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The distribution of the sample mean

Edexcel IAL Mathematics · WST03 Statistics 3

The mean of a random sample is a random variable. If the population is normal, the sample mean is exactly normal, with the same mean and a variance n times smaller.

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The result

If X1, ..., Xn is a random sample from N(μ, σ2) then

X̄ ~ N(μ, σ2/n), so Z = (X̄ - μ)/(σ/sqrt(n)) ~ N(0, 1).

This follows from the sum result: X1 + ... + Xn ~ N(nμ, nσ2), then divide by n (variance divided by n2).

Using it

X ~ N(50, 82) and n = 16: X̄ ~ N(50, 64/16) = N(50, 4), so the standard error is 2.

P(X̄ > 52) = P(Z > (52 - 50)/2) = P(Z > 1) = 0.1587.

A common error is to leave sqrt(n) out of the standardisation: always write (x̄ - μ)/(σ/sqrt(n)).

Choosing a sample size

How large must n be so that P(|X̄ - μ| < 1) ≥ 0.95, when σ = 8? Need 1.96 × 8/sqrt(n) ≤ 1, so sqrt(n) ≥ 15.68 and n ≥ 245.86. Round up: n = 246.

Worked example

X ~ N(50, 82). A random sample of 16 is taken. Find P(49 < X̄ < 51).

  1. X̄ ~ N(50, 4), standard error 2.
  2. z values: (49 - 50)/2 = -0.5 and (51 - 50)/2 = 0.5.
  3. P = 2Φ(0.5) - 1 = 2(0.6915) - 1

Answer: P = 0.3830

Key idea: X̄ ~ N(μ, σ2/n) for a normal population. Standardise with σ/sqrt(n), and round sample sizes up.

Check yourself

  1. X ~ N(50, 82). Find the variance of the mean of a random sample of 16.
    Show answer

    4. σ2/n = 64/16 = 4.

  2. X ~ N(50, 82). Find P(X̄ > 52) for a random sample of 16. Give your answer to 4 decimal places.
    Show answer

    0.1587. z = 2/2 = 1; P = 0.1587.

  3. X ~ N(50, 82). Find P(X̄ < 48.5) for a random sample of 16. Give your answer to 4 decimal places.
    Show answer

    0.2266. z = -1.5/2 = -0.75; P = 0.2266.

  4. X ~ N(50, 82). Find P(49 < X̄ < 51) for a random sample of 16. Give your answer to 4 decimal places.
    Show answer

    0.3829. 2Φ(0.5) - 1 = 0.3830.

  5. X ~ N(μ, 82). Find the smallest sample size n so that P(|X̄ - μ| < 1) ≥ 0.95.
    Show answer

    246. 1.96 × 8/sqrt(n) ≤ 1 gives n ≥ 245.86, so n = 246.

  6. A population has standard deviation 10. Find the standard deviation of the mean of a random sample of 25.
    Show answer

    2. 10/sqrt(25) = 2.

  7. X ~ N(μ, σ2) and X̄ is the mean of a random sample of size n. Which is true?
    • X̄ is normal only if n > 30
    • X̄ has variance σ2
    • X̄ is exactly normal for any n
    • X̄ has mean μ/n
    Show answer

    X̄ is exactly normal for any n. A linear combination of independent normal variables is normal, so X̄ is exactly normal.

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Last updated 8 November 2026. Spotted a mistake? Tell us.