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Math Help - Normal Distribution

  1. #1
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    Normal Distribution

    A random sample of the duration of 50 telphone calls handled by a local telephone company had a mean of 11.6 min and a standard devation of 3.8 Minutes. Find a 95% confidence interval for the true mean duration of calls handled by the company.
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  2. #2
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    Quote Originally Posted by harry View Post
    A random sample of the duration of 50 telphone calls handled by a local telephone company had a mean of 11.6 min and a standard devation of 3.8 Minutes. Find a 95% confidence interval for the true mean duration of calls handled by the company.
    The random variable:

    t = (SampleMean - PopulationMean)/(SampleSD/sqrt(n))

    where n is the sample size, has a t distribution

    With n=50, nu the number of degrees of freedom is 49, with this number of
    degrees of freedom a 95% interval for t is ~=(-2.01, 2.01). So the 95%
    confidence interval for the populaion mean is:

    (SampleMean - 2.01*SampleSD/sqrt(n), SampleMean + 2.01*SampleSD/sqrt(n))

    or: (10.52, 12.68).

    If you used the normal approximation here the 2.01 would be replaced by 1.95
    and the interval would be (10.55, 12.65)

    RonL
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