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Math Help - findings mean, SD and approximate distribution

  1. #1
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    findings mean, SD and approximate distribution

    Let X1,X2,....,X81 be a random sample from a distribution (not necessarily normal) with mean \mu=59 and SD \sigma=8. Let \bar{X}=\frac{1}{81}\Sigma^{81}_{i=1}X_{i}

    a) What is the mean of \bar{X}
    b) What is the standard deviation of \bar{X}<br />
    c) What is the approximate distribution of \bar{X}

    I'm not really sure what this question is asking. I know how to find means and standard deviations, but not from what is given in this question. Thank you.
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  2. #2
    Super Member Random Variable's Avatar
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    a)  \mu = 59

    b)  \frac {\sigma}{\sqrt{n}} = \frac{8}{9}

    c)  N\Big(\mu,\frac {\sigma^{2}}{n}\Big) = N\Big(59,\frac{64}{81}\Big)
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  3. #3
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    The mean is the same but the SD is divided by \sqrt{n}? I never would have known that. What is that SD equation from? Also, how can you figure it is a normal distribution? It is probably obvious but I can't find info like that at all in my book and would like to learn.
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  4. #4
    Super Member Random Variable's Avatar
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    If  X_{1}, X_{2}, ... , X_{n} are indepedent random variables with respective means  \mu_{1}, \mu_{2}. ... , \mu_{n} and variances  \sigma^{2}_{1}, \sigma^{2}_{2}, ... , \sigma^{2}_{n}

    then the mean and variance of  Y = \sum^{n}_{i=1} a_{i} X_{i} are  \mu_{Y} = \sum^{n}_{i=1} a_{i} \mu_{i} and  \sigma^{2}_{Y} = \sum^{n}_{i=1} a_{i}^{2} \sigma^{2}_{i} respectively

    You can prove the preceding directly from the definition.


    Therefore,  \mu_{\bar{X}} = \sum^{n}_{i=1} \frac {1}{n} \mu = \frac {1}{n} (n \mu) = \mu

    and  \sigma^{2}_{\bar{x}} = \sum^{n}_{i=1} (\frac {1}{n})^{2} \sigma^{2} = \frac{1}{n^{2}} (n \sigma^{2}) = \frac {\sigma^{2}}{n}


    part(c) is the result of the central limit theorem
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