life of a certain equipment follows an exponential law. The manufacturer wants to ensure that this equipment has a probability of less than $0.001$ of failing over 1 year. What is the minimum average lifespan in years of this electronic equipment?
I am not sure at all how to do this. I thought finding the minimum $\lambda$ in using the inequality $P(X < 1) < 0.001$, but it seems not right. Can you help me with that?
In using the distribution function, I got $λ=0.00100005$ years
 A: Your confusion stems from the relationship between the parameter of the distribution (specifically, the choice of parametrization) and the expected value of the random variable as a function of that parameter.
There are two common ways to parametrize an exponential distribution.  One is by rate, and the parameter is typically (but not always) denoted by $\lambda$:  $$\Pr[X \le x] = 1 - e^{-\lambda x}, \quad x, \lambda > 0.$$  With this parametrization, the expectation of $X$ is $$\operatorname{E}[X] = \frac{1}{\lambda}.$$  We call this parametrization by "rate" because in exponential lifetime models for which this distribution applies, $\lambda$ can be conceptualized as a rate at which failure (or more broadly, the event of interest) occurs; thus with a higher rate, the random lifetime $X$ tends to be shorter.
The second parametrization is by scale, and the parameter is often (but not always) denoted by $\theta$ or $\mu$:  $$\Pr[X \le x] = 1 - e^{x/\theta}, \quad x, \theta > 0.$$  The expectation of $X$ is unsurprisingly $$\operatorname{E}[X] = \theta,$$ as the obvious relationship between the two parametrizations is $\theta \lambda = 1$.  This is called a scale parametrization because the value of $\theta$ might be regarded as a scale factor of the random variable $X$.  If we plot the CDF for different choices of $\theta$, we find for instance that doubling $\theta$ will "stretch" the CDF horizontally by a factor of $2$.
All that said, it is straightforward to solve the criterion $$\Pr[X \le 1] \le 0.001$$ for either type of parameter; for instance, $$0.001 \le \Pr[X \le 1] = 1 - e^{-\lambda}$$ implies $$\lambda \le \log \frac{1000}{999} \approx 0.0010005,$$ and since $\theta \lambda = 1$, we also get $$\theta \ge \left(\log \frac{1000}{999}\right)^{-1} \approx 999.4999166,$$ which is the expectation.
A final note.  This represents the smallest expected (average) lifetime because while one could choose to have a more strict quality control  by choosing a $1$-year failure probability that is less than $0.001$, which would then increase the expected lifetime, and still satisfy the criterion.  But one cannot choose a larger failure probability.
