Science:Math Exam Resources/Courses/MATH103/April 2011/Question 05 (d)

MATH103 April 2011

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Question 05 (d)
In a room two different lightbulbs $A$ and $B$ are installed. According to the packaging, lightbulb $A$ fails before time $t$ with probability $F_{A}(t)=1-e^{-t}$ ($t$ in months). Lightbulb $B$ has a probability density function for the failure time given by $p_{B}(t)=Ce^{-2t}.$ Which light bulb has the longer expected lifetime? (i.e. longer average time to failure.) Show your reasoning.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The expected lifetime, or the expected time to failure, is the mean value of the random variable $t$ (the time to failure).

Hint 2
Recall from (a) and (b) that $p_{A}(t)=e^{-t}$ and $p_{B}(t)=2e^{-2t}.$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The expected lifetime of lightbulb $A$ is ${\begin{aligned}{\overline {t}}_{A}=\int _{0}^{\infty }tp_{A}(t)dt=\int _{0}^{\infty }te^{-t}dt\end{aligned}}$ Integrating by parts, ${\begin{aligned}{\overline {t}}_{A}&=\left[-te^{-t}\right]_{0}^{\infty }-\int _{0}^{\infty }(-e^{-t})dt=\int _{0}^{\infty }e^{-t}\,dt=\left.-e^{-t}\right|_{0}^{\infty }=1.\end{aligned}}$ The expected lifetime of lightbulb $B$ is ${\begin{aligned}{\overline {t}}_{B}=\int _{0}^{\infty }tp_{B}(t)dt=2\int _{0}^{\infty }te^{-2t}dt\end{aligned}}$ Again, integrating by parts, ${\begin{aligned}{\overline {t}}_{B}=2\left[-t{\frac {e^{-2t}}{2}}\right]_{0}^{\infty }-2\int _{0}^{\infty }{\frac {e^{-2t}}{(-2)}}dt=\int _{0}^{\infty }e^{-2t}dt=\left.{\frac {e^{-2t}}{-2}}\right|_{0}^{\infty }={\frac {1}{2}}.\end{aligned}}$ Hence, lightbulb $A$ has the longer expected lifetime.

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