# Approximately calculate the probability that the average lifetime of all the bulbs in a particular box exceeds $2500$ hours.

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.

• I think the variance is 2000^2x100 instead of 2000x(100^2). – velut luna Apr 23 '16 at 23:12
• When you sum identical variables, the variance is multiplied by $n^2$. When you sum independent but identically distributed variables, the variance is only multiplied by $n$ (because the covariances are all zero). – Ian Apr 23 '16 at 23:12
• You might have gotten a downvote because you did not use proper formatting. Formatting tips here. For future reference, continue to post your thoughts and attempts, and try to use proper spelling to the best of your abilities. – Em. Apr 24 '16 at 0:18

$\newcommand{\Var}{\operatorname{Var}}$ 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*}

• thanks :) your answers are correct! can u just explain how you knew to calculate the variance that way for both? – kale Apr 24 '16 at 21:43
• i just figured it out for 2) but i'm still confused for 3). – kale Apr 24 '16 at 22:00
• You are asked for the chance that a simple sum exceeds $220000$. So, you don't divide by anything since it is not an average, and so $\Var(X_1+\dotsb+X_{100}) = \text{Var}(X_1)+\dotsb+\text{Var}(X_{100}) = 100\text{Var}(X_1) = 100\cdot 2000^2$ where the first equality is true by independence and the second is true since they are identically distributed. – Em. Apr 24 '16 at 23:33
• Also, if you found this answer satisfactory, consider giving a check mark. – Em. Apr 24 '16 at 23:35
• but wouldnt 100Var(X) = (100^2)xVar(x)? I thought u squared the constant... – kale Apr 25 '16 at 15:33