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Exponential Distribution Probability

  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data
    The life times, Y in years of a certain brand of low-grade lightbulbs follow an exponential distribution with a mean of 0.6 years. A tester makes random observations of the life times of this particular brand of lightbulbs and records them one by one as either a success if the life time exceeds 1 year, or as a failure otherwise.

    Part a) Find the probability to 3 decimal places that the first success occurs in the fifth observation.

    Part b) Find the probability to 3 decimal places of the second success occurring in the 8th observation given that the first success occurred in the 3rd observation.

    Part c) Find the probability to 2 decimal places that the first success occurs in an odd-numbered observation. That is, the first success occurs in the 1st or 3rd or 5th or 7th (and so on) observation.

    2. Relevant equations
    Z= (x-mean)/stddev

    3. The attempt at a solution

    for part a)
    Assuming that each test is independent, I needP(Y1<1, Y2<1, Y3<1, Y4<1, Y5>1) = (P(Y<1))^4 P(Y>1) by independence.
    So
    I found E(X) = 0.6 and Var(X) = 0.36 so STDEV = 0.6. Then I found the Z score to be 0.667, which corresponded to P(Y<1) = 0.7486.

    So I then did (0.7486)4 * (1 -0.7486) which gave me 0.0789, but this is incorrect. Do you see where I made a mistake?

    Am I doing it completely wrong?

    Any help is appreciated, thanks
     
  2. jcsd
  3. Mar 1, 2015 #2

    RUber

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    You calculated normal distribution and not exponential.
     
  4. Mar 1, 2015 #3
    What formula am I meant to use? I only see F(x) = 1 - e^(lamda*t) but that doesn't look like the appropriate formula for this question
     
  5. Mar 1, 2015 #4

    RUber

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    Check that formula for how lambda relates to the mean.
    I think normally you want exp(-t/mean) for exponential.
     
  6. Mar 1, 2015 #5
    I realize that lamda is just 1/mean but I am unsure what to use for t, can I just use t = 5? This seems like the probability that the first success is WITHIN the first 5 tries rather than on the 5th try like required
     
  7. Mar 1, 2015 #6

    RUber

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    T is the time of the event. You are running the test for 1 year. Find the probability of failure before one year. Success is 1-p(fail).
    You can use the rest of your logic from above. P(first success on 5th) = p(fail)^4*p(success)
     
  8. Mar 1, 2015 #7

    got it, thanks!
     
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