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Thread: Probability of p(-4<X-u<4) from a list?

  1. #1
    Dec 2007

    Probability of p(-4<X-u<4) from a list?

    alright, bit of confusion, im given a decent sized list, but thanks to the TI83 ill just cut to the simple parts,

    i get a Sx of 16.244, n=28, xbar of 76.96, sample variance of 263.8875, population standard deviation of 15.951

    im asked what distribution would the mean of X bar follow (never could figure out which distribution i should use)

    lastly im asked what is the probability of P(-4≤X-u≤4) (x=xbar and u=mew)

    i guess where im really confused is the fact that its x- u, and not just a zscore or something

    Thanks in advance for any help...ill find some part of this forum where i can be helpful to return the favor.
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  2. #2
    Sep 2007
    It appears to me (assuming I'm reading your question right) that this is just an application of the central limit theorem (CLT). Basically the mean of any distribution follows a standard normal distribution if you standardize it (subtract the true mean and divide by the std) so that is the distribution you need to use.. the nice thing here is that they have already subtracted the mean for you so you don't have to do that... all you have to do is divide all three sides of the inequality by the standard deviation and use a standard normal table to obtain the probability.

    It would look like this...

    P(-4 < xbar - mu < 4) = P(-4/3.01 < (xbar - mu)/sigma/Sqrt(n) < 4/3.01)
    = P(-1.33 < Z < 1.33)

    = Phi(1.33) - Phi(-1.33) = 0.815

    (Look up Z of 1.33 in a normal table or use computer software to find the probability)

    Hope that helps!
    Last edited by jandrewross; December 7th 2007 at 09:03 AM. Reason: mistake in calculation
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