I have been solving the following exercise:
Lets say we have a chip with a critical component that needs to be replaced from time to time as soon as it breaks. Lets say that such a component is expected to break after 30 days. Now lets say we have an infinite amount of replacement parts, then the random variables $$ T_i\stackrel{iid}{\sim} Exp(\lambda=\tfrac{1}{30})\quad\text{for }i\in\{1,2,...\} $$ denote the lifetime of the $i$-th component. Now let $$ S=\sum_{i=1}^{39} T_i $$ be the time until we have replaced the component 38 times. What is the probability, that we will do more than 38 replacements within three years?
My textbook suggests the following solution:
In other words, we're looking for the probability $Pr[S<3\cdot365]=Pr[S<1095]$.
We know that $$ E[S]=E\left[\sum_{i=1}^{39}T_i\right] =\sum_{i=1}^{39}E[T_i]=39\cdot E[T_1] = 39\cdot 30 = 1170. $$ $$ Var[S]=Var\left[\sum_{i=1}^{39}T_i\right] =\sum_{i=1}^{39}Var[T_i] =39\cdot Var[T_1]=39\cdot 30^2=35100. $$ Now due to the central limit theorem we know that: $$ P[S<1095]=P\left[\frac{S-E[S]}{\sqrt{Var[S]}} <\frac{1095-E[S]}{\sqrt{Var[S]}}\right]\approx\Phi(-0.4)=1-\Phi(0.4)\approx 0.3346. $$
What I can't understand about the solution:
I know what the central limit theorem is about: In this case, $T_1,T_2,\ldots$ are i.i.d. with $Exp(\lambda)$ with expected value $\mu$ and variance $\sigma^2$. Now let $S_n$ denote $$ S_n=\sum_{i=1}^{n}T_i $$ Therefore $S_n$ has the expected value $n\mu$ and Variance $n\sigma^2$. So in this case with the central limit theorem we have: $$ \lim_{n\to\infty} P\left[\frac{S_n-n\mu}{\sigma\sqrt{n}}\leq z\right] = \Phi(z). $$ Which iswhat we have in the last line of the solution - except for the limit.
Now in this exercise we're not really taking the limit of $n$. Instead $n$ is just 39. Is the solution of the textbook just an approximation? If so, what are the reasons that one can take this approximation, even if $n$ is not going to infinity? Is this the best we can approximate the desired probability with? Or is the infinite sampling happening elsewhere, which would give us a valid reason to use the theorem?
To answer your questions in order...
Yes, the solution in your textbook is an approximation.
You rely on the Central Limit Theorem to make this approximation.
I would argue that, by and large, this is the best you can approximate the desired probability without using brute force methods where you are required to use a calculator to calculate this directly.
The infinite sampling isn't happening, per se. The Central Limit Theorem states that the sample mean ($\bar{X}=\frac{S_n}{n}$) of a population is Normally distributed, regardless of the distribution of the original variable ($X$), provided that $n$ is large enough. One can prove that, as $n \rightarrow \infty$, the sample mean of ($\bar{X}$) will become more and more Normally distributed such that if it were possible to "plug $\infty$ in for $n$," the sample mean would be exactly Normally distributed. What we see in practice, however, is that the sampling distribution of your sample mean will be close enough to a Normal distribution if $n$ is large enough. Practically, most people use $n\approx 30$ as a threshold where $n$ is large enough to apply the Central Limit Theorem. One point is that I use $\bar{X}$, the sample mean, instead of $S_n$ like you use above. The $\bar{X}$ form is the most common form of the Central Limit Theorem. If you take each term in $\frac{S_n-n\mu}{\sigma\sqrt{n}}$ and divide it by $n$, you will get $\frac{S_n/n - \mu}{\sigma/\sqrt{n}}=\frac{\bar{X} - \mu}{\sigma/\sqrt{n}}$. The explanations are equivalent.