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Help with use of Chebyshev's inequality and sample size

  1. Mar 14, 2014 #1
    1. The problem statement, all variables and given/known data

    4307a6c25e65fc6dc596f937b6482d60.png

    2. Relevant equations

    P (|Y - μ| < kσ) ≥ 1 - Var(Y)/(k2σ2) = 1 - 1/k2

    ??

    3. The attempt at a solution

    using the equation above

    1 - 1/k2 = .9

    .1 = 1/k2

    k2 = 10

    k = √10 = 3.162

    k = number of standard deviations. After this I don't know where to go.

    I don't have a solid understanding of Chebyshev's inequality either.
     
  2. jcsd
  3. Mar 14, 2014 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    First figure out what probability you need to find, then worry about how to find it using Chebyshev or some other method. So, if your measurements are ##M_1, M_2, \ldots, M_n##, what is their average, ##\bar{M}##? In terms of ##c## and ##U_1, U_2, \ldots, U_n##, what would be your formula for ##\bar{M}##? What would be the mean and variance of ##\bar{M}## (expressed in terms of ##c, \:n## and other given quantities)? Now how would you express the event "the average is within half a degree of ##c##"? At this point you are ready to apply some probability!
     
  4. Mar 14, 2014 #3

    I would say that the average ##\bar{M}## ##=## (##n c## + ##U_1, U_2, \ldots, U_n##) / ##n## since you are finding ##c## ##n## times and adding that to all the ##U_i## from each sample. Then to average it, you'd need to divide it by ##n## of course.

    If it is half a degree of ##c##, would that mean that ##|\bar{M}## - ##c| >= .5?##
     
  5. Mar 14, 2014 #4
    This would make c the mean, yes? Because it is the expected value/ the value you expect to be correct. Would that make ##U_n## the variance then? But if so, how will we find the variance if ##U_n## is not given? We only know that ##Var(U_n) = 3##

    But given what I know now, would this mean then that ##P (|\bar{M} - c| >= .5) = 1 -P (|\bar{M} - c| < .5) = 1 - σ^2 / nε^2## where ε = .5?
     
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