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)

Example:
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: October 31st, 2017 by Stephanie

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

  1. James

    what is the point of learning without understanding? There is no explanation of why N+0.5 or n-0.5

  2. Stephanie

    James,

    The vast majority of students using this site are struggling traditional students who have traditional, bulky textbooks (where you’ll find several hundred pages of explanation).

    The tables and charts on this site are for reference and quick answers to sticky problems without all of the textbook bloviation! I’m not attempting to replace learning and understanding…just enhancing it.

    Stephanie

  3. Nick

    James,

    I also get frustrated when I can’t get an explanation and got screwed up by this same concept. Here is a way I figured out to explain it. Let’s make a simple example that X is # of heads out of 50 tosses. Let’s use less-than or equal to for simplicity. Let’s say we want P (10 <= X <= 20). Numbers 10-20, inclusive, are "good outcomes." Since we are using a continuous variable (with decimals) to approximate the sum of discrete variables, anything between 9.5 and 20.5 will be rounded to 10 and 20 (discrete = whole #s only) so by EXPANDING the range to include leeway for rounding will give us a closer approximation since X = 9.7 is closer to meaning 10 good outcomes than 9. This idea of EXPANDING is at the heart of the whole matter.

    Now we come to the strictly less than case 10 < X < 20; so 10 and 20 are NO LONGER GOOD OUTCOMES. Remember that X is DISCRETE so 10 < X < 20 is the same as saying 11 <= X <= 19. Apply the same idea as before and your correction should again EXPAND this range to 10.5 <= X <= 19.5.

    Hope this helps, I could explain more, but I was trying to be brief (fail).

    -Nick

    P.S..if you're wondering what to do for a greater than/greater-than-or-equal-to…chances are you can flip your expression, ya dingus!

  4. Lona

    I have a question about continuity correction with the given values are discrete. How would you show/describe The Area between 36.5 and 37 or between 37 and 37.5? Is that even possible since the given value is known to be descrete?

  5. Muhammad Rafi

    I’ve seen the way in which the continuety correction is applied i.e. when to add 0.5 and when to subtract but I’ve still a question why to add 0.5 and why to subtract 0.5.
    Regards
    Muhammad Rafi
    Lecturer in Statistics

  6. saad saeed

    This was really comprehensive and precise! thank you so much to who ever who put this up!

  7. Emily

    Can I ask you something? Is this also correct?:

    If P(X=n) use P(n – 0.5 =< 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)

    What I mean is: Is P(X<5) the same as P(X=<5) (because P(X=5)=0 in a continuous distribution?)
    Please respond, because I see different things in different books and I'm confused!

  8. Lindsay Peterson

    Thank you! I finally found the right site to help me help me with my confusion. Very helpful!

  9. TB

    This site proved to be a godsend for me!
    I never expected to understand that part of the chapter through these simple notes!!!AWESOME!I’m greatly impressed!
    I thank you immensely for the precious help and just keep rocking!

  10. 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.

  11. 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!

  12. 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.

  13. 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.

  14. 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.

  15. Oscar M.

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