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Tsebyshev's probability

  1. Sep 16, 2016 #1
    1. The problem statement, all variables and given/known data
    The average lifetime of a product ##T=7.5## (years). The variance of the lifetime ##\sigma^{2} = 0.41##.

    Using Tsebyshev's inequality, determine the lower bound for the probability, that the product lasts at least 5 years.

    2. Relevant equations

    Tsebyshev's inequality:
    \begin{equation}
    P(|X-\mu | \geq t) \leq \frac{\sigma^{2}}{t^2}
    \iff
    P(|X-\mu | < t) \geq 1 - \frac{\sigma^{2}}{t^2}
    \end{equation}
    where ##\mu## is the expected value, ##\sigma^2## is the variance and ##t## is the deviation from the expected value.

    3. The attempt at a solution

    We want
    \begin{align*}
    P(T \geq 5)
    &\geq P(5 \leq T \leq 10)\\
    &= P(|T-7.5| \leq 2.5) &|t=2.5\\
    &\geq 1 - \frac{\sigma^2}{t^2} &| Tsebyshev\\
    &= 1-\frac{0.41^2}{2.5^2}\\
    &\approx 0.973
    \end{align*}
    This is apparently not the correct answer, and I'm not sure what I'm doing wrong. We obviously want the deviation from the mean to be less than 2.5; otherwise ##|T-\mu|## would produce a value that is below 5.
     
  2. jcsd
  3. Sep 16, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    Edit: Why did you square the 0.41?

    Some manual tuning: The worst case is some fraction p failing "just before" 5 years and all other machines failing after X years, X a bit larger than 7.5

    The condition for the mean is then 5p+(1-p)*X = 7.5.
    The condition for the variance is p*2.5^2 + (1-p)*(X-7.5)^2 = 0.41

    Solving gives p=0.062 and X=7.66.
     
    Last edited: Sep 16, 2016
  4. Sep 16, 2016 #3

    micromass

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    You can use it for both. One gives a lower bound, the other gives an upper bound.
     
  5. Sep 16, 2016 #4

    micromass

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    You are given ##\sigma^2 = 0.41##. But later on, you write ##0.41^2## at the end of your computation. There is no need for that square.
     
  6. Sep 16, 2016 #5
    So what I could do is calculate ##P(T<5)## and go from there?

    \begin{align*}
    P(T<5)
    &\leq P(T<5 \text{ and } T>10)\\
    &= P(|T-\mu| >= 2.5)\\
    &\leq \frac{\sigma^2}{2.5^2}\\
    &=\frac{0.41^2}{2.5^2}\\
    &= 0.026896
    \end{align*}
    But then I'm stuck with the shaded area shaded blue in this picture:
    S3_4.png
    and I need the one in the middle. At least that's my understanding.
     
  7. Sep 16, 2016 #6
    Oh. My. Gauss!

    Thank you. How typical of me...
     
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