# Given i.i.d random variable with mean $\mu$ and variance 1, $X_n = \frac{1}{n}(X_1+\cdots+X_n)$, approximate the probability by using CLT

Given i.i.d random variable with mean $\mu$ and variance 1, $\bar{X}_n = \frac{1}{n}(X_1+\cdots+X_n)$, use the CLT to approximate the following probability:

$$P(|\bar{X}_n - \mu| \ge \frac{2}{\sqrt {n}})$$

My attempt:

$$P(|\bar{X}_n - \mu| \ge \frac{2}{\sqrt {n}}$$ $$=P(\sqrt n|\bar{X}_n - \mu| \ge 2) \;\;\;\text{by multiplying both sides by \sqrt n}$$ $$=P(|\sqrt n \bar{X}_n - \sqrt n\mu| \ge 2)$$

Let $Z = \sqrt n \bar{X}_n - \sqrt n \mu$, by CLT, we know that $\sqrt n \bar{X}_n - \sqrt n \mu \rightarrow N(,)$

This is the part where I'm not sure how to figure out what $\sqrt n \bar{X}_n - \sqrt n \mu \rightarrow$ to

Is it $N(\mu, 1)$ since we are given i.i.d random variable with mean $\mu$ and variance 1 or $N(0, 1)$ by the definition of CLT?

If anyone could help me in determining how to find what $\sqrt n \bar{X}_n - \sqrt n \mu \rightarrow$ to or in general how to determine by CLT the approximation normal distribution for questions similar to this that would be really appreciated.

Suppose $X_1,\ldots,X_n$ are uncorrelated (a weaker assumption than independence) and all have the same expected value $\mu$ and the same variance $\sigma^2<\infty$ (a weaker assumption that identical distribution). Then $$\operatorname{E}\left( \frac{X_1+\cdots+X_n} {\sqrt n} - \mu \right) = 0 \text{ and } \operatorname{var}\left( \frac{X_1+\cdots+X_n} {\sqrt n} - \mu \right) = 1.$$ You don't need to know anything about the normal distribution, let alone the central limit theorem, in order to show that much.
If in addition $X_1,\ldots,X_n$ are normally distributed and independent, then so is $\dfrac{X_1+\cdots+X_n} {\sqrt n} - \mu$. And if they are not normally distributed but are independent, then the distribution of $\dfrac{X_1+\cdots+X_n} {\sqrt n} - \mu$ approaches a normal distribution as $n\to\infty$, and that normal distribution has the same expected value and the same variance.
• You meant the variance is equal to $\sigma^2$ – lcv Dec 18 '15 at 8:47
EDIT: Since last I read this, there has been a change in the value of $\mu$. So, I will let $\bar X = \bar X_n$. I think you can determine that $\mu = E[\bar X]$, and $\sigma^2 = \text{Var}[\bar X] = \frac{1}{n}$. Then \begin{align*} P\left[|\bar X - \mu| > \frac{2}{\sqrt n}\right] &= P\left[\bar X-\mu > \frac{2}{\sqrt n} \cup\bar X-\mu <\frac{-2}{\sqrt n}\right]\\ &=P\left[\frac{\bar X - \mu}{\sqrt{1/n}}>\frac{(2+\mu)-\mu}{\sqrt{1/n}\sqrt{n}}\right]+P\left[\frac{\bar X - \mu}{\sqrt{1/n}}<\frac{(-2+\mu)-\mu}{\sqrt{1/n}\sqrt{n}}\right]\tag{1}\\ &=P\left[\frac{\bar X - \mu}{\sigma}>2\right]+P\left[\frac{\bar X - \mu}{\sigma}<-2\right]\\ &=P(Z>2)+P(Z<-2)\\ &=0.04550026 \end{align*} where $(1)$ is true since the events are disjoint, and $Z$ is a standard normal distribution. The CLT comes into play since for large $n$ $$\frac{\bar X -\mu}{\sigma}\to Z.$$