Why the Direction of the Correction Matters
A normal random variable is continuous, while a count such as the number of claims or successes takes discrete values. A continuity correction improves a normal approximation by placing its boundary halfway between adjacent possible values of the count. The correction must go in the right direction: including a boundary value and excluding it lead to different halfway points.
This tutorial focuses on choosing those points from the inequality itself. First identify the integer values that satisfy the event; then put the normal boundary halfway between the last included value and the first excluded value. This method is aligned with the SOA VEE Mathematical Statistics syllabus.
For instance, a count with possible values \(0,1,2,\ldots\) has unit spacing. If instead \(X\) can take only even values, adjacent possible values differ by two, and the halfway point is one unit from either value. Applying a half-unit correction to that even-valued variable would put the boundary in the wrong place.
How to Choose the Boundary
For a unit-spaced integer variable, rewrite the event in terms of the integers it includes. The corrected normal event extends to the midpoint between the included and excluded integers. The table assumes that each cutoff shown is an integer that the variable can take.
| Discrete event | Integers included | Normal event after correction |
|---|---|---|
| \(X\leq c\) | At most \(c\) | \(Y\leq c+0.5\) |
| \(X<c\) | At most \(c-1\) | \(Y\leq c-0.5\) |
| \(X\geq c\) | At least \(c\) | \(Y\geq c-0.5\) |
| \(X>c\) | At least \(c+1\) | \(Y\geq c+0.5\) |
| \(a\leq X\leq b\) | From \(a\) through \(b\) | \(a-0.5\leq Y\leq b+0.5\) |
| \(a<X\leq b\) | From \(a+1\) through \(b\) | \(a+0.5\leq Y\leq b+0.5\) |
| \(a\leq X<b\) | From \(a\) through \(b-1\) | \(a-0.5\leq Y\leq b-0.5\) |
| \(a<X<b\) | From \(a+1\) through \(b-1\) | \(a+0.5\leq Y\leq b-0.5\) |
The signs in the corrected normal event can be written as inclusive inequalities because \(Y\) is continuous: the probability that \(Y\) equals any particular boundary is zero. The important choice is the boundary location, not whether the normal inequality is written as strict or inclusive.
Why the Midpoint Rule Works
A discrete probability at a lattice value \(x\) can be viewed approximately as the normal area in a small interval centered at \(x\). For a lattice spacing \(h\), that interval runs from \(x-h/2\) to \(x+h/2\). Its width is \(h\), and it reaches exactly halfway to each neighboring lattice value.
For example, the probability of a unit-spaced count being at most \(c\) is the sum of the probabilities at \(c,c-1,c-2,\ldots\). Under the normal approximation, assign each of these values its centered interval. The interval for \(c\) ends at \(c+0.5\), and the interval for \(c-1\) meets it at \(c-0.5\). Together the intervals for all included values extend to \(c+0.5\) on the upper side. This motivates replacing the discrete event by a normal area ending at \(c+0.5\). The other entries in the table follow by identifying the first and last included values in the same way.
This midpoint argument explains the direction; it does not guarantee that the normal approximation is accurate. The approximation still depends on the distribution and its parameters. A continuity correction adjusts the boundary, but cannot repair a poor normal approximation.
Worked Examples
Worked Example: A Strict Lower-Tail Event
In a hypothetical quality-control process, 100 independent components are inspected. Each component independently passes inspection with probability \(0.5\). Let \(X\) be the number that pass, so \(X\sim\operatorname{Bin}(100,0.5)\). Use a normal approximation with a continuity correction to estimate \(P(X<48)\).
Method and model. The count is a sum of 100 independent Bernoulli variables with common success probability \(0.5\). Thus \(\mu=np=100(0.5)=50\), and \(\sigma^2=np(1-p)=100(0.5)(0.5)=25\), so \(\sigma=5\). The event is strict: \(X<48\) means \(X\leq47\), not \(X\leq48\).
Correct the boundary. The last included count is 47 and the next excluded count is 48. Their midpoint is \(47.5\), so the normal approximation uses \(P(Y\leq47.5)\), where \(Y\sim N(50,25)\).
Standardize and calculate.
The estimated probability is \(0.3085\), rounded to four decimal places. Common mistake avoided: using \(48.5\) would include the integer 48, contrary to the strict inequality. The correction goes down from 48 because the event includes only values below 48.
Worked Example: An Inclusive Upper-Tail Event
A hypothetical monitoring system records 100 independent daily indicators, each equal to 1 with probability \(0.5\) and 0 otherwise. Let \(X\) be their sum. Estimate the probability that at least 56 indicators equal 1.
Method and model. Here \(X\sim\operatorname{Bin}(100,0.5)\), with mean \(50\), variance \(25\), and standard deviation \(5\). “At least 56” means \(X\geq56\): the value 56 is included.
Correct the boundary. The first included integer is 56, and the preceding excluded integer is 55. Their midpoint is \(55.5\). Thus the approximating normal event is \(Y\geq55.5\), with \(Y\sim N(50,25)\).
Standardize and calculate.
The estimated probability is \(0.1357\), rounded to four decimal places. Common mistake avoided: using \(56.5\) would omit the probability mass associated with the included value 56. An inclusive upper-tail event therefore moves its cutoff down by \(0.5\).
Worked Example: An Interval with a Strict Lower Endpoint
In a hypothetical independent-trial process, each of 400 trials has probability \(0.5\) of producing a specified outcome. Let \(X\) count the outcomes. Estimate \(P(190<X\leq210)\) using a normal approximation and a continuity correction.
Method and model. The count has distribution \(X\sim\operatorname{Bin}(400,0.5)\). Its mean is \(400(0.5)=200\), its variance is \(400(0.5)(0.5)=100\), and its standard deviation is \(10\).
Identify the included integers. The strict lower endpoint excludes 190, so the first included integer is 191. The inclusive upper endpoint includes 210, which is the last included integer. The event is therefore \(191\leq X\leq210\).
Correct both boundaries. The lower midpoint is between 190 and 191, giving 190.5. The upper midpoint is between 210 and 211, giving 210.5. The approximating event is \(190.5\leq Y\leq210.5\), where \(Y\sim N(200,100)\).
Standardize and calculate.
The estimate is \(0.6821\), rounded to four decimal places. The difference was evaluated before rounding. Common mistake avoided: using 189.5 as the lower boundary would treat 190 as included; using 210.5 as the upper boundary is correct because 210 is included.
Worked Example: A Lattice with Spacing Two
Let \(K\sim\operatorname{Bin}(100,0.5)\), and define \(X=2K\). Thus \(X\) counts twice the number of successes and can take only even values. Estimate \(P(X\leq98)\) using a normal approximation.
Method and model. Since \(E[K]=50\) and \(\operatorname{Var}(K)=25\), scaling by 2 gives \(E[X]=2(50)=100\) and \(\operatorname{Var}(X)=2^2(25)=100\). The approximating normal variable is \(Y\sim N(100,100)\), with standard deviation 10. The possible values of \(X\) are spaced by \(h=2\).
Correct the boundary. The event \(X\leq98\) includes the even values through 98; the next possible value is 100. The midpoint is \((98+100)/2=99\). Equivalently, the correction is \(h/2=1\), so the normal event is \(Y\leq98+1=99\).
Standardize and calculate.
The estimated probability is \(0.4602\), rounded to four decimal places. Common mistake avoided: adding \(0.5\) would assume unit spacing. Here the adjacent possible values are 98 and 100, so the midpoint correction is one, not one-half.
A Reliable Procedure and Common Traps
On a problem involving a continuity correction, do not begin by memorizing whether the sign “usually” goes up or down. Translate the event into its included lattice values first. The following steps keep strictness, endpoint inclusion, and standardization separate.
Identify the model, its mean and variance, and the spacing \(h\) between adjacent possible values.
Translate each strict or inclusive inequality literally. For example, \(X>c\) begins at \(c+1\) for a unit-spaced integer count.
Move halfway between the included and excluded lattice values, a distance \(h/2\) from a lattice value.
Use the corrected boundary with the normal mean and standard deviation, then calculate the appropriate tail or interval probability.
Several errors are especially easy to make under time pressure:
- Ignoring “strict.” \(X<c\) excludes \(c\), while \(X\leq c\) includes it. Their corrected upper boundaries differ by one unit.
- Correcting an upper tail as if it were a lower tail. For \(X\geq c\), the first included value is \(c\), so the lower boundary is \(c-0.5\), not \(c+0.5\).
- Correcting only one endpoint of an interval. Determine the first and last included values separately; each endpoint can be strict or inclusive independently.
- Assuming every integer-valued variable has unit spacing. Check which values are actually possible. A variable restricted to even integers has spacing two.
- Rounding normal areas too early. Subtract unrounded cumulative probabilities when possible, and round only the final result.
- Treating the correction as a cure-all. The midpoint adjustment addresses discreteness. It does not make a poor normal approximation reliable or remove problems caused by an unsuitable model.
Check Your Understanding
For each problem, identify the included values before applying the correction. The answers use normal approximations and are rounded to four decimal places.
- A hypothetical count \(X\sim\operatorname{Bin}(64,0.5)\). Use a continuity correction to estimate \(P(X\leq30)\). (Answer: \(0.3538\))
- A hypothetical count \(X\sim\operatorname{Bin}(100,0.5)\). Estimate \(P(X>54)\) using a normal approximation with a continuity correction. (Answer: \(0.1841\))
- For \(X\sim\operatorname{Bin}(100,0.5)\), estimate \(P(45\leq X<55)\). (Answer: \(0.6803\))
- Let \(K\sim\operatorname{Bin}(64,0.5)\) and \(X=2K\). Use the correct lattice spacing to estimate \(P(X\geq68)\). (Answer: \(0.3538\))
- For \(X\sim\operatorname{Bin}(400,0.5)\), estimate \(P(190\leq X\leq210)\). Apply the correction at both inclusive endpoints and do not round the cumulative probabilities before subtracting. (Answer: \(0.7063\))