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Math Help - Rate of decay and probability density function

  1. #1
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    Rate of decay and probability density function

    I'm having an issue because I can't connect my teacher's answer with the textbook's question.

    "Tritium is the basic fuel of hydrogen bombs and is used to increase the power for fission bombs. Tritium decays at a rate of 5.5 percent a year. This means that Prob[tritium atom dies <= t] follows an exponential distribution with mean = 1/Log(1-0.055)."

    It goes on to explain where it got that answer:
    "The formula for the mass at time t is f[t] = e^(r t). Therefore, the average percent growth rate is: [f(t+1)/f(t)] - 1. This simplifies to: e^r - 1. Setting this equation to -5.5, we get r = log (1-0.055)."

    I can't see the error in the logic, but the answer doesn't make sense to me. After all, if the rate of decay is -5.5%, why isn't r = -5.5%?

    The teacher says a decay of 5.5% means the tritium follows the exponential distribution with mean = 1/0.055.

    Can anyone help me make sense of this?
    Last edited by CrazyAsian; July 14th 2007 at 03:43 PM.
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  2. #2
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    Quote Originally Posted by CrazyAsian View Post
    I'm having an issue because I can't connect my teacher's answer with the textbook's question.

    "Tritium is the basic fuel of hydrogen bombs and is used to increase the power for fission bombs. Tritium decays at a rate of 5.5 percent a year. This means that Prob[tritium atom dies <= t] follows an exponential distribution with mean = 1/Log(1-0.055)."

    It goes on to explain where it got that answer by setting up a separable differential equation and solving for r. However, what doesn't make sense is 1/Log(1-0.055) is equal to -17.6771 which doesn't make sense.

    However, the teacher says a decay of 5.5% means the tritium follows the exponential distribution with mean = 1/0.055.

    Can anyone help me make sense of this?
    The exponential distribution with mean 1/\lambda has cumulative distribution function \Pr(X < t) = 1 - e^{-\lambda t}. Set \lambda = -\log(1-.055), the negative of what the book says. Then the probability the atom dies is in one year is \Pr(X < 1) = 1 - e^{\log(1-.055)} = .055. So the book is correct except for the sign.

    The book and your teacher are using different interpretations of a decay rate. The mean E(X) = 1/\lambda of the exponential distribution is interpreted as the expected waiting time to the event, here a decay. So \lambda is the average number of decays per unit of time. Your teacher is says a decay rate of 5.5% means the average number of decays per year is .055. The book says a decay rate of 5.5% means the probability of a decay in one year is .055. These interpretations are different because the average number of decays in one year is not the same the probability of a decay in one year.
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  3. #3
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    I think I figured it out...

    I think both the textbook and the teacher are wrong. My thinking now is...

    The teacher missed the fact that 5.5% is the annual rate of decay, not the continuous rate of decay.

    The textbook mistakenly uses 1/Log(1-0.055) as the mean survival time when it should actually be -1/Log(1-0.055).
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  4. #4
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    Thanks Jake! I wrote my response at the same time as yours. I think we are saying the same thing, but let me be sure...

    Quote Originally Posted by JakeD View Post
    The book and your teacher are using different interpretations of a decay rate. The mean E(X) = 1/\lambda of the exponential distribution is interpreted as the expected waiting time to the event, here a decay. So \lambda is the average number of decays per unit of time. Your teacher is says a decay rate of 5.5% means the average number of decays per year is .055. The book says a decay rate of 5.5% means the probability of a decay in one year is .055. These interpretations are different because the average number of decays in one year is not the same the probability of a decay in one year.

    Is what you are saying basically the same thing as what I'm saying about annual rate of decay vs. continuous?
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  5. #5
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    Quote Originally Posted by JakeD View Post
    The exponential distribution with mean 1/\lambda has cumulative distribution function \Pr(X < t) = 1 - e^{-\lambda t}. Set \lambda = -\log(1-.055), the negative of what the book says. Then the probability the atom dies is in one year is \Pr(X < 1) = 1 - e^{\log(1-.055)} = .055. So the book is correct except for the sign.

    The book and your teacher are using different interpretations of a decay rate. The mean E(X) = 1/\lambda of the exponential distribution is interpreted as the expected waiting time to the event, here a decay. So \lambda is the average number of decays per unit of time. Your teacher is says a decay rate of 5.5% means the average number of decays per year is .055. The book says a decay rate of 5.5% means the probability of a decay in one year is .055. These interpretations are different because the average number of decays in one year is not the same the probability of a decay in one year.
    Quote Originally Posted by CrazyAsian View Post
    I think both the textbook and the teacher are wrong. My thinking now is...

    The teacher missed the fact that 5.5% is the annual rate of decay, not the continuous rate of decay.

    The textbook mistakenly uses 1/Log(1-0.055) as the mean survival time when it should actually be -1/Log(1-0.055).
    Quote Originally Posted by CrazyAsian View Post
    Thanks Jake! I wrote my response at the same time as yours. I think we are saying the same thing, but let me be sure...

    Is what you are saying basically the same thing as what I'm saying about annual rate of decay vs. continuous?
    I'm also not sure we are saying the same thing. How do you define the annual rate of decay and continuous rate of decay?

    Let Y be the number of decays in one year and X be the time to decay. I'm distinguishing between the book's probability of decay in one year \Pr(Y \ge 1) = \Pr(X < 1) and your teacher's average number of decays in one year E(Y) = 1/E(X).

    If those are your definitions for annual and continuous rates of decay, then we agree.
    Last edited by JakeD; July 14th 2007 at 05:42 PM.
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