Unit 4 · Topic 4.1 · about 20 minutes

Sampling Distributions for Sample Means

Describe how sample means vary from sample to sample, check that the description applies, and find probabilities about a sample mean in context.

Predict first

Delivery times for a food app in one city vary a lot from order to order: their standard deviation is 15 minutes. Pick 25 deliveries at random and average their times. Typically, how far will that average land from the true mean delivery time for the city?

One sample mean is one draw from a distribution

Take a random sample of 50 deliveries and you get one sample mean, $\overline{x}$ . A different random sample of 50 would give a different $\overline{x}$ . The sampling distribution of the sample mean is the distribution of $\overline{x}$ over all possible samples of that size from that population. You met the idea, along with the central limit theorem, in Topic 2.12. This lesson puts numbers on it.

For a population with mean $μ$ and standard deviation $σ$ , when the sampled values are independent, the sampling distribution of $\overline{x}$ has

μ_{\overline{x}} = μ σ_{\overline{x}} = \frac{σ}{\sqrt{n}}

The first formula says sample means are centered on the population mean. Each one misses a little, but the misses balance out: on average, sample means land right on $μ$ . The second says how far they typically miss. It shrinks as $n$ grows, but only by the square root: quadruple the sample and the spread is cut in half.

How the spread of $\overline{x}$ shrinks when single delivery times have $σ = 15$ minutes
Sample size $n$	$σ_{\overline{x}} = 15 / \sqrt{n}$	What it describes
1	15 minutes	Single deliveries, which is the population itself
4	7.5 minutes	Means of 4 deliveries
16	3.75 minutes	Means of 16 deliveries
64	1.875 minutes	Means of 64 deliveries

Two conditions behind the formula

The formula $σ / \sqrt{n}$ assumes the sampled values are independent. When you sample without replacement, two conditions make that reasonable:

Randomization condition: the data come from a random sample.
10% condition: the population is at least 10 times the sample size, $n \leq 0.10 N$ .

The random sample keeps the sample from leaning toward one kind of value. The 10% condition is there because every value you remove changes what is left in the population. When the sample is a small slice of the population, that change is too small to matter and the standard deviation formula holds. Fifty deliveries out of the thousands the app handles in a year is no problem. Fifty students from a school of 300 is a problem, because 10% of 300 is only 30.

When is the shape normal?

The center and spread formulas work for any population. The shape depends on the population and on the sample size.

If the population distribution is approximately normal, the sampling distribution of $\overline{x}$ is approximately normal for any sample size, even $n = 5$ .
If the population is not normal, the sampling distribution of $\overline{x}$ is approximately normal provided $n \geq 30$ . This is the central limit theorem at work.
If the population is extremely skewed, 30 may not be enough, and you need a sample much larger than 30 before a normal model for $\overline{x}$ is trustworthy.

Delivery times are skewed right. Most orders arrive in 20 to 40 minutes, and a few take over an hour when a driver gets stuck. The distribution of single delivery times is not normal, and no sample size will change that. The mean of 50 delivery times is a different variable, and its distribution is close to normal.

Lab

Sampling Distribution Machine

Leave the population on Strongly skewed right and set the sample size to 2. Draw 1,000 samples and look at the shape of the sample means. Then try $n = 10$ and $n = 40$ . Each time, compare the SD of the statistics with the SD from the formula, which is $σ / \sqrt{n}$ . The simulation should land close to it at every sample size, even while the shape is still skewed.

Open the full Sampling Distribution Machine lab

Worked exampleThe chance a sample mean runs long

Delivery times for a food app in one city are skewed to the right with mean $μ = 34$ minutes and standard deviation $σ = 15$ minutes. A city inspector selects a random sample of 50 deliveries from last year's records. Find the probability that the sample mean delivery time is more than 37 minutes.

Center and spread. $μ_{\overline{x}} = 34$ minutes and $σ_{\overline{x}} = \frac{15}{\sqrt{50}} \approx 2.12$ minutes.
Conditions. Randomization: the inspector took a random sample of deliveries. 10%: 50 is far less than 10% of the city's deliveries last year, so the standard deviation formula applies.
Shape. The population is skewed, but $n = 50 \geq 30$ , so the sampling distribution of $\overline{x}$ is approximately normal.
Calculate. $z = \frac{37 - 34}{2.12} \approx 1.41$ . The area under the normal curve to the right of 37 is about 0.079. A calculator's normalcdf with lower bound 37, mean 34 and standard deviation 2.12 gives the same area without rounding $z$ first.
Interpret. In about 8% of random samples of 50 deliveries, the sample mean delivery time would be more than 37 minutes.

Answer.

About 0.079. A sample mean above 37 minutes happens in roughly 8 of every 100 random samples of 50 deliveries.

Sampling distribution of $\overline{x}$ for samples of 50 deliveries

The shaded area, about 0.079, is the probability that a random sample of 50 deliveries has a mean above 37 minutes.

Check your understanding

The weights of eggs from a large farm are approximately normal with mean 58 grams and standard deviation 4 grams. A worker selects a random sample of 8 eggs and finds their mean weight. Which describes the sampling distribution of the sample mean weight?

The time customers spend in a museum gift shop has a standard deviation of 9 minutes. For random samples of 36 customers, what is the standard deviation of the sampling distribution of the sample mean time, in minutes?

Your answer

The number of text messages sent per day by students at a large university is strongly skewed to the right, with a few students sending hundreds. A researcher takes a random sample of 12 students and records their mean number of texts. Which statement about the sampling distribution of this sample mean is correct?

Delivery times for a food app have mean 34 minutes and standard deviation 15 minutes. For random samples of 50 deliveries, the sampling distribution of the sample mean has standard deviation 2.12 minutes. Which is the best interpretation of 2.12 minutes?

A researcher plans to take a random sample from a large population. She increases the planned sample size from 25 to 100. What happens to the standard deviation of the sampling distribution of $\overline{x}$ ?

Practice

Practice until it is automatic

New numbers every time. Each one is checked the moment you answer, with the full working shown.

Sampling distribution of a sample mean practice page

Course alignment, for teachers

AP Statistics topic 4.1, Unit 4: Inference for Quantitative Data: Means.

Skill 3.D: Calculate means, standard deviations, and parameters for probability distributions.
Skill 4.D: Interpret statistical calculations and results to assess meaning or a claim.
Skill 4.E: Justify the use of a chosen statistical inference method by verifying conditions.