The life X, of the StayBrite light bulb is modelled by the probability density function\[f(x) = \left(\begin{matrix}2e^{-2x} \\ 0\end{matrix}\right)\]\[x \ge 0 -otherwise- \]Where x is measured in thousands of hours. (a) Sketch the graph of f(x). (b) Find the median life of these StayBrite bulbs (b) E.g. \[P(X \le M) = \int\limits_{-\infty}^{M} f(x)dx= {1\over2}\]

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The life X, of the StayBrite light bulb is modelled by the probability density function\[f(x) = \left(\begin{matrix}2e^{-2x} \\ 0\end{matrix}\right)\]\[x \ge 0 -otherwise- \]Where x is measured in thousands of hours. (a) Sketch the graph of f(x). (b) Find the median life of these StayBrite bulbs (b) E.g. \[P(X \le M) = \int\limits_{-\infty}^{M} f(x)dx= {1\over2}\]

Mathematics
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|dw:1328093870926:dw| There's nothing for x<0, because your function is defined for x>=0. Don't forget that! this is, by the way, a case of http://en.wikipedia.org/wiki/Exponential_distribution
Thanks! Do you have any idea how to solve the seoncd part of the questions?
You are looking for the median value of the exponential function, defined as above. Meaning, you have to solve that integral with the expo function PDF plugged in and integral boundaries [0,M]. Why not -infinity? The function is zero for negative values, so we don't need to consider these. So basically, once you've integrated, you have an expression that contains a M. Set it equal to 1/2 and proceed to find M. You should get \[λ^{−1} \ln(2)\] as a result. I hope this is clear ^^

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The answer is 347, but am not sure how to arrive at the answer? This isn't very clear. I know I have to set it at 1/2, but when I substitue the values, I don't get the answer... Also, I'm not sure how to work with the e (exponential).
I gave you the general answer above. You have here \[\lambda = 2\]. Plug it in the median function and done. I'll write you down the integration steps...
\[\int\limits^{M}_{-\infty}\lambda e^{-\lambda x}dx = [-e^{-\lambda x}]^{M}_{-\infty} = \lim_{a \rightarrow -\infty}[-e^{-\lambda M}+-e^{-\lambda a}] = -e^{-\lambda M}\] Okay, it was wrong to drop the - infinity, because otherwise you have a 1 that will make your life harder. now.. \[-e^{-\lambda M} = \frac{1}{2}\] use natural logarithm function on both sides \[ \ln(-e^{-\lambda M}) = \ln(\frac{1}{2})\] \[-\lambda M = -\ln(2)\] \[M = \lambda^{-1}\ln(2)\] Inserting 2 for lambda value yields: http://www.wolframalpha.com/input/?i=2%5E%28-1%29*ln%282%29

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