There are $n$ machines. Each has durability given by exponential distribution with $EX = 10$. If a dead machine is replaced with new one immediately, find minimal $n$ so we can say with $P = 0.99$ that total durability will be at least $1100$?
I know that i need to get Binomial distribution and approximate it by Normal distribution but i do not know how do get there. The way this task is written is also confusing me a bit.
Having $n$ different machines all break down can be interpreted as the same machine breaking down $n$ times, since the the time to breaking is exponentially distributed and thus memoryless.
Let $X \sim \operatorname{Poisson}(\lambda)$ be the random variable that models how many times the machine breaks down in one time step. We have $\lambda = \operatorname{E}(X) = \frac{1}{10}$, since the break down rate is $\frac{1}{10}$. Now we have $Y \sim \operatorname{Poisson}(1100\lambda)$, which models the number of machines that break down in $1100$ time steps.
We want to find $y$ such that $\Pr(Y \leq y) = 0.99$, where $y$ denotes the number of machines required. Now we can model a $\operatorname{Poisson}(\lambda)$ distribution with high $\lambda$ as a normal distribution with mean and variance equal to $\lambda$, so we approximate $Y\sim \mathcal{N}(110,110)$.
Since we know $Z = \frac{Y-\mu}{\sigma}\sim \mathcal{N}(0,1)$ and $\Pr(Z\leq 2.326)=0.99$, we need to solve for $$\frac{y-\mu}{\sqrt{\sigma^2}}=2.326 \implies y = 2.326 \sqrt{110}+110 = 134.4,$$ hence we need $135$ machines to assure a $99\%$ probability that the total durability will be at least $1100$ time steps.