Sampling Distribution When the Population Is Normal

When Is the Sample Mean Exactly Normal?

In “The 10% Condition for Sample Means,” we checked when observations from a sample selected without replacement can be treated as approximately independent for a standard-deviation calculation. Now we focus on the shape of the sampling distribution. If the population itself has a Normal distribution, then the sample mean is exactly Normally distributed for any positive sample size, provided the observations are independent.

This is different from relying on the Central Limit Theorem. The Central Limit Theorem explains why sample means can be approximately Normal for sufficiently large samples from many non-Normal populations. When the population is Normal, no large-sample requirement is needed for exact Normality: even a sample of size 2 has a Normal sampling distribution under the independence model.

We can write this as \(\bar{X}\sim N(\mu,\sigma/\sqrt{n})\), where the second parameter shown is the standard deviation. The sampling distribution is centered at the population mean, and its spread is the standard deviation of the sample mean, \(\sigma_{\bar{x}}=\sigma/\sqrt{n}\), as developed in “Standard Deviation of the Sample Mean.”

Why is the shape exactly Normal? Each individual measurement is Normally distributed. For independent observations from a Normal population, their sum is also Normally distributed. The sample mean is that sum divided by \(n\), so it is Normal as well. Its center is \(\mu\), and its standard deviation is \(\sigma/\sqrt{n}\). This result does not say that every sample mean equals \(\mu\); it describes the distribution of sample means across repeated samples.

The conditions matter. The population model must be Normal, and the observations must be independent. A random sample from a Normal population is commonly modeled as independent when the sampling design supports that assumption. If a sample is drawn without replacement from a finite population, observations are not truly independent. As explained in “The 10% Condition for Sample Means,” meeting the 10% condition can justify treating them as approximately independent for calculations; it does not make the sampling distribution exactly Normal by itself.

Using the Sampling Distribution

Once the sampling distribution is identified, a probability question about \(\bar{x}\) becomes a Normal probability calculation. First find the sampling distribution’s mean and standard deviation. Then standardize the requested sample-mean values or use a calculator’s normalcdf function with those parameters.

A common source of error is using \(\sigma\) as the standard deviation of \(\bar{x}\). The standard deviation \(\sigma\) describes individual observations; the standard deviation of their sample mean is smaller, \(\sigma/\sqrt{n}\). Also, use the sample size \(n\) in this formula, not the population size.

Exact Normality Does Not Require a Large Sample

The word exactly distinguishes this result from an approximation. If the population is Normal and observations are independent, the sampling distribution of \(\bar{x}\) is Normal whether \(n=2\), \(n=9\), or \(n=100\). The sample size still matters: increasing \(n\) reduces the standard deviation \(\sigma/\sqrt{n}\), so sample means cluster more closely around \(\mu\). But a large sample is not what makes the shape Normal in this setting.

By contrast, if the population is not Normal, a small-sample distribution of \(\bar{x}\) need not be Normal. The Central Limit Theorem, covered in the next tutorial, gives a reason that the sampling distribution often becomes approximately Normal for larger samples. Do not use that later result to justify exact Normality here: exactness comes from the Normal population model and independence.

How Sample Size Changes the Spread

The formula \(\sigma_{\bar{x}}=\sigma/\sqrt{n}\) shows that sample size affects the spread of sample means, not the population mean. For example, multiplying \(n\) by 4 divides the standard deviation of \(\bar{x}\) by 2. When the population is Normal, the distribution remains exactly Normal as this spread changes.

Be careful about what a probability describes. A probability such as \(P(9.6<\bar{x}<10.4)\) is about the mean of a sample of a specified size. It is not the probability that one individual measurement falls in that interval. The individual distribution and the sampling distribution generally have different standard deviations.

Common Mistakes and AP Exam Tip

• Requiring a large sample: A large sample is not required for exact Normality when the population itself is Normal and observations are independent. A small \(n\) is acceptable under those conditions.
• Confusing exact and approximate Normality: The Normal population model gives an exactly Normal sampling distribution. A large-sample approximation for a non-Normal population is a different justification.
• Using the individual standard deviation: For sample means, use \(\sigma/\sqrt{n}\), not \(\sigma\), as the standard deviation of the sampling distribution.
• Ignoring the sampling design: A Normal population alone does not establish independent observations. For sampling without replacement, refer to “The 10% Condition for Sample Means” and describe the result as an independence approximation when appropriate.
• Leaving out the context or units: State whether the probability concerns an individual measurement or a sample mean, and include units such as inches, minutes, or hours.

For full-credit communication, identify the Normal population model and the independence assumption, state the sampling distribution of \(\bar{x}\) with its mean and standard deviation, show the probability calculation, and give a conclusion in context. Do not claim that the sample size must be at least 30 when the population is already modeled as Normal.