What Is the Sampling Distribution of a Sample Mean?
When we take a sample, the sample mean \(\bar{x}\) summarizes the values in that sample. A different sample of the same size from the same population may produce a different mean. The collection of all those possible sample means, together with their probabilities, is called the sampling distribution of \(\bar{x}\).
This is a probability distribution for a statistic, not a list of values from one sample. To build one, first specify the population, the sample size, and the sampling method. Then list every possible sample allowed by that method, calculate \(\bar{x}\) for each sample, and find the probability of each possible mean. As in earlier probability work, the probabilities for a complete distribution must add to 1.
It helps to keep three related ideas separate. The population distribution describes the values of the individuals in the population. A sample is one group of individuals selected from that population. The sampling distribution describes the values of a statistic, such as \(\bar{x}\), across all possible samples of the stated size.
For this tutorial’s first example, we will use samples of size 2 drawn without replacement. That means the same population value cannot be selected twice in a sample. We treat the samples as simple random samples, so each distinct pair has the same chance of being selected. The sampling method matters: changing it can change which samples are possible and their probabilities.
Build a Sampling Distribution by Listing Samples
Consider the small population \(\{2,4,6,8\}\). There are four population values and we will select a sample of size 2 without replacement. Since selecting 2 and then 4 gives the same pair as selecting 4 and then 2, we list each unordered pair only once. The number of possible pairs is 6.
Calculate the mean for each pair. For example, the mean of the sample \(\{2,4\}\) is \((2+4)/2=3\). The complete list is:
| Sample | Sample mean \(\bar{x}\) |
|---|---|
| \(\{2,4\}\) | 3 |
| \(\{2,6\}\) | 4 |
| \(\{2,8\}\) | 5 |
| \(\{4,6\}\) | 5 |
| \(\{4,8\}\) | 6 |
| \(\{6,8\}\) | 7 |
The sample means are not all different: two of the six possible samples have a mean of 5. Since the six pairs are equally likely, each pair has probability \(1/6\). Therefore, the probability that \(\bar{x}=5\) is \(2/6\), because either of two pairs produces that mean. For every other mean in the table, just one pair produces it, so its probability is \(1/6\).
| Possible value of \(\bar{x}\) | Probability |
|---|---|
| 3 | \(1/6\) |
| 4 | \(1/6\) |
| 5 | \(2/6\) |
| 6 | \(1/6\) |
| 7 | \(1/6\) |
| Total | \(6/6=1\) |
The total is \(1\) because the numerators add to \(1+1+2+1+1=6\), giving \(6/6=1\). This final table is the sampling distribution of \(\bar{x}\) for samples of size 2 drawn without replacement from \(\{2,4,6,8\}\). It shows both what values the sample mean can take and how likely each value is.
Worked Example: Enumerating Every Pair
A small population consists of the values \(2,4,6,8\). A sample of size 2 is selected at random without replacement. Build the sampling distribution of \(\bar{x}\).
Step 1: List the possible samples. The six equally likely pairs are \(\{2,4\}\), \(\{2,6\}\), \(\{2,8\}\), \(\{4,6\}\), \(\{4,8\}\), and \(\{6,8\}\). Each pair has probability \(1/6\).
Step 2: Calculate each sample mean. The means, in the same order, are \(3,4,5,5,6,7\). For instance, \(\{2,8\}\) has mean \((2+8)/2=5\), and \(\{4,6\}\) also has mean \((4+6)/2=5\).
Step 3: Assign probabilities to the distinct means. One sample produces 3, one produces 4, two produce 5, one produces 6, and one produces 7. Thus the probabilities are \(1/6,1/6,2/6,1/6,1/6\), respectively. As a check, their sum is \((1+1+2+1+1)/6=6/6=1\).
The distribution is therefore: \(\bar{x}=3\) with probability \(1/6\), \(\bar{x}=4\) with probability \(1/6\), \(\bar{x}=5\) with probability \(2/6\), \(\bar{x}=6\) with probability \(1/6\), and \(\bar{x}=7\) with probability \(1/6\). The value 5 is more likely than any single other listed mean because two distinct samples produce it.
Read the Distribution as Probabilities About a Statistic
A sampling distribution lets us answer questions about what a sample mean could be before a particular sample is selected. In the population \(\{2,4,6,8\}\) example, the probability that a sample of size 2 has mean 5 is \(2/6=1/3\). The probability that its mean is at least 6 is \(1/6+1/6=2/6=1/3\), since the possible means 6 and 7 each have probability \(1/6\).
These are probabilities about \(\bar{x}\), the statistic that would result from the random sample. They are not probabilities that a particular population value is 5 or 6. Nor does the table predict which mean one specific sample must have. It describes the outcomes and probabilities across the full set of possible samples under the stated sampling method.
Notice also that a possible mean can occur more often than another even though all samples are equally likely. The samples \(\{2,8\}\) and \(\{4,6\}\) are each just as likely as \(\{2,4\}\), but both of the first two produce a mean of 5. Probabilities for a mean are found by grouping the samples that produce it.
Worked Example: Three Possible Sample Means
A practice exercise uses a population consisting of \(3,9,15\). A sample of size 2 is drawn at random without replacement. Find the sampling distribution of \(\bar{x}\), then find the probability that \(\bar{x}\) is at least 9.
There are three equally likely samples: \(\{3,9\}\), \(\{3,15\}\), and \(\{9,15\}\). Their means are:
Each sample produces a different mean, so the sampling distribution assigns probability \(1/3\) to each of 6, 9, and 12. The probabilities total \(1/3+1/3+1/3=3/3=1\).
The event \(\bar{x}\ge 9\) includes means 9 and 12. Therefore, \(P(\bar{x}\ge 9)=1/3+1/3=2/3\). In context, two out of the three equally likely samples of two practice values have a mean of at least 9.
The Sampling Method Changes the Distribution
Always say whether sampling is with or without replacement. Without replacement, a population value cannot appear twice in the same sample. With replacement, each selected value is returned before the next selection, so the same value can be selected again. These methods can produce different possible samples and different probabilities for \(\bar{x}\).
When sampling with replacement, it is useful to keep track of the order of selections. For example, selecting 4 and then 10 is a different ordered outcome from selecting 10 and then 4, even though both outcomes have the same sample mean. When building the distribution, count both outcomes if both can occur.
Worked Example: Sampling With Replacement
A population consists of the values 4 and 10. Two values are selected with replacement, and each selection is equally likely to be 4 or 10. Build the sampling distribution of \(\bar{x}\).
There are four equally likely ordered outcomes: \((4,4)\), \((4,10)\), \((10,4)\), and \((10,10)\). Each has probability \(1/4\). Their sample means are:
The mean 4 occurs for one outcome, the mean 7 for two outcomes, and the mean 10 for one outcome. Thus the sampling distribution is \(P(\bar{x}=4)=1/4\), \(P(\bar{x}=7)=2/4=1/2\), and \(P(\bar{x}=10)=1/4\). The probabilities add to \(1/4+2/4+1/4=4/4=1\).
The two mixed outcomes have the same mean, so they are combined in the distribution. This example also shows why it is important to count all outcomes under the actual sampling method rather than assume every distinct mean has the same probability.
A Reliable Construction Process
For a small population, direct enumeration is a clear way to construct a sampling distribution. A table can keep the samples, their means, and their probabilities organized. Use this process whenever a question asks you to build the distribution from all possible samples.
Record the population, sample size, and whether sampling is with or without replacement.
Include every sample allowed by the method, without counting an unordered pair twice when order is irrelevant.
Find \(\bar{x}\) for each sample by adding its values and dividing by the sample size.
Combine samples with the same mean and add their probabilities to find the probability of that mean.
Confirm that every possible sample was included and that the probabilities of all possible means sum to 1.
Common Mistakes and AP Exam Tip
- Confusing the three distributions: The population distribution is about individual population values; a sample is one selected group; the sampling distribution is about a statistic across possible samples.
- Listing only the means without probabilities: A sampling distribution includes both possible values of \(\bar{x}\) and the probability of each value.
- Giving every distinct mean the same probability automatically: Count how many possible samples produce each mean. In the first example, two samples produce 5, so its probability is \(2/6\), not \(1/6\).
- Forgetting the sampling method: Sampling with replacement permits repeated values in a sample; sampling without replacement does not. State which method is being used before listing samples.
- Counting outcomes inconsistently: If samples are unordered pairs, do not list both \(\{2,4\}\) and \(\{4,2\}\). If the sampling method treats selections as ordered, as in the replacement example, count both orders.
- Leaving out the probability check: Add the probabilities for all possible values of \(\bar{x}\). A complete distribution must have total probability 1.
- Claiming one sample must have a particular mean: The distribution describes probabilities across possible samples. It does not determine the result of a specific sample before that sample is selected.
For full-credit communication, identify the population, sample size, and sampling method; show the possible samples and their means; and explain how the probabilities were counted. When multiple samples produce the same mean, combine their probabilities explicitly.
Check Your Understanding
Use the ideas in this tutorial to answer each question. When a question asks you to build a distribution, show how you counted its probabilities.
- In your own words, distinguish a population distribution from the sampling distribution of \(\bar{x}\).
- For samples of size 2 without replacement from \(\{1,3,5\}\), list all possible samples and calculate each sample mean.
- In Question 2, write the sampling distribution of \(\bar{x}\) and verify that its probabilities add to 1.
- For the population \(\{2,10\}\), two values are selected with replacement. List the ordered outcomes and find the probability of each possible sample mean.
- Why might two different samples have the same mean, and how does that affect the probability assigned to that mean?