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Math Help - Finding probability that sample mean differs from true mean?

  1. #1
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    Finding probability that sample mean differs from true mean?

    I have the following problem:

    "The breaking strength of a safety belt is known to be approximately normally distributed with a true mean (mu) of 8100 pounds and a true standard deviation (sigma) of 105 pounds. If a simple random sample of 40 belts is chosen, what is the probability that the sample mean (x-bar) will differ from the true mean (mu) by more than 30 pounds (in either direction)?"

    I have an abnormally high z score, so I don't think I am doing it correctly.
    n = 40 mu = 8100 sigma = 105
    mu x-bar = 40
    sigma x-bar = sigma/mu x-bar = 105/40 = 2.625

    P(8070 < x-bar < 8130) = P((8070-40)/2.625 < z < (8130-40)/2.625) =
    P(3059.05 < z < 3081.9)

    Where have I gone wrong?
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  2. #2
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    Quote Originally Posted by oxyron View Post
    I have the following problem:

    "The breaking strength of a safety belt is known to be approximately normally distributed with a true mean (mu) of 8100 pounds and a true standard deviation (sigma) of 105 pounds. If a simple random sample of 40 belts is chosen, what is the probability that the sample mean (x-bar) will differ from the true mean (mu) by more than 30 pounds (in either direction)?"

    I have an abnormally high z score, so I don't think I am doing it correctly.
    n = 40 mu = 8100 sigma = 105
    mu x-bar = 40 Mr F says: Are you seriously trying to say that the mean of the sample mean is 40!!!

    sigma x-bar = sigma/mu x-bar = 105/40 = 2.625

    P(8070 < x-bar < 8130) = P((8070-40)/2.625 < z < (8130-40)/2.625) =
    P(3059.05 < z < 3081.9)

    Where have I gone wrong?
    \bar{X} ~ Normal \left(\mu = 8100, \sigma = \frac{105}{\sqrt{40}}\right).

    Calculate \Pr(\bar{X} < 8070) + \Pr(\bar{X} > 8130) = 2\Pr(\bar{X} > 8130).
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