Topic 5.7 Sampling Distributions for Sample Means: describe the mean, standard deviation, and shape of the sampling distribution of a sample mean, using the central limit theorem and the standard deviation formula sigma over root n.

What this topic is asking

The College Board (Topic 5.7) wants you to describe the mean, standard deviation, and shape of the sampling distribution of a sample mean $\bar{x}$, using the central limit theorem for shape and the formula $\dfrac{\sigma}{\sqrt{n}}$ for the standard deviation.

Center and spread of the sample mean

The mean result, $\mu_{\bar{x}} = \mu$, says $\bar{x}$ is unbiased, centered on the true mean. The standard deviation $\dfrac{\sigma}{\sqrt{n}}$ is smaller than the population's $\sigma$ by a factor of $\sqrt{n}$, because averaging several observations cancels out some of the individual variation. This is the precise statement of why a mean is more stable than a single measurement: a sample of $25$ has a sample mean with one-fifth the spread of the raw population.

Shape: when is the sample mean normal?

The two routes to normality are worth keeping straight. A normal population gives a normal $\bar{x}$ immediately, so even a small sample works and the CLT is not needed. A non-normal population requires a large $n$ for the CLT to make $\bar{x}$ approximately normal. On the exam, you justify the shape by either citing the normal population or citing the CLT with a large $n$, and a complete answer states which justification applies.

Why the spread shrinks with sample size

The $\dfrac{\sigma}{\sqrt{n}}$ formula is the quantitative heart of the topic and explains a deep practical fact: bigger samples give more reliable means. Because the standard deviation of $\bar{x}$ is inversely proportional to $\sqrt{n}$, quadrupling the sample size halves the spread of the sample mean (since $\sqrt{4} = 2$), and to cut the spread to a tenth you need a hundred times the data. This "diminishing returns" of sample size, precision improves with $\sqrt{n}$, not $n$, recurs throughout inference, where it controls how the margin of error of a confidence interval shrinks. It also clarifies the difference between $\sigma$ and $\sigma_{\bar{x}}$: $\sigma$ describes how individual values vary in the population (fixed, unaffected by sampling), while $\sigma_{\bar{x}}$ describes how the average of $n$ values varies from sample to sample (shrinking as $n$ grows). Confusing these two, using $\sigma$ where $\dfrac{\sigma}{\sqrt{n}}$ is needed, is the most common error, because it overstates the variability of the mean by a factor of $\sqrt{n}$.

The template for mean inference

Topic 5.7 is the sampling-distribution foundation for inference about a mean (Unit 7). A confidence interval for $\mu$ is centered at $\bar{x}$ with a width built from the standard error (the $\dfrac{\sigma}{\sqrt{n}}$ idea, with $\sigma$ usually estimated by $s$, which is why a $t$-distribution appears later), and a significance test for a mean computes a standardized statistic using exactly this spread. So the workflow established here, state the center $\mu$, the spread $\dfrac{\sigma}{\sqrt{n}}$, and justify the shape (normal population or CLT), then standardize and find the area, is the template every mean procedure follows. A full exam response checks the $10\%$ condition for the standard deviation, justifies normality, standardizes with $z = \dfrac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, and interprets the probability in context. Mastering this for a single mean prepares you both for the difference of two means (Topic 5.8) and for all of Unit 7.

Try this

Q1. A population has $\sigma = 10$. Find the standard deviation of $\bar{x}$ for $n = 100$. [1 point]

• Cue. $\sigma_{\bar{x}} = \dfrac{10}{\sqrt{100}} = \dfrac{10}{10} = 1$.

Q2. A population is skewed and $n = 9$. Can you assume $\bar{x}$ is normal? Explain. [2 points]

• Cue. No; the population is not normal and $n$ is small, so the CLT approximation is not good; only a normal population or a large $n$ would justify normality.