Minimum sample size required question The lifetime (in years) of a device, $X$ can be modelled by an $Exp(1)$ distribution.  What is the minimum number of devices I need so that, with probability no smaller than 0.95, at least 30% of these devices will last beyond 1 year?
I don't really know how to start with this question. But here's what I've tried:
The pdf of X: $$P(X) = e^{-x}$$
We are solving for $x$ in $P(e^{-x}>0.3)>0.95$.
And I am stuck. I doubt that my method is correct. Would strongly appreciate any sort of help on this question. Thank you.
 A: It rather depends on what tools you have available.  There are also curiosities: if the minimum is $n$ then $n+1$ turns out not to work, so approximations may not be sufficiently precise.  But here are some hints:

*

*What is the probability $p$ that a particular item will last more than a year?  What is the probability $q$ that it will fail in the first year?

*If you have $n$ independent items, the number lasting (or failing) has a binomial distribution with parameters $n$ and probability you just calculated, so you can work out the probability that the number lasting is greater than or equal to $0.3n$ (or equivalently the number failing is less than or equal to $0.7n$, which might be simpler to handle if you have a cumulative distribution function for the binomial)

*One way of doing this is to test sufficient $n$ and find these binomial probabilities: you want the smallest $n$ where this gives at least $0.95$.

If you wanted an approximation, then you could think:

*

*the expected number lasting is $np$, and the variance is $np(1-p)$ so standard deviation $\sqrt{np(1-p)}$


*in a normal distribution used as an approximation to the binomial, $95\%$ of the probability is above the point about $1.645$ standard deviations below the mean


*so you want to find $n$ such that $np - 1.645\sqrt{np(1-p)} \ge 0.3n$, which you can solve for $n$
but I think this gives an approximate answer which, though of the right order of magnitude, is noticeably too large.  As a check, I think the answer here is between $100$ and $150$.
