Sampling Distribution of the Sample Mean

Scope Label

Core 9758. This branch develops the distribution of the sample mean $\bar{X}$, including exact normal cases, the central limit theorem, and standardising with the standard error.

Use it with the hub Sampling and Estimation and the normal-distribution topic Special Continuous Random Variables.

From One Sample Mean to a Sampling Distribution

One observed sample gives one value of $\bar{x}$.

Before the sample is taken, however, many possible random samples could occur. Each sample may give a different sample mean. Therefore the sample mean has its own distribution.

The notation is:

• $X$ is one randomly chosen value from the population;
• $\bar{X}$ is the random variable representing the mean of a random sample;
• $\bar{x}$ is one observed value of $\bar{X}$ after sampling.

This distinction is essential. A question about $X$ is about one item. A question about $\bar{X}$ is about an average of $n$ items.

Caption: Repeated random samples produce different values of $\bar{x}$; these possible values form the sampling distribution of $\bar{X}$.

Mean and Variance of $\bar{X}$

The standard H2 formulas below assume a proper random sample whose observations can be treated as independent and identically distributed. This is automatic for a sample from an infinite population, and it is also the usual model for sampling from a finite population with replacement.

For finite sampling without replacement, the exact variance can need a finite-population adjustment. H2 questions using the formula below normally signal the standard random-sample model, so do not add an extra correction unless the question explicitly requires it.

For a random sample of size $n$ from a population with mean $\mu$ and variance ${\sigma }^2$,

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

and

$$\mathrm{Var}(\bar{X})=\frac{{\sigma }^2}{n}.$$

Hence

$$\mathrm{SD}(\bar{X})=\frac{\sigma }{\sqrt{n}}.$$

This standard deviation is called the standard error of the sample mean.

Interpret the two results separately:

• $E(\bar{X})=\mu$ says the sample mean is centred correctly;
• $\mathrm{Var}(\bar{X})={\sigma }^2/n$ says sample means become less variable as $n$ increases.

Caption: As sample size increases, the distribution of the sample mean stays centred at $\mu$ but becomes less spread out.

Distribution of $\bar{X}$: The Exact Normal Case

If the population itself is normal, then the sample mean is exactly normal.

If

$$X\sim N(\mu ,{\sigma }^2),$$

then for a random sample of size $n$,

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

This is exact for any sample size $n$.

For example, if

$$X\sim N(50,12^2)$$

and $n=36$, then

$$\bar{X}\sim N\left(50,\frac{12^2}{36}\right)=N(50,4).$$

The variance of $\bar{X}$ is $4$, so the standard deviation of $\bar{X}$ is $2$.

Distribution of $\bar{X}$: The Central Limit Theorem

If the population is not necessarily normal, the sample mean may still be approximately normal when the sample size is large.

If the population has mean $\mu$ and variance ${\sigma }^2$, then for large $n$,

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

This is the central limit theorem.

In H2 work, the usual guide is

$$n\ge 30.$$

The important distinction is:

• $X$ itself may be skewed, discrete, or non-normal;
• $\bar{X}$ may still be approximately normal when $n$ is large.

Caption: The central limit theorem connects many population shapes to an approximately normal distribution for the sample mean when $n$ is large.

Choosing the Distribution of $\bar{X}$

Use this decision process:

1. Is the population normal?
2. If yes, use the exact result $\bar{X}\sim N(\mu ,{\sigma }^2/n)$.
3. If no, is $n$ large enough?
4. If yes, use the approximation $\bar{X}\approx N(\mu ,{\sigma }^2/n)$.
5. If no, the usual normal model for $\bar{X}$ is not justified from the given information.

Caption: To choose the distribution of $\bar{X}$, first ask whether the population is normal; if not, check whether the sample size is large enough for the central limit theorem.

Standardising $\bar{X}$

Once the distribution of $\bar{X}$ is known or approximated, probability statements can be standardised.

For one observation,

$$Z=\frac{X-\mu }{\sigma }.$$

For a sample mean,

$$Z=\frac{\bar{X}-\mu }{\sigma /\sqrt{n}}.$$

The denominator is the standard error, not the population standard deviation. This is the most common procedural error in this branch.

Caption: Standardising one observation uses $\sigma$; standardising a sample mean uses $\sigma /\sqrt{n}$.

Worked Example 1: Exact Normal Distribution of $\bar{X}$

The diameter of a metal rod is normally distributed with mean $1.00$ cm and standard deviation $0.01$ cm. A random sample of $4$ rods is selected.

Find

$$P(\bar{X}>0.99).$$

Let $X$ be the diameter of one rod. Then

$$X\sim N(1.00,0.01^2).$$

Since the population is normal,

$$\bar{X}\sim N\left(1.00,\frac{0.01^2}{4}\right).$$

The standard deviation of $\bar{X}$ is

$$\frac{0.01}{\sqrt{4}}=\mathrm{0.005.}$$

Therefore

$$P(\bar{X}>0.99)=P\left(Z>\frac{0.99-1.00}{0.005}\right)=P(Z>-2).$$

Hence

$$P(\bar{X}>0.99)\approx \mathrm{0.977.}$$

Worked Example 2: Central Limit Theorem

Suppose

$$X\sim B(10,0.6).$$

A random sample of size $40$ is taken. Find approximately

$$P(\bar{X}>5.5).$$

First find the population mean and variance of $X$:

$$E(X)=10(0.6)=6,$$

and

$$\mathrm{Var}(X)=10(0.6)(0.4)=\mathrm{2.4.}$$

Since $n=40$ is large, by the central limit theorem,

$$\bar{X}\approx N\left(6,\frac{2.4}{40}\right).$$

The standard deviation of $\bar{X}$ is

$$\sqrt{\frac{2.4}{40}}=\sqrt{0.06}.$$

Therefore

$$P(\bar{X}>5.5)=P\left(Z>\frac{5.5-6}{\sqrt{0.06}}\right).$$

Since

$$\frac{5.5-6}{\sqrt{0.06}}\approx -2.041,$$

we get

$$P(\bar{X}>5.5)=P(Z>-2.041)\approx \mathrm{0.979.}$$

No continuity correction is needed here because the central limit theorem is being applied to the sample mean $\bar{X}$, not directly to a discrete count.

Worked Example 3: Symmetric Interval for $\bar{X}$

Suppose $X$ has mean $80$ and standard deviation $15$. A random sample of size $64$ is taken, and the central limit theorem is applicable.

Find

$$P(77<\bar{X}<83).$$

The sample mean has approximate distribution

$$\bar{X}\approx N\left(80,\frac{15^2}{64}\right).$$

The standard error is

$$\frac{15}{\sqrt{64}}=\mathrm{1.875.}$$

Therefore

$$P(77<\bar{X}<83)=P\left(\frac{77-80}{1.875}<Z<\frac{83-80}{1.875}\right).$$

So

$$P(77<\bar{X}<83)=P(-1.6<Z<1.6)\approx \mathrm{0.890.}$$

Link to Hypothesis Testing

Hypothesis testing uses this branch directly. A test for a population mean assumes a value of $\mu$ under $H_0$, uses the distribution of $\bar{X}$ under that assumption, and checks whether the observed $\bar{x}$ is unusually extreme.

So the sampling distribution is not just another probability calculation. It is the foundation of inference about a population mean.

Common Pitfalls

• Treating $\bar{X}$ and $\bar{x}$ as the same object.
• Using $\sigma$ instead of $\sigma /\sqrt{n}$ when standardising $\bar{X}$.
• Forgetting that the exact normal result requires the population to be normal.
• Applying the central limit theorem when $n$ is small and no normal population is given.
• Thinking the central limit theorem makes $X$ normal. It applies to $\bar{X}$.
• Adding a continuity correction when the question is about the sample mean rather than a discrete count.

Revision Checklist

• Can you explain why $\bar{X}$ is random before sampling?
• Can you state and interpret $E(\bar{X})$?
• Can you state and interpret $\mathrm{Var}(\bar{X})$?
• Can you identify the standard error $\sigma /\sqrt{n}$?
• Can you decide whether the distribution of $\bar{X}$ is exact normal or approximate normal?
• Can you standardise probability statements involving $\bar{X}$ correctly?
• Can you explain how this branch prepares for hypothesis testing?