Step 1. Assume $E[X_i] = \mu$ and $\text{Var}(X_i) = \sigma^2 < \infty$. Define $Z_i = (X_i - \mu)/\sigma$, so $E[Z_i] = 0$ and $\text{Var}(Z_i) = 1$.

Step 2. The moment generating function of $\bar{Z} = \frac{1}{n}\sum Z_i$ is $M_{\bar{Z}}(t) = \left[M_Z(t/n)\right]^n$.

Step 3. Taylor-expand: $M_Z(s) = 1 + s^2/2 + o(s^2)$ as $s \to 0$. Thus $M_{\bar{Z}}(t) = \left[1 + \frac{t^2}{2n^2} + o(1/n^2)\right]^n \to e^{t^2/2}$.

Step 4. Since $e^{t^2/2}$ is the MGF of $N(0,1)$, by Lévy's continuity theorem, $\sqrt{n}\,\bar{Z} \xrightarrow{d} N(0,1)$.

The CLT approximation is generally valid when $n \geq 30$. For highly skewed or heavy-tailed populations, larger $n$ may be needed. The Berry–Esseen theorem quantifies the rate: $|F_n(x) - \Phi(x)| \leq \frac{C\,\rho}{\sigma^3 \sqrt{n}}$ where $\rho = E[|X-\mu|^3]$.

Unbiasedness: $E[\bar{X}] = E\left[\frac{1}{n}\sum X_i\right] = \frac{1}{n}\sum E[X_i] = \frac{n\mu}{n} = \mu$.

Variance: $\text{Var}(\bar{X}) = \frac{1}{n^2}\sum \text{Var}(X_i) = \frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n}$ by independence. Any other linear combination $\sum a_i X_i$ with $\sum a_i = 1$ has variance $\sigma^2 \sum a_i^2 \geq \sigma^2/n$ by Cauchy–Schwarz, with equality iff all $a_i = 1/n$.