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Math Help - Normal distribution

  1. #1
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    Normal distribution

    In an analysis of healthcare data, ages have been rounded to the nearest multiple of 5 years. The difference between the true age and the rounded age is assumed to be uniformly distributed on the interval from −2.5 years to 2.5 years. The healthcare data are based on a random sample of 48 people.

    What is the approximate probability that the mean of the rounded ages is within

    0.25 years of the mean of the true ages?


    Here's my solution:

    \text{Looking for: }Pr[-2.5 \leq\bar{x}\leq 2.5]

    E[X] = \frac{2.5+(-2.5)}{2}=0

    var[X]=\frac{(2.5-(-2.5))^2}{12}=\frac{25}{12}

    E[\bar{x}]=0*48

    var[\bar{x}]=\frac{48*25}{2}=600

    \sigma_{\bar{x}}=\frac{600}{\sqrt{48}}





    The solution in the book as \sigma_{\bar{x}} as \frac{\frac{25}{12}}{\sqrt{48}}

    I don't understand why
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  2. #2
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    Re: Normal distribution

    Quote Originally Posted by downthesun01 View Post
    In an analysis of healthcare data, ages have been rounded to the nearest multiple of 5 years. The difference between the true age and the rounded age is assumed to be uniformly distributed on the interval from −2.5 years to 2.5 years. The healthcare data are based on a random sample of 48 people.

    What is the approximate probability that the mean of the rounded ages is within

    0.25 years of the mean of the true ages?


    Here's my solution:

    \text{Looking for: }Pr[-2.5 \leq\bar{x}\leq 2.5]

    E[X] = \frac{2.5+(-2.5)}{2}=0

    var[X]=\frac{(2.5-(-2.5))^2}{12}=\frac{25}{12}

    E[\bar{x}]=0*48

    var[\bar{x}]=\frac{48*25}{2}=600

    How did you get this? What happened to the "12" in the denominator before?

    \sigma_{\bar{x}}=\frac{600}{\sqrt{48}}



    The solution in the book as \sigma_{\bar{x}} as \frac{\frac{25}{12}}{\sqrt{48}}

    I don't understand why
    Thanks from downthesun01
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  3. #3
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    Re: Normal distribution

    Sorry, it should be

    \frac{48*25}{12}=100

    To find the the standard deviation of \bar{x} don't I have to multiply Var[X] by n? Then, take the square root of that, and divide it by the square root of n?
    Last edited by downthesun01; September 19th 2013 at 04:30 PM.
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  4. #4
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    Re: Normal distribution

    I figured out my mistake. Thanks
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