Why a Continuous Approximation Needs a Correction
A binomial count can take only integer values, while a normal random variable can take any real value. When a normal distribution is used to approximate a binomial distribution, treating an integer count as if it were a point on a continuous scale can omit or misallocate probability. A continuity correction accounts for the width of the discrete count categories by moving a boundary by half a count.
This adjustment is useful both when approximating binomial probabilities and when constructing a continuity-corrected approximate interval for a proportion. It improves the alignment between the discrete count scale and the continuous normal scale; it does not make the approximation exact or remove the need to check its conditions.
Choosing the Correct Half-Count Boundaries
Think of the probability at each integer count \(k\) as occupying the interval from \(k-0.5\) to \(k+0.5\) on a continuous number line. The normal approximation assigns probability to that interval rather than to the single point \(k\). This gives the following boundary rules.
| Binomial event | Normal event with continuity correction |
|---|---|
| \(X\leq k\) | \(Y\leq k+0.5\) |
| \(X<k\) | \(Y\leq k-0.5\) |
| \(X\geq k\) | \(Y\geq k-0.5\) |
| \(X>k\) | \(Y\geq k+0.5\) |
| \(a\leq X\leq b\) | \(a-0.5\leq Y\leq b+0.5\) |
Here \(Y\) is a normal random variable with mean \(np\) and variance \(np(1-p)\). For example, the event \(10\leq X\leq15\) includes the six integer outcomes 10, 11, 12, 13, 14, and 15. Their combined continuous-scale region runs from 9.5 through 15.5, not from 10 through 15.
Worked Example: Approximating a Binomial Probability
A hypothetical inspection finds a conforming part with probability \(p=0.30\), independently for each part. For \(n=40\) inspected parts, estimate the probability that the number \(X\) of conforming parts is between 10 and 15 inclusive, using a normal approximation with continuity correction.
Method and model. The count has distribution \(X\sim\operatorname{Bin}(40,0.30)\). The mean and variance are \(np=40(0.30)=12\) and \(np(1-p)=40(0.30)(0.70)=8.4\), so the standard deviation is \(\sqrt{8.4}\approx2.8983\). These parameters permit a normal approximation as a working estimate; the correction accounts for the integer boundaries.
Correct the event boundaries. For the inclusive event \(10\leq X\leq15\), use \(9.5\leq Y\leq15.5\). Thus
Standardize and evaluate. The upper standardized boundary is \(3.5/\sqrt{8.4}\approx1.2076\). The lower one is \(-2.5/\sqrt{8.4}\approx-0.8626\). Using standard normal probabilities to four decimals gives \(\Phi(1.2076)\approx0.8864\) and \(\Phi(-0.8626)\approx0.1942\). Therefore the estimated probability is \(0.8864-0.1942=0.6922\).
Check and common mistake. The estimate lies between 0 and 1, as a probability must. Using \(10.5\) and \(14.5\) instead would shrink the event and would correspond to excluding part of the probability for the endpoint counts. For an inclusive range, extend outward to \(a-0.5\) and \(b+0.5\).
One-Sided Events and the Direction of the Correction
The direction depends on whether the event includes or excludes its endpoint. For \(X\leq k\), the count \(k\) is included, so its continuous-scale cell extends up to \(k+0.5\). For \(X\geq k\), the cell at \(k\) extends down to \(k-0.5\). Reversing either direction changes the event being approximated.
Worked Example: A Lower-Tail Probability
A hypothetical quality-control process produces an acceptable item with probability \(p=0.10\), independently from item to item. If \(n=20\) items are inspected, estimate \(P(X\leq3)\), where \(X\) is the number of acceptable items.
Method. Under the model \(X\sim\operatorname{Bin}(20,0.10)\), the mean is \(np=2\), and the variance is \(np(1-p)=20(0.10)(0.90)=1.8\). The standard deviation is \(\sqrt{1.8}\approx1.3416\). Because the event includes 3, its corrected upper boundary is \(3.5\).
The standardized boundary is \((3.5-2)/\sqrt{1.8}\approx1.1180\). Consequently, \(P(X\leq3)\approx\Phi(1.1180)\approx0.8686\), to four decimals.
Check and common mistake. The event is a lower tail, so the approximation is a single cumulative probability. Using 2.5 as the boundary would exclude the count \(X=3\) and approximate \(P(X\leq2)\) instead. Also note that \(np(1-p)=1.8\) is small: this normal estimate should be treated cautiously. A half-count adjustment cannot compensate for a poor normal approximation.
Continuity Correction for a Proportion Interval
Suppose \(X\sim\operatorname{Bin}(n,p)\) and the observed number of successes is \(x\). The sample proportion is \(\hat p=x/n\). The ordinary Wald interval uses the estimated standard error \(\sqrt{\hat p(1-\hat p)/n}\) and a normal critical value \(z_{1-\alpha/2}\), where the desired confidence level is \(1-\alpha\). A continuity-corrected Wald interval adds half a count, expressed as a proportion, to the usual margin.
The adjustment has a count-scale explanation. The usual Wald margin corresponds to \(z_{1-\alpha/2}\sqrt{n\hat p(1-\hat p)}\) counts. Adding half a count gives an adjusted count margin; dividing by \(n\) converts it to a proportion margin of \(0.5/n\). The interval therefore extends by \(0.5/n\) at each end compared with the ordinary Wald interval. Its total width increases by \(1/n\).
This formula is a practical plug-in correction: it uses the estimated proportion in the standard error rather than solving a binomial probability equation for each possible value of \(p\). It is therefore an approximate Wald-type procedure, not a unique consequence of exact inversion. The correction can improve the normal approximation's treatment of the discrete count scale, but it does not guarantee exact or uniformly better coverage.
Worked Example: A Continuity-Corrected 95% Proportion Interval
In a hypothetical poll, \(x=45\) of \(n=150\) respondents support a proposal. Calculate the continuity-corrected 95% Wald interval, using \(z_{0.975}=1.96\). Round the endpoints to four decimal places.
Method and estimate. The sample proportion is \(\hat p=45/150=0.30\). The estimated standard error is \(\sqrt{0.30(0.70)/150}=\sqrt{0.0014}\approx0.0374166\).
Calculate the margin. The usual normal margin is \(1.96(0.0374166)\approx0.0733365\). The half-count adjustment on the proportion scale is \(0.5/150\approx0.0033333\). The corrected margin is therefore \(0.0733365+0.0033333=0.0766698\).
Calculate and report the endpoints. The lower endpoint is \(0.30-0.0766698=0.2233302\), and the upper endpoint is \(0.30+0.0766698=0.3766698\). Rounded to four decimal places, the interval is \((0.2233,\,0.3767)\).
Check and common mistake. The interval is symmetric about \(0.30\), and its margin is larger than the uncorrected Wald margin. Omitting \(0.5/n\) would instead give approximately \((0.2267,\,0.3733)\), an interval that is too narrow for the specified corrected procedure.
What the Correction Does—and Does Not Do
For a probability calculation, the half-count adjustment matches the edges of integer-valued outcomes to the continuous regions used by the normal distribution. For an approximate proportion interval, adding \(0.5/n\) at both ends makes the interval wider than the corresponding uncorrected Wald interval. These changes can improve approximate calibration when discreteness matters, especially when the sample size is not very large.
The word coverage refers to the probability, under repeated sampling from the stated model, that a random interval contains the true parameter \(p\). A nominal 95% interval is designed to have coverage near 0.95 under its approximation and assumptions. Since a binomial count is discrete, its attainable tail probabilities change in steps; an approximate interval's actual coverage can consequently differ from its nominal level. The continuity correction acknowledges the count spacing but does not ensure exact 95% coverage for every \(n\) and \(p\).
Before using a normal approximation for a binomial count, check that the sample size and success probability provide adequate expected counts; a common working check is that both \(np\) and \(n(1-p)\) are at least 10. For an interval, the corresponding check is often made with \(n\hat p\) and \(n(1-\hat p)\). A continuity correction does not repair a failed condition. Nor does it prevent a Wald interval from extending below 0 or above 1. Near a boundary, the earlier material on Wilson intervals is relevant because the ordinary Wald construction can behave poorly there.
Worked Example: A Corrected 90% Interval With Few Successes
A hypothetical service survey records 5 favorable responses among 50 respondents. Calculate the continuity-corrected 90% Wald interval using \(z_{0.95}=1.645\), and round the endpoints to four decimal places.
Estimate and standard error. Here \(\hat p=5/50=0.10\). The estimated standard error is \(\sqrt{0.10(0.90)/50}=\sqrt{0.0018}\approx0.0424264\).
Corrected margin. The normal part of the margin is \(1.645(0.0424264)\approx0.0697914\). The half-count adjustment is \(0.5/50=0.0100\), so the total margin is \(0.0697914+0.0100=0.0797914\).
The endpoints are \(0.10-0.0797914=0.0202086\) and \(0.10+0.0797914=0.1797914\). To four decimal places, the interval is \((0.0202,\,0.1798)\). It remains within \([0,1]\), but only five observed successes make a normal-based interval questionable; the corrected arithmetic does not override the approximation check.
Common Exam Traps
- Moving an inclusive range inward. For \(a\leq X\leq b\), use \(a-0.5\) and \(b+0.5\), not \(a+0.5\) and \(b-0.5\).
- Using the wrong side for a tail. \(X\leq k\) uses \(k+0.5\); \(X\geq k\) uses \(k-0.5\).
- Forgetting the scale change in an interval. A half-count becomes \(0.5/n\) on the proportion scale, not \(0.5\).
- Claiming exact coverage. The continuity-corrected Wald interval is an approximation. Its actual coverage need not equal its nominal confidence level.
- Assuming correction cures every weakness. It does not ensure adequate expected counts or keep endpoints inside \([0,1]\).
Check Your Understanding
For probability questions, use the normal approximation and continuity correction. Round probabilities to four decimals and interval endpoints to four decimals.
- A hypothetical batch has \(n=30\) items, each conforming independently with probability \(p=0.40\). Estimate \(P(8\leq X\leq15)\) using a normal approximation with a continuity correction. (Answer: approximately \(0.8571\))
- For a hypothetical independent Bernoulli process with \(n=60\) and \(p=0.20\), estimate \(P(X\leq8)\) using a continuity correction. (Answer: approximately \(0.1293\))
- In a hypothetical sample, \(x=32\) of \(n=200\) respondents select an option. Using \(z_{0.975}=1.96\), calculate the continuity-corrected 95% Wald interval for \(p\). (Answer: approximately \((0.1067,\,0.2133)\))
- A hypothetical inspection records \(x=6\) conforming parts among \(n=80\). Using \(z_{0.95}=1.645\), calculate the continuity-corrected 90% Wald interval. (Answer: approximately \((0.0203,\,0.1297)\))
- A count \(X\) is approximated by a normal distribution with mean 50 and standard deviation 5. Use a continuity correction to estimate \(P(45\leq X\leq55)\). (Answer: approximately \(0.7287\))