How can I use the central limit theorem to calculate the probability of exceeding a certain profit? 
A manufacturer of Cameras has to offer a two year warranty for his products. For simplicity, assume that each camera consists of $8$ components and each of them has a chance of breaking during the two year period of $3\%$. If a component breaks and is replaced/repaired, the component will not break again during the two year period. It costs $50\$$ to produce a component and $50\$$ to repair/replace a broken component.
The company assumes it will sell $800$ cameras at launch. At what price should the company sell the camera so that there is a $98\%$ probability of making a profit of $120000\$$? (Hint:Use the normal approximation)

I am not necesarily looking for complete solutions. I am just stuck at certain parts of the calculation and would be very grateful for any hints.
Here is what I have done so far:
First of all, the central limit theoreom states: Let $X_1,\ldots,X_n$ be i.i.d random variables with finite mean $\mu$ and variance $\sigma ^2$. Let $S_n=X_1+\ldots+X_n$. Then we can define a standardized random variable $Z_n=\frac{S_n-n\mu}{\sqrt{n}\sigma}$ such that:
$$\lim_{n \to \infty}P\left(Z_n\le z \right)=P(Z \le z)$$
where $Z$ is a standard normal variable.

In this problem I basically want to find the price $y$ the company needs to sell the cameras at in order to have a $98\%$ chance of making a profit $P_n$ of at least $120000$. In other words:
$$P(P_n\ge120000)\approx0.98$$
I am going to need to find an expression for the random variable $P_n$ so that I can standardize it and then use a normal table to find my unknown $y$. For the profit, I was thinkink of the following formula:
$$\begin{aligned} \text{Profit}&=\text{Revenue}-\text{Costs}\\[10pt] &=\text{Revenue}-\text{Production Costs}-\text{Expected Repair Costs} \\[10pt] &=800y-8 \cdot800\cdot 50 $\ -\text{Expected repair costs} \end{aligned}$$
For the repair costs: Let $X_i$ be the state (broken or not broken) of the $i$-th component of a camera. Then:
$$X_i= \begin{cases}1 & \text{if} & \text{broken} \\ 0 & \text{if} & \text{not broken} \end{cases}$$
with $P(X_i=1)=0.03$ and $P(X_i=0)=0.97$.
Since the camera consists of $8$ components, I need some random variable $S_n$:
$$S_n=X_1+\ldots X_n$$
The mean of $S_n$ is $E[S_n]=\mu_{S_n}=np$ and the variance $\text{Var}(S_n)=\sigma^2_{S_n}=np(1-p)$.
Here is where I am stuck. I somehow need to integrate this random variable into the formula for profit. Do I then need to calculate the expected value of the profit again? It just seems like I am stuck in this loop of endlessly calculating expected values of random variables and I have a feeling that I am making a mistake somewhere. I am also not sure if I have even set up the problem correctly so in case I am making a large error somewhere please tell me.
 A: 
assume that each camera consists of 8 components and each of them has a chance of breaking during the two year period of 3%.

First note that there is an error in the exercise...to go on with the solution you have to assume also independence among the chance of breaking during the given period.
Assumed this, note that the only rv of the exercise is the Chance of breaking which is a bernulli as you stated. To forecast a certain profit with probability of $98\%$ you have to forecast which amount of cost you will have at $98\%$ confidence, thus your forecasted # of repairs is
$$\mathbb{P}\left[Z<\frac{x-800\times8\times0.03}{\sqrt{800\times8\times0.03\times0.97}}\right]=0.98$$
$$\frac{x-800\times8\times0.03}{\sqrt{800\times8\times0.03\times0.97}}=2.054$$
That is a forecasted repair / substitution # of repairs $x=221$ that means a forecasted cost of $\$11,050$
thus a forecasted selling price of
$$\frac{320,000+120,000+11,050}{800}=563.81$$
is needed, considering an extra cost of about $\$11,050$
