Assume that the lifetime of a hard disk is distributed according to an exponential law. The manufacturer wants to guarantee that the hard drive has a probability of less than 0.001 of failing over one year. What is the minimum average life of the hard drive?
The average life is the same thing as the expected value, so by minimizing the average life we want to minimize the expected value, which means we want to minimize $1/\lambda$. Hence we have to maximize $\lambda$.
But I didn't really know how to maximize it.
Also, we have that the owner wants to make the probability of the hard disk failing less than 0.001, this can be translated to: $P(x \leq1)=0.001$, so $\int_0^1 \lambda e^{-\lambda x} = 0.001$, solving this for lambda we find $\lambda = 0.001$
But I didn't really know what to do with this information or what does this information mean! How can we approach this problem exactly?
Let $X$ be the random lifetime (measured in years) of such a hard drive. Thus the assumption that the lifetime is distributed according to an exponential law implies $$X \sim \operatorname{Exponential}(\mu), \\ \Pr[X \le x] = 1 - e^{-x/\mu}, \quad x \ge 0, \tag{1}$$ where $\mu = \operatorname{E}[X]$ is the mean lifetime.
The requirement imposed by the manufacturer, that the $1$-year failure probability is not to exceed $0.001$, is expressed as $$\Pr[X \le 1] \le 0.001. \tag{2}$$ Therefore, $(1)$ and $(2)$ combined gives us
$$1 - e^{-1/\mu} = \Pr[X \le 1] \le 0.001, \tag{3}$$
and solving this inequality for $\mu$ yields the desired minimum lifetime condition; i.e., if $\mu$ is at least this number of years, the probability that such a hard drive will fail within $1$ year will not exceed $0.001$.
As a follow-up question, what is the median lifetime of such a hard drive? That is to say, if you had a large number of such hard drives, after how many years would you expect to see half of them fail?