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Exponential distribution/hydra population

  1. Oct 31, 2012 #1

    Somebody asked me for some help with his calc II homework, and this was one of the questions:

    "A hydra is a small freshwater animal and studies have shown that its probability of dying does not increase with the passage of time. The lack of influence of age on mortality rates for this species indicates that an exponential distribution is an appropriate model for the mortality of hydra. A biologist studies a population of 500 hydra and observes that 200 of them die within the first 2 years. How many of the hydra would you expect to die within the first six months?"

    Now, I saw a problem like this on a past actuarial exam and ran straight to the following process. But I'm about ready to smack the instructor in the head for believing this is a calc II problem, so I wanted to check to make sure I'm not missing anything in the wording of the problem that makes it much simpler than this before I do so.

    My process: First, the probability that a given hydra will die within 2 years is 0.4. Therefore, set 0.4 equal to the integral from 0 to 2 of (lambda * e^(-lambda * t)) dt. Integrating this gives 1 - e^-2(lambda). Then set e^-2(lambda) equal to 0.6 and solve, giving lambda = -ln(0.6)/2. Next, find P(T<0.5) which is integral from 0 to 0.5 of (lambda * e^(-lambda * t))dt, or 1 - e^[ln(0.6)/4], or 1 - (0.6)^(1/4) = 0.11988. Then multiply that by 500 to get 59.944, rounding up to 60.
  2. jcsd
  3. Nov 1, 2012 #2


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    That isn't how I would expect a CalcII student to work the problem. It is likely in the exponential decay section of his book and wouldn't be interpreted as a probability problem. The student would assume from the given information that the population at time t is given by an expontial decay$$
    P(t) = P_0e^{-kt} = 500e^{-kt}$$Then it is given that $$P(2) = 300 = 500e^{-2k}$$Solving as you did gives ##k=\frac {\ln {.6}}{-2}=.2554## approximately. So is equation is$$
    P(t) = 500e^{-.2554t}$$This gives ##P(1/2) = 440.04##, or approximately 60 dead hydras.

    So it is in fact an appropriate CalcII problem.
    Last edited: Nov 1, 2012
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