Because different samples can yield different sample means, the sample mean is a random variable with a distribution.

The key question: What is the distribution of the sample mean \(\bar{X}\)?

The answer depends on two things:

• Is the original population normally distributed?
• What is the sample size n?

Quantities describing a distribution or population are called parameters:

Property 1: Expected value of the sample mean

\(E(\bar{X}) = \mu_{\bar{X}} = \mu\)

The mean of all possible sample means equals the population mean

Property 2: Variance of the sample mean

\(V(\bar{X}) = \sigma_{\bar{X}}^2 = \frac{\sigma^2}{n}\)

The variance of sample means equals the population variance divided by n

(this property holds only for a random sample)

Property 3: Standard Error

\(\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}\)

The standard deviation of the sample mean is called the "standard error"

💡 Key insight: There is an inverse relationship between sample size and variance of sample means.

Larger sample → smaller variance → means more concentrated around μ

Conclusion: As sample size increases, the sampling distribution of the mean becomes:

• more narrow (smaller variance)
• more concentrated around the population mean μ

If: we sample from a population where the variable is normally distributed with mean μ and variance σ²

Then: the sample mean is also normally distributed!

\(\bar{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right)\)

💡 Note: In this case the sample mean is normally distributed for any sample size n, even if n is small!

Theorem:

If a population has any distribution (not necessarily normal!) with mean μ and variance σ²,

then for a sufficiently large sample, the sample mean is approximately normally distributed:

\(\bar{X} \xrightarrow{n \to \infty} N\left(\mu, \frac{\sigma^2}{n}\right)\)

🎯 This is one of the most important theorems in statistics!

Rule of thumb: usually sufficient \(n \geq 30\)

But it depends on the original distribution:

• Symmetric distribution: even relatively small n (15–20) may suffice
• Asymmetric distribution: need larger n (30+)
• Highly asymmetric distribution: need very large n (50+)

\(Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}}\)

💡 Note the difference:

• For a single observation X:   \(Z = \frac{X - \mu}{\sigma}\)
• For a sample mean \(\bar{X}\):   \(Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}\)

Question: Birth weight is distributed with μ = 3.2 kg and σ = 0.5 kg.

A sample of 36 babies is taken. What is the probability the sample mean exceeds 3.35 kg?