1
$\begingroup$

Possible Duplicate:
A question about sampling distribution

We defind a variance of a random variable $X$ as $E(X-X_{\text{mean}})^2$=$E(X^2)-(E(X))^2$ and the mean = $E(X)$. However, for the case of sampling , Why would find out the sample mean as $\frac{X_1+\cdots+X_n}{n}$ for the case $X_1, \dots,X_n$ are iid and the sample variance are $\frac{\sum_{i=1}^{n} (X_i-X_{\text{mean}})^2}{n-1}$.

  • 0
    I have inserted a power of 2 in your sample variance formula2012-01-19
  • 0
    what i mean is like why the sample mean = $\frac{X_1+\cdots+X_n}{n}$2012-01-19

2 Answers 2