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Homework Help: Central Limit Theorem and probability

  1. Sep 10, 2009 #1
    1. The problem statement, all variables and given/known data

    What is the probability that the average of 150 random points from the interval (0,1) is within .02 of the midpoint of the interval?

    2. Relevant equations

    3. The attempt at a solution

    I need to determine P(.48<((X1....X150)/150)<.52). I think I need to compute the variance and ultimately work out the two limits to integrate using the standard integral for a normal distribution. However, I have no idea how to accomplish this.
  2. jcsd
  3. Sep 10, 2009 #2


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    Homework Helper

    you shoudl be able to find an initial mean & variance for your unifromly distributed single random variable, start from the definition of mean & variance

    use this single random variable mean & variance & the number of samples to compute a mean and variance for your average, assumming a normal distribution under CLT
  4. Sep 12, 2009 #3
    It looks like the mean and variance of a uniformly distributed single random variable over the interval of (0,1) is .5 and 1/12 respectively. However, I am still stumped on how to use this information to compute the required probability. Any further suggestions on how to get
  5. Sep 12, 2009 #4
    Mean of sample = mean of population

    Standard deviation of sample = standard deviation of population divided by square root of n

    So, mean of sample = 0.5
    S.d. of sample = (1/12)/square root of 150 = 0.006804

    So, perform the standardization on P(.48<X bar)<.52) and you will get the probability. ;) Hope that helps. ;)
  6. Sep 14, 2009 #5
    Thanks, but I am stilll stuck. The limits I am coming up with are +_.006 which don't result in the book answer which is .6046.
  7. Sep 14, 2009 #6


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    can you be a bit clearer with what you have done so far? will help to understand where you may be going wrong
  8. Sep 15, 2009 #7
    In computing P(.48<Xbar<.52), I come up with,
    +-.02/(sqrt(150)times(1/((sqrt(12))=+-.00565; I don't think that these are the correct limits for the integral.
  9. Sep 17, 2009 #8
    I finally figured it out, guys. The Z value is between +-.848 which, assuming a normal distribution, gives P(.48<Xbar<.52) =.6046, which I think is correct. Thanks for the help!
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