The sample mean \(\overline{X}_n\) is defined as the average of the random variables \(X_1, X_2, ..., X_n\). Since each \(X_i\) is an i.i.d. standard normal random variable, we can calculate the mean and variance of \(\overline{X}_n\) using properties of the normal distribution.
Mean of \(\overline{X}_n\):
The mean of \(\overline{X}_n\) is equal to the mean of the individual random variables \(X_i\), which is 0. This can be seen from the linearity of expectations:
\(\mathbb{E}[\overline{X}_n] = \frac{1}{n}\left(\mathbb{E}[X_1] + \mathbb{E}[X_2] + ... + \mathbb{E}[X_n]\right) = \frac{1}{n} \cdot n \cdot \mathbb{E}[X_1] = \mathbb{E}[X_1] = 0\).
Variance of \(\overline{X}_n\):
The variance of \(\overline{X}_n\) can be calculated using the fact that the variance of the sum of independent random variables is equal to the sum of their variances. Since each \(X_i\) is i.i.d. with variance 1, we have:
\(\text{Var}[\overline{X}_n] = \text{Var}\left(\frac{1}{n}\left(X_1 + X_2 + ... + X_n\right)\right) = \frac{1}{n^2}\left(\text{Var}[X_1] + \text{Var}[X_2] + ... + \text{Var}[X_n]\right) = \frac{1}{n^2} \cdot n \cdot \text{Var}[X_1] = \frac{\text{Var}[X_1]}{n} = \frac{1}{n}\).
Therefore, the mean of \(\overline{X}_n\) is 0 and the variance of \(\overline{X}_n\) is \(\frac{1}{n}\).