This section will comprehensively explore how the sample mean behaves, especially when dealing with normal distributions or large sample sizes, and its implications in statistical analysis.

## Introduction

The sample mean $(\bar{X})$ is the average of a set of data from a population. It's key for understanding how averages of samples distribute, particularly with normal distributions or large samples.

## Normal Distribution for Sample Mean

When samples come from a normally distributed population, $\bar{X}$ also has a normal distribution.

### Key Points:

**Mean:**Same as the population mean $(\mu)$.**Variance:**Population variance divided by sample size $(\sigma^2/n)$.**Standard Deviation:**$\sigma/\sqrt{n}$.

## Central Limit Theorem (CLT)

**CLT **shows that $\bar{X}$ becomes normally distributed as sample size grows, regardless of the population's distribution.

### Understanding CLT:

**Large Samples:**Typically, 30 or more.**Distribution:**$\bar{X}$ becomes normal.**Mean and Variance:**Mean is $\mu$, and variance is $\sigma^2/n$.

## Examples

### Example 1: Normal Population

A population is normally distributed with a mean $( \mu )$ of 50 and a standard deviation $( \sigma )$ of 10. We are to find the distribution of the sample mean $( \bar{X} )$ for a sample size $( n )$ of 25.

#### Solution:

**Population Parameters:**$\mu = 50$, $\sigma = 10$.**Sample Size:**$n = 25$.**Calculating the Standard Deviation of ( \bar{X} )**:- Formula: $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$
- Calculation: $\sigma_{\bar{X}} = \frac{10}{\sqrt{25}} = \frac{10}{5} = 2$.

**Result:**- The sample mean $\bar{X}$ follows a normal distribution.
- Mean of $\bar{X} = 50$ (same as the population mean).
- Standard Deviation of $\bar{X} = 2$.

### Example 2: Applying the CLT

Given a population with an unspecified distribution, a mean of 30, and a standard deviation of 6, we need to determine the characteristics of the sample mean's distribution for a sample size of 50.

**Solution:**

**Population Parameters:**$\mu = 30$, $\sigma = 6$.**Sample Size:**$n = 50$.**Applying the CLT:**Given the large sample size, the CLT suggests that $\bar{X}$ will be approximately normally distributed.**Calculating the Standard Deviation of ( \bar{X} ):**- Formula: $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$
- Calculation: $\sigma_{\bar{X}} = \frac{6}{\sqrt{50}} \approx 0.85$ (rounded off to two decimal places).

**Result:**- The sample mean $\bar{X}$ is approximately normally distributed.
- Mean of $\bar{X}$ $= 30$ (same as the population mean).
- Standard Deviation of $\bar{X} ≈ 0.85$.

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.