Distribution of Sample Mean

Bigger Samples Concentrate the Mean

Sample size n16

👀 See It

①The sample mean X̄ is itself a random variable that varies by sample

②As n grows, the distribution of X̄ narrows around the population mean

③The width is proportional to σ/√n — quadruple n and the width halves

Mean and Standard Deviation of X̄

Expectation and SD of the sample mean

E(X̄) = m, σ(X̄) = σ/√n

Population mean m, population SD σ, sample size n — the mean stays, the SD scales by 1/√n

Normal approximation

if n is large, X̄ ~ N(m, σ²/n)

Even if the population is not normal, for large n the sample mean approaches a normal distribution

The Effect of √n

📏 Width Is Inversely Proportional to √n

①Since σ(X̄)=σ/√n, the SD is inversely proportional to √n

②Quadrupling n halves the SD

③To double the precision you must quadruple the sample

Compute It Directly

Example 1

From a population with mean 50 and SD 8, a sample of size 16 is drawn. Find the mean and SD of the sample mean.

Substitute into E(X̄)=m, σ(X̄)=σ/√n.

E(X̄) = 50, σ(X̄) = 8/√16

Compute with √16=4.

σ(X̄) = 8/4 = 2

▸ mean 50, SD 2

The mean stays equal to the population mean; only the SD is divided by √n.

Example 2

From the same population, if the sample size grows to 64, what is the SD of the sample mean?

σ(X̄)=8/√64.

Compute with √64=8.

σ(X̄) = 8/8 = 1

▸ 1 (n ×4 → SD ×1/2)

Growing n from 16 to 64 (×4) halves the SD from 2 to 1.

Wrap-up

Key result

E(X̄)=m, σ(X̄)=σ/√n, (large n) X̄ ~ N(m, σ²/n)

The sample mean has mean m and SD σ/√n — for large n it is normal

2021 CSAT Math type, adapted

From a population with SD 10, a sample of size 25 is drawn. What is the SD of the sample mean?

①0.4

②2

③5

④10

⑤50

▸ ② 2

Put σ=10, n=25 into σ(X̄)=σ/√n.

σ(X̄) = 10/√25

Compute with √25=5.

σ(X̄) = 10/5 = 2

🎯 Exam Points

①E(X̄)=m (same as population mean)

②σ(X̄)=σ/√n (divide by √n)

③for large n, X̄~N(m, σ²/n)

④double precision needs 4× the sample

⑤do not confuse the variance V(X̄)=σ²/n

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