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Chebyshevs theorem : find k so that at most 10%

  1. Aug 9, 2010 #1
    the test scores for a large statistics class have an unknown distribution with a mean of 70 and a standard deviation of 10

    find k so that at most 10% of the scores are more than k standard deviations above the mean.

    I'm a bit confused by the question it self.
    does the question means :
    1-1/k^2 = 0.1
    k = 1.05


    1-1/k^2 = 0.2
    k = 1.12


    1-1/k^2 = 0.8
    k = square root 5 = 2.23

    pls help, thanks in advance.
    Last edited: Aug 9, 2010
  2. jcsd
  3. Aug 9, 2010 #2
    to avoid being flamed for homework type of question, I added my own opinion towards this question. your help is very much appreciated.
  4. Aug 9, 2010 #3
    I misread your question. The minimal value for k given "at most" 10% k SDs above the mean would be [tex]1-1/k^2=0.8[/tex] so [tex]k=2.236[/tex]. This assumes your distribution is perfectly symmetrical.
    Last edited: Aug 9, 2010
  5. Aug 10, 2010 #4
    'at most' 10%.

    why take the other remaining 80%?
    it asked for 'at most' 10%.
    doesn't this mean anything not greater than 10%?
    if it asked 'at least' 10% then anything greater than 10%
  6. Aug 10, 2010 #5
    80% of the test scores are within 2.236 SD of the mean. 10% are at least 2.236 SD above the mean, 10% at least 2.236 SD below the mean. This is the minimal value of k. If you want to be sure that at most 10% are more than k SD above the mean, take any arbitrary value of k greater than 2.236. That's the way I read it anyway.
    Last edited: Aug 10, 2010
  7. Aug 11, 2010 #6
    ok thank you. I got it.
    the keyword is above the mean.
    the value of k before the last 10% of each side(20%). so 80% is in between the 20%
    is how I interpreted it as it is.
  8. Aug 11, 2010 #7
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