I'm trying to solve the following questions
Suppose that the lifetimes of light bulbs are independent, exponentially distributed random variables with a mean of $2000$ hours each.
1) Calculate the probability that a randomly chosen light bulb survives for more than $2500$ hours.
For this I got $e^{-2500/2000}$ (I don't know if this is needed for the other parts but thought I'd mention it).
I'm stuck on the other two parts
Suppose that the bulbs come in boxes of $100$.
2) Approximately calculate the probability that the average lifetime of all the bulbs in a particular box exceeds $2500$ hours.
3) Approximately calculate the probability that the sum of the lifetimes of all the bulbs in a particular box exceeds $220,000$ hours.
I know I need to use the central limit theorem somehow for both so I need to calculate the mean and variance for both parts. For number 2, I have no clue what to do. For number 3, I calculated:
$\text{mean} = 2000\times100=200000$
$\text{variance} = 2000\times100^2=20000000$
Then I did
$$P\left(z>\frac{220000-\text{mean}}{\text{sq. root of variance}}\right) = p(z>4.47)$$
but since you can't find probabilities for values greater than $3$ in the $z$ table, I got stuck and assumed I did it wrong.
So I really don't know how to solve either. Any help would be great! Thanks.
Best Answer
$\newcommand{\Var}{\operatorname{Var}}$ I call the life time of a particular bulb $X_i$
1) Is fine.
2) There are 100 bulbs in the box, and hence the average is $$\bar X = \frac{X_1+\dotsb+X_{100}}{100}.$$
Then $$E[\bar X] =\frac{1}{100}E[X_1+\dotsb+X_{100}] = \frac{1}{100}\cdot100E[X_1] = 2000$$ and \begin{align*} \Var(\bar X) &= \Var\left(\frac{X_1+\dotsb+X_{100} }{100}\right) \\ &= \frac{1}{100^2}\left[\Var(X_1)+\dotsb+\Var(X_{100})\right]\\ &= \frac{2000^2}{100}\\ &= 40000. \end{align*} Then the problem becomes \begin{align*} P\left(\frac{X_1+\dotsb+X_{100}}{100}>2500\right)&=1-P\left(\frac{X_1+\dotsb+X_{100}}{100}\leq 2500\right)\\ &\approx1-P\left(Z\leq\frac{2500-2000}{\sqrt{40000}}\right)\tag 1\\\ &=1-\Phi(2.5)\tag 2\\ &=0.006209665 \end{align*} where in $(1)$ $Z$ is a standard normal and we are using a normal approximation, and in $(2)$, $\Phi$ is the standard normal cdf.
3) Let $S = X_1+\dotsb+X_{100}$. Then $$E[S] = E[X_1+\dotsb+X_{100}] =100\cdot 2000 = 200000$$ and $$\Var(S) = \Var(X_1+\dotsb+X_{100}) = 100\cdot 2000^2 =20000^2.$$
Then the problem becomes \begin{align*} P(S>220000)&=1-P(S\leq 220000)\\ &\approx 1-P\left(Z\leq \frac{220000-200000}{\sqrt{20000^2}}\right)\\ &=1-\Phi(1)\\ &=0.1586553 \end{align*}