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a certain type of light bulb has lifetimes that follows an exponential distribution with mean 1000 hours. Find the median lifetime. (ie the lifetime x such that 50% of the light bulbs fail before x.) The answer is 1000log2.I do not know to to get the answer.please help. Thank you!

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Hint: What does your book say is the cumulative probability distribution function or CDF of this random variable? For what value of $x$ does $F(x)$ equal $\frac{1}{2}$? If your book does not mention the CDF of exponential random variables at all, does it at least tell you that $P\{X > t\} = e^{-\lambda t}$? –  Dilip Sarwate Dec 4 '13 at 4:36
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1 Answer 1

up vote 1 down vote accepted

Let random variable $X$ be the lifetime of the bulb.

The density function of $X$ is $\frac{1}{1000}e^{-x/1000}$ (for $x\ge 0$). We want to find $m$ so that $$\Pr(X\gt m)=\frac{1}{2}.$$ So we want $$\int_m^\infty \frac{1}{1000}e^{-x/1000}\,dx=\frac{1}{2}.$$ Integrate. We get $$e^{-m/1000}=\frac{1}{2}.$$ Take the natural logarithm of both sides. We get $$-\frac{m}{1000}=\ln(1/2)=-\ln 2.$$ It follows that $m=1000\ln 2$.

Remark: The exponential distribution fits the lifetime of real-life incandescent bulbs quite poorly.

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Thank you very much! –  Jing Dec 4 '13 at 5:10
    
You are welcome. I left some details (the integration) to you. –  André Nicolas Dec 4 '13 at 5:11
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