$ \bar{X} = \frac{1}{n}\sum{X_i} $ Let the sample mean of a [[Random Variable|random variable]] be the [[Mean|mean]] of a [[Random Sample|random sample]] from that variable. Since the mean of a random sample is $n\mu$, the sample mean is the mean of the random variable $\mu$. Since the [[Variance|variance]] of a random sample is $n\sigma^2$, the variance of the sample mean is $Var(\frac{1}{n}\bar{X}) = \frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n}$. Since a [[Linear Combination|linear combination]] of [[Probabilistic Independence|independent]] [[Normal Distribution|normal distributions]] is also a normal distribution, the sample mean of a random sample for the distribution $N(\mu, \sigma^2)$ is $N(\mu, \frac{\sigma^2}{n})$.