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Old June 30th, 2009, 10:28 AM
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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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