Sketching the Sampling Distribution of x-bar

What a Sampling-Distribution Sketch Shows

In “Sampling Distribution of a Sample Mean Defined,” we described the sampling distribution of \(\bar{x}\) as the distribution of sample means from all possible samples of a fixed size. A sketch is a compact way to show three features of that distribution: its center, its spread, and its shape. These describe sample means across repeated samples—not the individual observations in one sample.

For a random sample from a population with mean \(\mu\) and standard deviation \(\sigma\), the sampling distribution is centered at \(\mu\). Its standard deviation, often called the standard error of \(\bar{x}\), is \(\sigma/\sqrt{n}\) when observations are independent. As covered in “The 10% Condition for Sample Means,” for a simple random sample without replacement from a finite population, check the 10% condition when using this formula.

Formula: For independent observations, the center and standard deviation of the sampling distribution of \(\bar{x}\) are:

\mu_{\bar{x}}=\mu \qquad\text{and}\qquad \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}

The center and spread use the same measurement units as the original variable. For example, if individual measurements are in minutes, then both \(\mu_{\bar{x}}\) and \(\sigma_{\bar{x}}\) are in minutes. The spread of sample means is not the same as the spread of individual observations: for \(n>1\), \(\sigma/\sqrt{n}\) is smaller than \(\sigma\).

The numbers \(\mu\), \(\sigma\), and \(n\) determine the center and spread, but they do not always determine the shape. As covered in “Sampling Distribution When the Population Is Normal,” if the population is Normal, then the sampling distribution of \(\bar{x}\) is exactly Normal for any sample size. Otherwise, use the population’s shape and the CLT reasoning from “Central Limit Theorem Explained” and “Is n = 30 Large Enough for the CLT” to decide whether an approximately Normal shape is reasonable.

A Reliable Way to Build the Sketch

Start with the horizontal axis and label it with the quantity \(\bar{x}\), including its units. Mark the center at \(\mu\), then calculate \(\sigma_{\bar{x}}\). To make the spread visible, mark locations one and two standard deviations to either side of the center. These are useful reference points for drawing, not additional parameters of the distribution.

Sketching steps:

Label the horizontal axis with \(\bar{x}\) and its units.
Mark the center at \(\mu_{\bar{x}}=\mu\).
Calculate the standard deviation of sample means, \(\sigma_{\bar{x}}=\sigma/\sqrt{n}\), and label it in the same units.
Choose a shape: Normal if exact Normality is established; approximately Normal if the CLT provides reasonable support; otherwise, do not draw an unjustified bell curve.
If you draw a Normal curve, mark \(\mu-\sigma_{\bar{x}}\), \(\mu+\sigma_{\bar{x}}\), \(\mu-2\sigma_{\bar{x}}\), and \(\mu+2\sigma_{\bar{x}}\) when useful.

A bell-shaped sketch should be centered over \(\mu\), with most of its width described by \(\sigma_{\bar{x}}\). If the sampling distribution is not known to be Normal or approximately Normal, still label its center and spread, but show the shape cautiously and explain what is known. Do not force every sampling distribution into a bell curve simply because you can calculate its mean and standard deviation.

Worked Example: An Exact Normal Sampling Distribution

A device’s operating lifetime has a Normal population distribution with mean \(\mu=12\) hours and standard deviation \(\sigma=4\) hours. A random sample of \(n=16\) devices is selected independently. Describe and label a sketch of the sampling distribution of \(\bar{x}\).

State. The variable of interest is the sample mean lifetime, \(\bar{x}\), measured in hours.

Plan. Find the sampling distribution’s center and standard deviation. Since the population is Normal and the observations are independent, the sampling distribution of \(\bar{x}\) is exactly Normal, even though the sample size is 16.

Do. The center is the population mean:

\mu_{\bar{x}}=\mu=12\text{ hours}

The standard deviation of the sample means is:

\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} =\frac{4}{\sqrt{16}} =\frac{4}{4} =1\text{ hour}

As a check, the variance of \(\bar{x}\) is \(\sigma^2/n=4^2/16=16/16=1\) square hour, so its standard deviation is \(\sqrt{1}=1\) hour. Useful axis labels are one standard deviation below and above the center, \(12-1=11\) and \(12+1=13\), and two standard deviations below and above, \(12-2=10\) and \(12+2=14\).

Conclude. Draw an exactly Normal curve centered at 12 hours. Label the horizontal axis in hours, place \(\bar{x}=12\) at the center, and note that the standard deviation of the sample means is 1 hour. The reference values 10, 11, 12, 13, and 14 hours make the curve’s center and spread clear.

Choose the Shape From the Population and Sample Size

For a population known to be Normal, the shape decision is direct: the sampling distribution is exactly Normal. When the population is not Normal, the CLT gives a basis for an approximately Normal sketch as the sample size grows. How large a sample needs to be depends on the population shape. A roughly symmetric population without outliers is more favorable than a strongly skewed population with extreme values.

A careful sketch distinguishes “exactly Normal” from “approximately Normal.” Use “exactly” only when the population is Normal and the sampling assumptions hold. Use “approximately” when the population is not Normal but the CLT and information about the population support a Normal model. If the sample is small and the population is skewed, the shape may remain skewed; the information may not support a bell-shaped sketch.

A curve’s center and spread do not specify its shape. Many different distributions can share the same mean and standard deviation. So if a question gives only \(\mu\), \(\sigma\), and \(n\), you can calculate and label the center and spread, but you need information about the population or sample size relative to its shape before claiming that the sampling distribution is Normal.

Worked Example: A Reasonable Normal Approximation

A city’s daily water use per household comes from a moderately right-skewed population with mean \(\mu=42\) gallons and standard deviation \(\sigma=24\) gallons. Suppose a random sample of \(n=64\) households is selected independently, and the population has no extremely influential outliers. Describe a reasonable sketch of the sampling distribution of \(\bar{x}\).

State. We are sketching the distribution of sample mean daily water use, \(\bar{x}\), in gallons.

Plan. Calculate its center and standard deviation. Then judge the shape using the population description and sample size. The population is right-skewed, so the sample mean is not automatically Normal; the fairly large sample gives the CLT a reasonable basis for an approximate Normal sketch in this stated setting.

Do. The center is:

\mu_{\bar{x}}=\mu=42\text{ gallons}

The standard deviation of sample means is:

\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} =\frac{24}{\sqrt{64}} =\frac{24}{8} =3\text{ gallons}

Checking through the variance gives \(\sigma^2/n=24^2/64=576/64=9\) square gallons, and \(\sqrt{9}=3\) gallons. One- and two-standard-deviation reference values are \(42\pm3\), or 39 and 45 gallons, and \(42\pm6\), or 36 and 48 gallons.

Conclude. Sketch an approximately Normal curve centered at 42 gallons, with standard deviation 3 gallons. Label 39, 42, and 45 gallons at the center and one standard deviation away; 36 and 48 gallons are useful two-standard-deviation reference points. Describe the shape as approximately Normal, not exactly Normal, because the population itself is right-skewed.

When the Sketch Should Not Be a Bell Curve

If the population is skewed and the sample size is small, do not assume the sampling distribution has already become approximately Normal. Its center and standard deviation can still be calculated under the appropriate sampling assumptions. But those two facts do not justify drawing a symmetric bell curve. When the population is strongly skewed, the sampling distribution for a small sample may also be skewed; state that cautiously rather than claiming an exact shape you have not been given.

This is a useful distinction in written answers: “The sampling distribution is centered at 18 and has standard deviation about 4.47” is a statement about center and spread. “The sampling distribution is Normal” is a separate claim about shape, and it needs separate justification.

Worked Example: Small Samples From a Skewed Population

A population of individual service times is strongly right-skewed. Its mean is \(\mu=18\) minutes and its standard deviation is \(\sigma=10\) minutes. A random sample of \(n=5\) independent service times is selected. Describe the center, spread, and shape to show in a sketch.

State. The variable is the sample mean service time, \(\bar{x}\), in minutes.

Plan. Calculate the center and standard deviation using the sampling-distribution formulas. For shape, consider that the population is strongly right-skewed and the sample size is small. These facts do not justify a Normal approximation.

Do. The center is:

\mu_{\bar{x}}=\mu=18\text{ minutes}

The standard deviation of sample means is:

\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} =\frac{10}{\sqrt{5}} \approx4.4721\text{ minutes}

As a check, the variance is \(\sigma^2/n=10^2/5=100/5=20\) square minutes, and \(\sqrt{20}\approx4.4721\) minutes. To place reference marks, one standard deviation from 18 is approximately \(18-4.4721=13.53\) and \(18+4.4721=22.47\) minutes. Two standard deviations from 18 are approximately \(18-8.9442=9.06\) and \(18+8.9442=26.94\) minutes. These are scale markers, not a claim that the distribution is bell-shaped.

Conclude. Mark the center at 18 minutes and label the standard deviation of the sample means as approximately 4.47 minutes. Because \(n=5\) is small and the population is strongly right-skewed, do not draw a symmetric Normal curve. Show a cautious right-skewed shape or state that a Normal shape is not justified by the information given.

Common Mistakes and AP Exam Tip

Using \(\sigma\) as the spread of sample means: The sampling distribution’s standard deviation is \(\sigma/\sqrt{n}\), not \(\sigma\). Label the spread in the distribution of \(\bar{x}\).
Putting the curve’s center at a sample statistic: The sampling distribution is centered at the population mean \(\mu\), not at a particular observed sample mean.
Forgetting units: If \(\mu\) is measured in gallons, label both the center and \(\sigma_{\bar{x}}\) in gallons. The standard deviation is not in squared units.
Drawing a Normal curve automatically: The mean and standard deviation do not determine shape. Give a reason for exact or approximate Normality, or say the shape is not established.
Calling an approximation exact: A large sample from a non-Normal population can support an approximately Normal shape under CLT reasoning; it does not make that shape exact.
Confusing sample means with observations: The sketch describes \(\bar{x}\) across repeated samples. It does not claim that individual observations have the same center, spread, or shape as the sampling distribution.

For a strong AP response, state the center, show the standard-deviation calculation, include units, and justify the shape. For example: “The sampling distribution of \(\bar{x}\) is centered at 42 gallons and has standard deviation 3 gallons. Since \(n=64\) and the population has no extremely influential outliers, an approximately Normal sketch is reasonable.” If the population is strongly skewed and \(n\) is small, say that the center and spread can be labeled but a Normal shape is not justified.

Key takeaway: Sketch the sampling distribution of \(\bar{x}\) by placing its center at \(\mu\) and its standard deviation at \(\sigma/\sqrt{n}\), with units. Choose the shape separately: it is exactly Normal for a Normal population, may be approximately Normal when CLT reasoning is appropriate, and is not determined by \(\mu\), \(\sigma\), and \(n\) alone.

Check Your Understanding

For each situation, identify what belongs on the sketch. Explain the shape decision as well as the center and spread.

A Normal population has \(\mu=80\) centimeters and \(\sigma=12\) centimeters. Independent samples of size \(n=36\) are taken. Find and label the center and standard deviation, and state whether the shape is exact or approximate.
A population has \(\mu=25\) minutes and \(\sigma=15\) minutes. A random sample of \(n=25\) is taken from a population described as strongly right-skewed with extreme values. Find the center and standard deviation. Is a Normal sketch justified from this information?
A population has \(\mu=6.5\) units and \(\sigma=2.4\) units. Samples of size \(n=64\) are taken, and the population is roughly symmetric with no outliers. Find the standard deviation of \(\bar{x}\) and describe a reasonable shape.
Explain why knowing \(\mu=40\), \(\sigma=8\), and \(n=16\) is enough to calculate center and spread but not always enough to determine the shape.
A student labels the sampling distribution’s center as \(\mu\) and its spread as \(\sigma\). Identify the correction needed and explain what each quantity describes.