I have come across this problem:

A room is lit by $10$ light bulbs. The lifetime, X, of the light bulbs follows an exponential distribution with mean $ \mu$ = $1000$ hours. In a time window of $2000$ hours:

  1. what is the expected number of failures?

I did this as: For time window of 2000 hours, and 10 bulbs, expected value= $\frac{1}{1000}$ * $2000$ * $10$ = $20$

  1. what is the probability that there will be more than $5$ failures?

The light bulbs are replaced immediately upon failure.

I calculated the probability that a light bulb will not fail (X> $2000$) is:

$ \mu$ =$1000$ , $\lambda$ = $\frac{1}{1000}$

P(X>2000) = $e^{-\lambda x}$ = $e^{-2000/1000}$ = 0.13 ( 13% )

and tried to incorporate that in to solving the questions but I think I'm heading the wrong way.
Edit: I also used Cumulative Poisson Probability to find upper/lower limits but I don't know how to apply it to multiple objects. Also I am curious as to how the problem would change if the light bulbs were not being replaced upon failure.

  • $\begingroup$ You need to use the poisson distribution with parameter $\lambda=2000/\mu$ to answer the more involved questions. $\endgroup$ – kodlu Nov 17 '15 at 9:06
  • $\begingroup$ So let me see if i get this. I will use P(x; μ) = (e-μ) (μx) / x! whith $\lambda$ instead of $ \mu$ and multiply the results by 10 for the number of lught bulbs? $\endgroup$ – apot Nov 17 '15 at 21:32
  • $\begingroup$ Twice I cleaned up your very very clumsy use of MathJax and both times you undid my edits. $\endgroup$ – Michael Hardy Nov 18 '15 at 4:39
  • $\begingroup$ Yes I am very sorry for that. I did it while adding some stuff to the question and accidentally erased yours. I am very new to this plus the timing was terrible. Doing my best to learn, thanks for your edits. $\endgroup$ – apot Nov 18 '15 at 4:42

If there were just one light bulb with an average lifetime of $1000$ hours, then the number of failures in time $t$ would have a Poisson distribution with expected value $t/1000$. With $10$ light bulbs functioning independently of each other, that expected value is simply multiplied by $10$. The reason is that the distribution of the sum of independent Poisson-distributed random variables is Poisson-distributed. I think the question of why that is so has been answered here a number of times.

  • $\begingroup$ Thank you. If it's not to much to ask can you explain how the approach would change if the lightbulbs were not being replaced upon failure? Is that sentence there just as a clarification or it would actually made a difference? I suspect the latter but I cannot find an example anywhere $\endgroup$ – apot Nov 18 '15 at 5:37

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