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Confidence intervals for a normal mean (known variance)

Edexcel IAL Mathematics · WST03 Statistics 3

A single sample mean is unlikely to equal μ exactly. A confidence interval gives a range of plausible values for μ, built so that a stated percentage of such intervals contain it.

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The formula

For a random sample of n from N(μ, σ2) with σ known, a confidence interval for μ is

x̄ ± z × σ/sqrt(n)

where z comes from the percentage points table (4 d.p.).

Confidence levelzTail area each side
90%1.64490.05
95%1.96000.025
98%2.32630.01
99%2.57580.005

Example

σ = 4, n = 25, x̄ = 31.2. Standard error = 4/5 = 0.8.

95% interval: 31.2 ± 1.96 × 0.8 = 31.2 ± 1.568 = (29.632, 32.768).

Show the substitution: a correct interval with no working is not given full credit. Using the wrong n (for example 150 instead of 200) loses the accuracy mark.

Width and sample size

Width = 2 × z × σ/sqrt(n). Higher confidence means a larger z and a wider interval; a larger sample makes it narrower.

To find n for a maximum width, solve 2zσ/sqrt(n) ≤ w and round up.

The centre of an interval is x̄; half its width is zσ/sqrt(n). These two facts let you work backwards from a given interval.

Worked example

A random sample of 25 from N(μ, 42) has mean 31.2. Find a 99% confidence interval for μ.

  1. Standard error = 4/sqrt(25) = 0.8
  2. z = 2.5758
  3. 31.2 ± 2.5758 × 0.8 = 31.2 ± 2.0606

Answer: (29.14, 33.26)

Key idea: x̄ ± zσ/sqrt(n) with z = 1.6449, 1.96, 2.3263 or 2.5758. Wider for more confidence, narrower for larger n.

Check yourself

  1. A random sample of 25 from N(μ, 42) has mean 31.2. Find the lower limit of a 95% confidence interval for μ. Give your answer to 3 decimal places.
    Show answer

    29.632. 31.2 - 1.96 × 0.8 = 29.632.

  2. A random sample of 25 from N(μ, 42) has mean 31.2. Find the upper limit of a 99% confidence interval for μ. Give your answer to 3 significant figures.
    Show answer

    33.3. 31.2 + 2.5758 × 0.8 = 33.2606.

  3. A random sample of 25 from N(μ, 42) is taken. Find the width of a 90% confidence interval for μ. Give your answer to 3 significant figures.
    Show answer

    2.63. 2 × 1.6449 × 0.8 = 2.6318.

  4. A population is N(μ, 52). Find the smallest sample size for which a 95% confidence interval for μ has width at most 2.
    Show answer

    97. 2 × 1.96 × 5/sqrt(n) ≤ 2 gives sqrt(n) ≥ 9.8, n ≥ 96.04, so n = 97.

  5. A confidence interval for μ is (12.3, 15.7). Find the sample mean used.
    Show answer

    14. The interval is centred on x̄: (12.3 + 15.7)/2 = 14.

  6. A 95% confidence interval for μ from a random sample of 16 is (48.04, 51.96). Find the population standard deviation σ.
    Show answer

    4. Half-width = 1.96 = 1.96σ/4, so σ = 4.

  7. A random sample of 36 from N(μ, 62) gives a confidence interval for μ of width 3.2898. Find the confidence level, as a percentage.
    Show answer

    90. Half-width 1.6449 = z × 6/6, so z = 1.6449, which leaves 0.05 in each tail: 90%.

  8. How does a 99% confidence interval compare with a 95% interval from the same sample?
    • It is narrower
    • It is wider
    • It has the same width but a different centre
    • It has a different centre and the same width
    Show answer

    It is wider. More confidence needs a larger z (2.5758 against 1.96), so the interval is wider; the centre x̄ is unchanged.

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