Here is the question:
An inspector selects every nth item in a production line for a detailed inspection. Suppose that the time between item arrivals is an exponential random variable with mean 1 minute,and suppose that it takes 2 minutes to inspect an item. Find the smallest value of $n$ such that with a probability of 90% or more, the inspection is completed before the arrival of the next item that requires inspection.
I really don't understand the question. If $Z$ is the time between two arrivals, then we want $\mathbb{P}[Z\gt 2]\gt 0.9$ if $t$ is in minutes. So what is the importance of $n$ here? Where does it affect? According to the question, is it true that $\mathbb{P}[Z = t] = e^{-t}$ for all $t\in \mathbb{R}$? I'm confused about this question and also the Poisson distribution which I think is related to the question.
Let $T_k$ bet the arrival time of the n-th item. Then according to the text, the time between the arrivals $$\delta_k = T_k-T_{k-1} \sim \text{Exp}(1),$$ is exponentially distributed. If every n-th item is inspected, then an inspection of the K-th item means, that the next item inspected is the K+n-th.
The inspection of the K-th item starts at arrival (i.e. at $T_K$). Since the inspection takes 2 minutes, the inspection finishes at $T_K+2$. So the question is, what is the smallest n, such that the next item arrives after the inspection finish with 90% probability? I.e. find $$n^* = \min\{n\in\mathbb{N}: P(T_{K+n} >T_K+2)>0.9\}$$
And $$\begin{align}P(T_{K+n} >T_K+2) &= P\left(T_K+\sum_{k=K+1}^n T_k-T_{k-1} > T_K+2\right)\\ &=P\left(\sum_{k=K+1}^n \delta_k > 2\right) \end{align}$$
Where $\sum_{k=K+1}^n \delta_k \sim \Gamma(n, 1)$ since sums of exponential distributed random variables are gamma distributed.