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 $\bar{X} ≈ 0.85$.