Discussion_8

Discussion 8

11. A company buys 100 lightbulbs, each of which has an exponential lifetime of 1,000 hours. What is the expected time for the first of these bulbs to burn out?

I tried two approaches on this. I think the second one is correct, but the first one has a certain intuitive appeal. Which is probably not good:

A. All the bulbs are iid in the exponential distribution. So the probability of any bulb burning out is:

Using the cumulative distribution function,

\(P(X>k)\quad =\quad { e }^{ -k/\lambda }\)

and

\(P(X\le k)\quad =\quad 1-{ e }^{ -k/\lambda }\) = 11

So, 11 hours to lights out.

In R:

x <- 1-log(exp(-10))
x

## [1] 11

Answer 2: Based on a bunch of other research, the pdf of minX is:

\(P(minX)\quad =\quad \frac { n }{ \lambda } e^{ \frac { -n }{ \lambda } (x) }\quad\)

which translates to an expected value of:

\(E[X]\quad =\frac { \lambda }{ n }\)

which results in:

\(E[X]\quad =\frac { \lambda }{ n } =\quad \frac { 1000 }{ 100 } \quad =\quad 10\)