# Central limit theorem Question about working light bulbs

A person has $100$ light bulbs whose lifetimes are independent exponentials with mean $5$ hours. If the bulbs are used one at a time, with a failed bulb being replaced immediately by a new one, approximate the probability that there is a working bulb after $525$ hours.

When I tried to solve the question, I tried to find $P\{X\geq1\}$, where $X$ is the number of working bulbs after $525$ hours, and $X = \sum_{i=1}^{100} X_i$ where $X_i$ is an indicator variable for a bulb whose lifetime is greater than $525$ hours. If the bulb has a lifetime greater than $525$ then its value is $1$, otherwise $0$. I then used the central limit theorem with mean $= 5$ and variance $= 5$. My answer seems to be incorrect.

The solution manual says that $X$ must represent the total lifetime of all the bulbs, and we must find $P\{X\geq525\}$ using the central limit theorem. I don't understand how the total lifetime of all the bulbs (sum of all the lifetimes) being greater than $525$ means that there is a working lightbulb after $525$ hours. Can't the sum of the lifetimes of all the bulbs be greater than $525$ even if the lifetime of each bulb is less than $525$?

“...there is a working bulb after $525$ hours” means that the sum of lifetimes of all $100$ bulbs exceeds $525$ hours and not that an individual bulb had that lifetime. And the sum of the lifetimes may exceed $525$: after the first bulb failes, it is immediately replaced by a new one, if that second one failes by a third one and so on. Hence we want to calculate $P(X_1+\cdots X_{100})>525$.
As the expected value of $X_i$ is $5$ and variance is $25$ we know that the expected value of $X$ is $100\cdot5$ and its variance is $100\cdot 25$. Now $X$ is approximately normal distributed, just calculate $P(X>525)$.