Es gilt $$ \text{Var}(z^{\star}) = E \left\{ [z^\star - E(z^\star)]^2 \right\} = E \left\{ \left( \frac{1}{n} \sum_{i=1}^n z_i - E \left\{ \frac{1}{n} \sum_{i=1}^n z_i \right\} \right)^2 \right\} \\ = E \left\{ \left( \frac{1}{n} \sum_{i=1}^n z_i - \frac{1}{n} \sum_{i=1}^n E \{ z_i \} \right)^2 \right\} = E \left\{ \left( \frac{1}{n} \sum_{i=1}^n (z_i - \mu) \right)^2 \right\} = \frac{1}{n^2} \sum_{i=0}^n E\{ (z_i - \mu)^2 \} \\ = \frac{1}{n^2} n \sigma^2 = \frac{\sigma^2}{n} $$
Also ist die Standardabweichung von \( z^\star \) gleich \( \frac{\sigma}{\sqrt{n} } \)