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Find percentage of a sample given the mean and standard deviation

  1. Aug 31, 2013 #1
    1. The problem statement, all variables and given/known data
    People drink on average, 78L of beer per year with a standard deviation of 25L. at least what percentage of people would drink between 20 and 136L per year?

    3. The attempt at a solution

    I found the z-scores of both and they came up to be +/- 2.32. Compared these values to the z-table and got P(Z<2.32)=0.0102 and P(Z<2.32)=0.9898. Then i found the percentage as followed... 0.9898-0.0102 which is around 98%. Which is wrong. Is my assumption that the graph is symmetrical correct?
     
    Last edited: Aug 31, 2013
  2. jcsd
  3. Aug 31, 2013 #2

    Curious3141

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    Sure, this is implicitly given to be a normal distribution, which is symmetrical.

    First of all, your left-tail probability makes no sense as you wrote it. Shouldn't it be ##p(z<-2.32) = 0.0102##?

    When you subtract those two, the answer isn't "around 97%", it's a lot closer to 98%. Did you try being a lot more precise with your answer?

    Because of the symmetry, you could just as easily have taken ##p(-2.32<z<2.32) = 1 - 2\times 0.0102##.
     
  4. Aug 31, 2013 #3
    You're correct. The left tail is wrong as you stated. But even with more precision in my numbers, it's still a fair bit off from the actual answer which is 81.40%
     
  5. Aug 31, 2013 #4

    Curious3141

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  6. Aug 31, 2013 #5
    I wouldn't implicitly consider it normal unless stated. Anyway, one can exploit some of available probabilistic inequalities. Since the Chebyshev's inequality is usually taught, try this one, maiad. Most likely, you will succeed to get your "correct" value :-)
     
  7. Aug 31, 2013 #6

    Curious3141

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    That actually does work out to the expected answer. And I think you're quite correct - considering the phrase "at least" in the question, they aren't expecting any assumption about the underlying distribution.

    Maiad - please take Camillio's advice, Chebyshev's inequality seems to be what's expected.
     
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