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Actuarial Statistics · Central limit theorem

Continuity Correction in Normal Approximation

Updated 11 October 2026 · Fact-checked

Continuity correction adjusts a discrete integer-valued probability before you use a normal approximation. Treat each integer k as the interval from k − 0.5 to k + 0.5. So P(X ≤ k) becomes P(Y < k + 0.5) and P(X ≥ k) becomes P(Y > k − 0.5). Then standardise and use normal tables.

Understand Continuity Correction

A binomial or Poisson variable takes only whole-number values. A normal variable is continuous. It has zero probability at any single point. If you approximate directly, you lose part of the probability at the boundary value.

The fix is to picture each integer k as a bar of width 1, running from k − 0.5 to k + 0.5. The area of that bar is the probability P(X = k). The normal curve should be integrated over the same bar. This is the continuity correction.

So you move the boundary by 0.5 so that the right integers are included or excluded. For P(X ≤ 30), the bar for 30 must be fully included. The cut-off is therefore 30.5. For P(X < 30), the bar for 30 must be excluded. That is P(X ≤ 29), so the cut-off is 29.5.

The correction matters most when the standard deviation is small. When n or the mean is large, the effect shrinks. In exams, you are usually told to use it, or the question is clearly about a discrete variable approximated by a normal. Use it whenever the variable takes integer values with spacing 1.

The correction is not needed when the variable is already continuous, such as a sample mean of continuous claim sizes. It is also not needed when you approximate one continuous distribution by another.

Key rules to remember

Normal approximation to the binomial
X ~ Bin(n, p) ⇒ Y ~ N(np, np(1 − p))
A common guide is that both np and n(1 − p) should be reasonably large, for example above 5. Check what the question says.
Normal approximation to the Poisson
X ~ Poisson(λ) ⇒ Y ~ N(λ, λ)
Works better for larger λ. The standard deviation is √λ, not λ.
Single value
P(X = k) ≈ P(k − 0.5 < Y < k + 0.5)
Valid for integer-valued X with spacing 1.
Lower tail, inclusive
P(X ≤ k) ≈ P(Y < k + 0.5)
Add 0.5 to the upper limit.
Lower tail, strict
P(X < k) = P(X ≤ k − 1) ≈ P(Y < k − 0.5)
Convert to ≤ first, then correct.
Upper tail, inclusive
P(X ≥ k) ≈ P(Y > k − 0.5)
Subtract 0.5 from the lower limit.
Upper tail, strict
P(X > k) = P(X ≥ k + 1) ≈ P(Y > k + 0.5)
Convert to ≥ first, then correct.
Range
P(a ≤ X ≤ b) ≈ P(a − 0.5 < Y < b + 0.5)
Widen the interval by 0.5 at each end. For strict inequalities, convert to inclusive integer limits first.
Standardising
Z = (Y − μ) ÷ σ
Apply it to the corrected limit, not the original integer.

How to solve Continuity Correction questions

Use this method for any question that approximates a discrete integer-valued distribution by a normal.

  1. 1Identify the exact distribution of X and its parameters, for example Bin(n, p) or Poisson(λ).
  2. 2Write down the approximating normal: find the mean and variance, then take the square root for σ.
  3. 3Rewrite the required probability using only ≤ or ≥ on integers. Turn X < k into X ≤ k − 1 and X > k into X ≥ k + 1.
  4. 4Apply the correction: for ≤ b use b + 0.5, for ≥ a use a − 0.5. For a single value use ±0.5 around it.
  5. 5Standardise the corrected limits with Z = (Y − μ) ÷ σ.
  6. 6Read the probabilities from the normal table, using symmetry for negative z values: Φ(−z) = 1 − Φ(z).
  7. 7Combine the tail areas, then state the final answer to a sensible number of decimal places and say it is an approximation.

Quickest way: Half-unit boundary shortcut

When to use it: Use it in multiple-choice questions and when time is short, once you are comfortable with the integer-bar idea.

  1. Convert the wording to integers: X < 30 means X ≤ 29, X > 30 means X ≥ 31.
  2. Move each boundary outward by 0.5 so that the boundary integer stays included: ≤ 29 gives 29.5 and ≥ 31 gives 30.5.
  3. Both X ≤ 30 and X < 31 land on the same cut-off, 30.5. This is a quick check.
  4. Compute z once with the corrected cut-off, then read the table.

Common mistakes in Continuity Correction

  • Moving the boundary in the wrong direction, for example using 29.5 for P(X ≤ 30).

    Students memorise ±0.5 without thinking about which integers must stay in.

    Fix: Draw the integer bars. For ≤ 30, the bar for 30 must be included, so go up to 30.5.

  • Treating X < k as X ≤ k and correcting to k + 0.5.

    Strict inequalities look like non-strict ones, but for discrete variables they differ.

    Fix: Convert to integers first: X < k is X ≤ k − 1, so the cut-off is k − 0.5.

  • Using the variance instead of the standard deviation when standardising.

    The normal is written N(μ, σ²), and np(1 − p) or λ is the variance.

    Fix: Always take the square root before dividing.

  • Applying the correction to a continuous variable such as a sample mean.

    Students link CLT questions with the correction automatically.

    Fix: Use it only when the original variable is integer-valued, such as a count. Then check whether the question is about a count or a mean of continuous values.

  • Forgetting to correct both ends of a range.

    Students correct one limit and then rush the other.

    Fix: For P(a ≤ X ≤ b), write a − 0.5 and b + 0.5 on separate lines.

  • Using the approximation when the parameters are too small, then quoting the result as exact.

    The method is mechanical and gives a number whatever the parameters.

    Fix: Check np and n(1 − p), or λ, are reasonably large. State that the answer is approximate.

Worked examples

Example 1

X ~ Bin(100, 0.4). Use a normal approximation with continuity correction to estimate P(35 ≤ X ≤ 45).

Show the solution
  1. Mean = np = 100 × 0.4 = 40. Variance = np(1 − p) = 100 × 0.4 × 0.6 = 24. So σ = √24 ≈ 4.899.
  2. Approximating normal: Y ~ N(40, 24).
  3. Apply the correction to the range: P(35 ≤ X ≤ 45) ≈ P(34.5 < Y < 45.5).
  4. Standardise: z for 34.5 = (34.5 − 40) ÷ 4.899 = −1.123. z for 45.5 = (45.5 − 40) ÷ 4.899 = 1.123.
  5. Φ(1.123) ≈ 0.8692, so Φ(−1.123) ≈ 0.1308.
  6. Probability ≈ 0.8692 − 0.1308 = 0.7384.

Answer: P(35 ≤ X ≤ 45) ≈ 0.738

Example 2

The number of claims N in a year on a portfolio is Poisson with mean 25. Use a normal approximation with continuity correction to estimate P(N > 30).

Show the solution
  1. Approximating normal: Y ~ N(25, 25), so σ = 5.
  2. Convert to integers: N > 30 means N ≥ 31.
  3. Apply the correction: P(N ≥ 31) ≈ P(Y > 30.5).
  4. Standardise: z = (30.5 − 25) ÷ 5 = 1.1.
  5. P(Z > 1.1) = 1 − Φ(1.1) = 1 − 0.8643 = 0.1357.

Answer: P(N > 30) ≈ 0.136

Exam tips

  • Write the integer form of the inequality before you do anything else. Most marks are lost at this step.
  • Show the corrected limit explicitly, such as 30.5, so the examiner can award method marks even if the table reading is off.
  • Say which normal you are using, with its mean and variance, and state any assumption such as independence.
  • In written questions, comment briefly on why the approximation is reasonable by checking np and n(1 − p), or λ.
  • In a computer-based paper, you can compare the approximation with the exact binomial or Poisson probability from software to see the effect of the correction.

Practice questions from Central limit theorem

Continuity Correction: frequently asked questions

When should I use continuity correction?

Use it when you approximate an integer-valued discrete distribution, such as binomial or Poisson, by a normal distribution. Do not use it when the original variable is continuous.

Why is the correction 0.5?

Integer values are spaced 1 apart. Each integer therefore occupies an interval of width 1, centred on it, which extends 0.5 on each side.

How do I handle P(X < k) with continuity correction?

Rewrite it as P(X ≤ k − 1) for integer X. Then the cut-off is k − 0.5, so P(X < k) ≈ P(Y < k − 0.5).

Does the correction matter for large n?

Its effect shrinks as the standard deviation grows, but it still improves accuracy. Exam questions usually expect you to apply it unless told otherwise.