Statistics How To

Continuity Correction Factor: What is it?

Statistics Definitions > Continuity Correction Factor

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What is the Continuity Correction Factor?

A continuity correction factor is used when you use a continuous function to approximate a discrete one. For example, when you want to approximate a binomial with a normal distribution. According to the Central Limit Theorem, the sample mean of a distribution becomes approximately normal if the sample size is large enough. The binomial distribution can be approximated with a normal distribution too, as long as n*p and n*q are both at least 5.

The continuity correction factor a way to account for the fact that a normal distribution is continuous, and a binomial is not. When you use a normal distribution to approximate a binomial distribution, you’re going to have to use a continuity correction factor. It’s as simple as adding or subtracting .5 to the discrete x-value: use the following table to decide whether to add or subtract.

Continuity Correction Factor Table

If   P(X=n) use   P(n – 0.5 < X < n + 0.5)
If   P(X>n) use   P(X > n + 0.5)
If   P(X≤n) use    P(X < n + 0.5)
If    P (X<n) use   P(X < n – 0.5)
If    P(X ≥ n) use   P(X > n – 0.5)

If  P(X≥351), use P (X≥351-0.5)= P (X≥350.5)

Continuity Correction Factor Example

Sample problem: If n=20 and p=.25, what is the probability that X ≥ 8?

continuity correction factor

Step 1: Work out np and nq:
np = 20 * .25 = 5 (note: this is also the mean of the binomial distribution)
nq = 20 * .75 = 15
These are both over 5, so we can use the continuity correction factor.

Step 2: Figure out the variance of the binomial distribution:
n*p*q = 20 * .25 * .75 = 3.75
Set this number aside for a moment.

Step 3: Use the continuity correction factor on the X value.
X ≥ 8 becomes X ≥ 7.5.

Step 4: Find the z-score.
z score formula
z = 7.5 – 5 / √3.75 = 1.29

Step 5: Look up Step 4 in the z-table.
1.29 = .4015.

Step 6: Subtract Step from 1 to get the area (as we are looking for the right tail):
1 – .4015 = 0.5985.
The probability that X ≥ 8 is 0.5985.

Why is the continuity correction factor used?

While the normal distribution is continuous (it includes all real numbers), the binomial distribution can only take integers.

continuity correction factor

A normal distribution curve.

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Continuity Correction Factor: What is it? was last modified: February 18th, 2018 by Stephanie

10 thoughts on “Continuity Correction Factor: What is it?

  1. Andale

    What you’re doing is trying to compare one distribution to another. For example, in order to get your binomial to “fit” a normal, you have to either add or subtract 0.5.

  2. Naomi

    Hi, this was very helpful in understanding what I am doing, however…

    I understand that “this procedure can be validly employed only if both samples satisfy the standard binomial requirement: that n(p) and n(1—p) must both be equal to or greater than 5.”

    I am trying to understand if the proportion of Ss who succeeded on my test is different between two groups. In one group 4/21 Ss succeeded, and in the other it is 9/20. If only the numbers were 5 and 10, I could run the test, and it would be significant – is there any correction I can employ to be able to examine the significance of the difference?

    Thank you!

  3. Andale

    Some sources allow 10 (instead of 5). I’d say you were safe. That said, be aware that the higher you go, the less accurate your results are likely to be.

  4. Andale

    Do you mean the probability (area)? It would depend on what you’re looking for. For example, if you were looking up a right tailed test, you would need to subtract from 1 as the z table only gives you left hand tails.

  5. Andrey Novikov

    I hope the following will help to understand when to add 0.5 and when to subtract 0.5.
    Let us see, for example, the case P(X>=x) supposing x is integer.
    The thing is that P(X>=x)=P(X>x-1) because X takes only integer values.
    BUT applying the normal approximation to these two probabilities we obtain two DIFFERENT results for the SAME probability:
    P(Z>=(x-m)/sigma) and P(Z>(x-1-m)/sigma)!. Because there is no reason to prefer one or another, we better choose an “intermediate” value between the two, namely, P(Z>(x-0.5-m)/sigma), which gives the same recommendation TO subtract 0.5 from x.

  6. Oscar M.

    There is typo in step 5. The z-score probability is .4015, not .9015. The probability should be much lower.