
#1
Apr612, 06:33 PM

P: 288

1. The problem statement, all variables and given/known data
3. The attempt at a solution I think I'm just evaluating integrals in this problem, not so? For part a) [itex]\int_0 ^1 λe^{λt}dt = \int_0 ^1 e^{t}dt = e^{t} ^1 _0 = \frac{1}{e} + 1[/itex] For part b) [itex]\int_0 ^3 e^{t}dt = e^{t} ^3 _0 = \frac{1}{e^3} + 1[/itex] For part c) [itex]\int_3 ^4 e^{t}dt = e^{t} ^4 _3 = \frac{1}{e^4} + \frac{1}{e^3}[/itex] For part d) Consider the limit as R goes to ∞ of [itex]\int_4 ^R e^{t}dt = e^{t}  ^R _4 = 0 + \frac{1}{e^4}[/itex] Does that seem right? 



#2
Apr612, 06:52 PM

P: 288

This is a very similar question that I'd like my work checked on.
[itex]\int_0 ^T λe^{λt}dt = \int_0 ^T (.01)e^{(.01)t}dt =e^{(.01)t} ^T _0 =\frac{1}{e^{\frac{T}{100}}} + 1 = Q[/itex]. I think this is the probability that the lightbulb will fail within the T hours. Thus, the probability that it will not fail should be given by [itex]1Q[/itex]; that is, the probability of all outcomes minus the probability that the lightbulb fails. For part b, I can solve the equation [itex]\frac{1}{e^{\frac{t}{100}}} + 1 = \frac{1}{2}[/itex] for t. Write [itex]e^{\frac{t}{100}} = \frac{1}{2}[/itex] [itex]\frac{t}{100} = ln(2)[/itex] [itex]t = 100ln(2)[/itex] So 100*ln(2) hours is the time when the reliability of the bulb is 1/2. Is that right? 



#3
Apr712, 02:48 AM

HW Helper
Thanks
P: 4,664

The question is very badly worded. I think what they are saying is that the times between decays are iid exponential random variables with rate λ. That means that the count of the number of decays in a time interval [a,b] is a Poisson random variable with mean λ(ba). In particular, the probability that a decay occurs after time t = 4 is 1, since the probability of 0 decays in the interval [4,T] is exp(λ(T4)), which → 0 at T → ∞. RGV 



#4
Apr712, 09:46 AM

P: 288

Probability Density Functions
How about these integrals then:
a) [itex]\int_0 ^1 (10)e^{(t)} dt = 1  \frac{1}{e}[/itex] b) [itex]\int_0 ^1 (3)e^{3t} dt = 1  \frac{1}{e^3}[/itex] c) [itex]\int_0 ^1 (43)e^{(43)t} dt = 1  \frac{1}{e}[/itex] d) [itex]\int_0 ^1 (R4)e^{(R4)t} dt = 1 e^{4R}[/itex], which, as you indicated, goes to one as R goes to infinity. How is that? 



#5
Apr712, 10:05 AM

P: 288

I guess this changes my answer to question six slightly. Would I write instead:
Calculate the probability that it will burn out: [itex]\int_0 ^1 (.01)(T0)e^{(.01)(T0)t} dt = e^{(.01)T}  1[/itex] Thus, the probability that it will not burn out is given by [itex]1  e^{(.01)T} + 1 = 2  e^{(.01)T}[/itex] We can set this expression equal to 2 and solve for T, which we find to be ln(3/2)/.01 Is that better? 



#6
Apr712, 11:16 AM

HW Helper
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P: 4,664

So, I would go with your original answers to the machine question. Let me ask you a serious question here: do you actually know what the different models and methods are about, or are you just guessing? You seem very uncertain. I would recommend that you do additional reading about these topics, to eliminate your sources of apparent confusion. If your textbook does not have a decent discussion, the web certainly does. The source http://www.ssc.upenn.edu/~rwright/courses/poisson.pdf has a fairly gentle introduction to this area. RGV 



#7
Apr712, 02:20 PM

P: 288

I am just guessing. My textbook is just a pdf, so just out of curiosity I did a search for "poisson." The results that came up were in chapter five called "Important Distributions and Densities." The chapter that I pulled these questions from were from chapter two called "Continuous Probability Densities." It's certainly clear how the two are related, but poisson isn't mentioned in my text for another two chapters! So when I sat down to these problems, I thought they were just integral evaluations, and when you told me that I had done the first problem wrong, I assumed that I had misunderstood how to properly integrate these functions. That is, that I have to readjust the bounds and include (ba) terms. That's why I reworked the second problem: it seemed very similar, and at this point in the text, I only knew one way to properly solve, and it made no mention of poisson.




#8
Apr712, 05:23 PM

HW Helper
Thanks
P: 4,664

My interpretation could be all wrong, but in that case I have no clue at all what the author could possible want, and the problem would be describing something very different from actual, realworld radioactivity. The link I gave you in my previous reply explains the material in a nottoocomplicated way. Have you tried reading it? RGV 


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