The Sampling Distribution of a Sample Mean

You’ll be able to

State the mean and standard deviation of the sampling distribution of x-bar
Apply the Central Limit Theorem to determine the shape of the distribution of x-bar
Compute probabilities involving a sample mean

Center and spread of x-bar

If we draw an SRS of size n from a population with mean μ and standard deviation σ, the sampling distribution of the sample mean x-bar has mean μ (unbiased) and standard deviation σ/sqrt(n). The standard deviation of x-bar is smaller than the population’s by a factor of sqrt(n) — averaging cancels out extremes, so means vary less than individuals.

The Central Limit Theorem (CLT)

The shape of the sampling distribution of x-bar depends on the population and n. If the population is Normal, x-bar is Normal for any n. If the population is not Normal, the Central Limit Theorem says the distribution of x-bar becomes approximately Normal when n is large enough — the common rule of thumb is n ≥ 30. The CLT is what makes inference about means possible even for skewed populations.

Sampling distribution of x-bar

μ_(x-bar) = μ · σ_(x-bar) = σ / sqrt(n) · shape ≈ Normal if population Normal OR n ≥ 30 (CLT)

The standard deviation of x-bar (the "standard error" once σ is estimated) shrinks with the square root of n. Requires the sample be no more than 10% of the population for the formula to hold.

Worked example

A population has μ = 100 and σ = 15. For an SRS of size n = 25, find the mean and standard deviation of x-bar, and P(x-bar > 106) if the population is Normal.

1.Mean of x-bar: μ_(x-bar) = μ = 100.
2.Standard deviation of x-bar: σ_(x-bar) = σ/sqrt(n) = 15/sqrt(25) = 15/5 = 3.
3.Since the population is Normal, x-bar is Normal. Standardize: z = (106 − 100)/3 = 6/3 = 2.
4.P(x-bar > 106) = P(Z > 2) ≈ 0.025.

Answer: x-bar has mean 100 and standard deviation 15/5 = 3. Standardizing, z = 2, so P(x-bar > 106) ≈ 0.025 — only about a 2.5% chance a sample of 25 averages above 106.

Checkpoint

A population is strongly right-skewed with μ = 50 and σ = 12. For samples of size n = 100, the sampling distribution of x-bar is:

On the exam

Before finding a probability for x-bar, justify the Normal shape: either state the population is Normal, or invoke the CLT with n ≥ 30. Free-response answers that skip this justification lose the condition point even if the arithmetic is perfect.

Checkpoint

A population has standard deviation σ = 20. If the sample size increases from 25 to 100, the standard deviation of x-bar changes from 4 to:

Answer the 2 checkpoints as you read.