Answer: If $X_1, X_2, \ldots, X_n$ is a sequence of iid random variables with a finite mean $\mu$ and finite non-zero variance $\sigma^2$, then the distribution of $\frac{\hat{\mu} - \mu}{\sqrt{n}}$ tends to a standard normal distribution as $n \to \infty$.

The Central Limit Theorem (CLT) states that if $X_1, X_2, \ldots, X_n$ is a sequence of iid random variables with a finite mean $\mu$ and finite non-zero variance $\sigma^2$, then the distribution of $\frac{\hat{\mu} - \mu}{\sigma / \sqrt{n}}$ tends to a standard normal distribution as $n \to \infty$. Put simply: $$\frac{\hat{\mu} - \mu}{\sigma / \sqrt{n}} \to N(0, 1)$$ Where $\hat{\mu} = \bar{X}$ = Sample Mean. **Analysis of options:** - **Option A**: Correct - This accurately describes the CLT with the proper denominator ($\sqrt{n}$) and limit as $n \to \infty$ - **Option B**: Incorrect - The limit is $n \to 0$ instead of $n \to \infty$ - **Option C**: Incorrect - The denominator is $\sigma\sqrt{n}$ instead of $\sqrt{n}$ and the limit is $n \to 1$ instead of $n \to \infty$ - **Option D**: Incorrect - Since option A is correct **Key points:** 1. The CLT requires independent and identically distributed (iid) random variables 2. The variables must have finite mean ($\mu$) and finite non-zero variance ($\sigma^2$) 3. The standardized sample mean converges to a standard normal distribution as sample size approaches infinity 4. The standardization is: $\frac{\bar{X} - \mu}{\sigma/\sqrt{n}} \sim N(0,1)$ as $n \to \infty$

Author: Nikitesh Somanthe