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Unformatted text preview: AMS 311 (Fall, 2009) Joe Mitchell PROBABILITY THEORY Homework Set # 5 Solution Notes (1). [Related to Example 5b (Ross, Chap 5), and exercises 32, 33 (Ross, Chap 5).] (15 points) The lifetime (in hours) of a lightbulb is an exponentially distributed random variable with parameter = 0 . 10 (units of hours 1 ). (a). What is the probability that the light bulb is still burning one day after it is installed? (b). Assume that the bulb was installed at noon today and assume that at 3:00pm today you notice that the bulb is still working. (i). What is the chance that the bulb will burn out at some time between 4:30pm and 6:00pm today? (ii). What is the expected time when the bulb burns out? (again, given that it was still working at 3:00pm today) Let X be the lifetime (in hours); X is exp(0.1). (a). P ( X > 24) = integraltext 24 (0 . 1) e . 1 x dx = e 2 . 4 (b). (i) P (4 . 5 < X < 6  X > 3 . 0) = P (1 . 5 < X < 3 . 0) = F (3 . 0) F (1 . 5) = e . 15 e . 3 by the memoryless property of exponential; or do it directly from the definition: P (4 . 5 < X < 6  X > 3 . 0) = P (4 . 5 < X < 6 ,X > 3 . 0) P ( X > 3 . 0) = P (4 . 5 < X < 6) P ( X > 3 . 0) = e (0 . 1)(4 . 5) e (0 . 1)(6) e (0 . 1)(3 . 0) = e . 15 e . 3 (ii). Given that the bulb is still working at 3:00pm, the memoryless property of the exponential tells us that it is as if the bulb were newly installed at 3:00pm. Thus, the expected time it burns out is 10 hours later (at 1:00am)....
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 Fall '08
 Tucker,A

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