A continuity correction is applied when you want to use a continuous distribution to approximate a discrete distribution. Typically it is used when you want to use a normal distribution to approximate a binomial distribution.
Recall that the binomial distribution tells us the probability of obtaining x successes in n trials, given the probability of success in a single trial is p. To answer questions about probability with a binomial distribution we could simply use a Binomial Distribution Calculator, but we could also approximate the probability using a normal distribution with a continuity correction.
A continuity correction is the name given to adding or subtracting 0.5 to a discrete x-value.
For example, suppose we would like to find the probability that a coin lands on heads less than or equal to 45 times during 100 flips. That is, we want to find P(X ≤ 45). To use the normal distribution to approximate the binomial distribution, we would instead find P(X ≤ 45.5).
The following table shows when you should add or subtract 0.5, based on the type of probability you’re trying to find:
| Using Binomial Distribution | Using Normal Distribution with Continuity Correction |
|---|---|
| X = 45 | 44.5 < X < 45.5 |
| X ≤ 45 | X < 45.5 |
| X < 45 | X < 44.5 |
| X ≥ 45 | X > 44.5 |
| X > 45 | X > 45.5 |
Note:
It’s only appropriate to apply a continuity correction to the normal distribution to approximate the binomial distribution when n*p and n*(1-p) are both at least 5.
For example, suppose n = 15 and p = 0.6. In this case:
n*p = 15 * 0.6 = 9
n*(1-p) = 15 * (1 – 0.6) = 15 * (0.4) = 6
Since both of these numbers are greater than or equal to 5, it would be okay to apply a continuity correction in this scenario.
The following example illustrates how to apply a continuity correction to the normal distribution to approximate the binomial distribution.
Example of Applying a Continuity Correction
Suppose we want to know the probability that a coin lands on heads less than or equal to 43 times during 100 flips. In this case:
n = number of trials = 100
X = number of successes = 43
p = probability of success in a given trial = 0.50
We can plug these numbers into the Binomial Distribution Calculator to see that the probability of the coin landing on heads less than or equal to 43 times is 0.09667.
To approximate the binomial distribution by applying a continuity correction to the normal distribution, we can use the following steps:
Step 1: Verify that n*p and n*(1-p) are both at least 5.
n*p = 100*0.5 = 50
n*(1-p) = 100*(1 – 0.5) = 100*0.5 = 50
Both numbers are greater than or equal to 5, so we’re good to proceed.
Step 2: Determine if you should add or subtract 0.5
Referring to the table above, we see that we’re supposed to add 0.5 when we’re working with a probability in the form of X ≤ 43. Thus, we will be finding P(X< 43.5).
Step 3: Find the mean (μ) and standard deviation (σ) of the binomial distribution.
μ = n*p = 100*0.5 = 50
σ = √n*p*(1-p) = √100*.5*(1-.5) = √25 = 5
Step 4: Find the z-score using the mean and standard deviation found in the previous step.
z = (x – μ) / σ = (43.5 – 50) / 5 = -6.5 / 5 = -1.3.
Step 5: Use the Z table to find the probability associated with the z-score.
According to the Z table, the probability associated with z = -1.3 is 0.0968.
Thus, the exact probability we found using the binomial distribution was 0.09667 while the approximate probability we found using the continuity correction with the normal distribution was 0.0968. These two values are pretty close.
When to Use a Continuity Correction
Before modern statistical software existed and calculations had to be done manually, continuity corrections were often used to find probabilities involving discrete distributions. Today, continuity corrections play less of a role in computing probabilities since we can typically rely on software or calculators to calculate probabilities for us.
Instead, it’s simply a topic discussed in statistics classes to illustrate the relationship between a binomial distribution and a normal distribution and to show that it’s possible for a normal distribution to approximate a binomial distribution by applying a continuity correction.
Continuity Correction Calculator
Use the Continuity Correction Calculator to automatically apply a continuity correction to a normal distribution to approximate binomial probabilities.
Excellent notes about the issue that often is confused for students.
It would be nice to add rules and the examples of computing about + – 0.5 when the interest are probabilities for random variable being in different intervals (open, closed, half-open), even though they can be appropriately set using provided rules.
Hello,
I am currently taking a statistics class and i have an okay grasp on everything except what to do in the case the question asks for “exactly or equal to” during continuity corrections. I understand i do both +.5 and -.5 but what do i do with the 2 results aftwards? do i add them or subtract them or what?
I appreciate your help thank you.
find the probability in between that two point
Hello,
I wanted to know UNTIL WHEN we do continuity correction when having for instance a binomial distribution. In some exercise, the correction does it, but not always, especially if n (the number of trials) is very large…
Thanks a lot in advance!
Hi Elie…Continuity correction is applied when approximating a discrete distribution (such as a binomial distribution) with a continuous one (such as the normal distribution). The key question is when to apply it and when it becomes unnecessary, particularly in cases where \( n \), the number of trials, is large.
Here’s a detailed explanation:
—
### **What is Continuity Correction?**
The continuity correction accounts for the fact that the binomial distribution is discrete, while the normal distribution is continuous. To bridge the gap between these two, a correction of \( \pm 0.5 \) is applied to the discrete value when using the normal approximation.
For example:
– If you’re approximating \( P(X \leq k) \), you calculate \( P(X \leq k + 0.5) \) under the normal approximation.
– If approximating \( P(X > k) \), you calculate \( P(X \geq k + 0.5) \).
—
### **When to Use Continuity Correction?**
1. **Small or Moderate \( n \):**
– When \( n \) (number of trials) is small or moderate, and the binomial distribution is noticeably discrete (with visible “jumps” between possible values), the continuity correction is important to improve the approximation.
– Rule of thumb: Apply the correction if \( n \) is small enough that the binomial probabilities are not smoothly distributed.
2. **Large \( n \):**
– When \( n \) is large, the binomial distribution becomes nearly indistinguishable from a continuous normal distribution due to the Central Limit Theorem. In such cases, the continuity correction becomes negligible because the approximation is already very close.
– Rule of thumb: If \( n \) is large (e.g., \( n \geq 30 \)), the correction has minimal impact and is often skipped.
—
### **Key Factors Affecting Continuity Correction**
1. **\( np \) and \( n(1-p) \):**
– The normal approximation to the binomial is valid when \( np \geq 5 \) and \( n(1-p) \geq 5 \). If these conditions are met, the correction may not significantly affect the results.
– For smaller \( np \) or \( n(1-p) \), the correction is more important because the normal approximation may not fit well without it.
2. **Precision Required:**
– For practical purposes, continuity correction is often skipped in large sample scenarios, as the results are sufficiently close without it.
– However, in exercises or highly precise statistical contexts, you might still apply it to refine the approximation.
3. **Nature of the Problem:**
– If the discrete nature of the binomial distribution plays a critical role (e.g., in exact probability calculations or hypothesis testing), apply the correction.
– If the focus is on general trends or approximations, it may be omitted.
—
### **Illustrative Example**
#### Small \( n \):
Suppose \( n = 10 \), \( p = 0.5 \). To approximate \( P(X \leq 5) \) using the normal distribution:
– Mean \( \mu = np = 5 \)
– Standard deviation \( \sigma = \sqrt{np(1-p)} = 1.58 \)
– Without correction: \( P(X \leq 5) \approx P(Z \leq (5 – 5)/1.58) = P(Z \leq 0) = 0.5 \)
– With correction: \( P(X \leq 5) \approx P(Z \leq (5.5 – 5)/1.58) = P(Z \leq 0.32) = 0.6255 \)
The continuity correction significantly improves the approximation.
#### Large \( n \):
Suppose \( n = 100 \), \( p = 0.5 \). To approximate \( P(X \leq 50) \):
– Mean \( \mu = 50 \)
– Standard deviation \( \sigma = 5 \)
– Without correction: \( P(X \leq 50) \approx P(Z \leq (50 – 50)/5) = P(Z \leq 0) = 0.5 \)
– With correction: \( P(X \leq 50) \approx P(Z \leq (50.5 – 50)/5) = P(Z \leq 0.1) = 0.5398 \)
The difference is minimal, and in practice, the correction might be skipped.
—
### **Summary**
– **Apply continuity correction**: For small or moderate \( n \), or when \( np \) and \( n(1-p) \) are small.
– **Skip continuity correction**: For large \( n \) (e.g., \( n \geq 30 \)) or when the difference is negligible for your context.
– Always check \( np \) and \( n(1-p) \) conditions to determine the appropriateness of the normal approximation itself.