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Exam P · Central Limit and Normal Approximation · Tutorial 840 of 915

Continuity Corrections for Count Tails

Use the correct half-unit boundary for each count tail, then standardize and evaluate the corresponding normal probability.

Advanced 10 min read

What You'll Learn

  • Translate inclusive and strict count inequalities into the correct continuity-corrected boundaries.
  • Calculate lower-tail and upper-tail probabilities using a normal approximation.
  • Convert a count interval or two-sided tail event into one or more normal intervals.
  • Check whether a tail calculation should use a CDF or its complement.
  • Recognize why using the integer cutoff itself changes the estimated probability.

Why Count Tails Need Careful Boundaries

A normal approximation replaces an integer-valued count with a continuous random variable. That change is useful for calculation, but it creates a boundary question: should an event such as \(X\leq k\) be represented by a normal area ending at \(k\), or somewhere nearby? The continuity correction answers that question. A half-unit adjustment lets the continuous normal curve represent the whole probability mass associated with each integer count.

This tutorial focuses on using that adjustment for lower tails, upper tails, and two-sided events. In every example, assume the count has mean \(\mu\) and standard deviation \(\sigma\), and that the normal approximation is appropriate for the setting. Let \(Y\sim N(\mu,\sigma^2)\) be the approximating normal variable, and let \(\Phi(z)\) denote the standard normal cumulative distribution function.

Definition: For an integer-valued count \(X\), a continuity correction replaces the integer values included in an event by adjacent continuous intervals whose boundaries lie halfway between integers. Thus the count value \(j\) is represented by the interval from \(j-0.5\) to \(j+0.5\) on the continuous scale.

Boundary Rules for Count Tails

For an inclusive lower tail, the event \(X\leq k\) includes every integer through \(k\). Its continuous counterpart therefore extends through \(k+0.5\), the point halfway between \(k\) and \(k+1\). An inclusive upper tail \(X\geq k\) starts halfway between \(k-1\) and \(k\), at \(k-0.5\). A strict inequality changes which integer is the first or last included value.

$$ \begin{aligned} P(X\leq k) &\approx P(Y\leq k+0.5) =\Phi\!\left(\frac{k+0.5-\mu}{\sigma}\right),\\ P(X<k) &\approx P(Y\leq k-0.5) =\Phi\!\left(\frac{k-0.5-\mu}{\sigma}\right),\\ P(X\geq k) &\approx P(Y\geq k-0.5) =1-\Phi\!\left(\frac{k-0.5-\mu}{\sigma}\right),\\ P(X>k) &\approx P(Y\geq k+0.5) =1-\Phi\!\left(\frac{k+0.5-\mu}{\sigma}\right). \end{aligned} $$

These formulas can be remembered by first identifying the included integers, rather than memorizing four separate expressions. If the smallest included integer is \(L\) and the largest is \(U\), the corresponding continuous interval is from \(L-0.5\) to \(U+0.5\). For example, \(a\leq X\leq b\) becomes \(a-0.5\leq Y\leq b+0.5\). The normal distribution assigns probability zero to a single endpoint, so it makes no difference whether the corrected continuous endpoint is written with \(<\) or \(\leq\).

Why the Half-Unit Shift Works

The correction follows from assigning each possible count its own unit-width interval. The integer \(j\) is represented by the interval \([j-0.5,j+0.5]\). These intervals meet without gaps or overlap, and the interval for \(j\) is centered at the count value itself.

1
Identify the included integers.
For \(X\leq k\), the included values are all integers up to and including \(k\).
2
Join their unit-width intervals.
The interval for the largest included value \(k\) ends at \(k+0.5\). The intervals for all smaller integers join it, giving the continuous region to the left of that boundary.
3
Evaluate the normal area.
Replacing the count probabilities by the corresponding normal area gives \(P(X\leq k)\approx P(Y\leq k+0.5)\). Standardizing that boundary produces the lower-tail formula.

The same reasoning applies to any set of included integers. In a two-sided event, translate each tail or interval separately and combine the resulting normal areas. This is safer than trying to adjust a final z-score from memory.

Worked Example: An Inclusive Lower Tail

A hypothetical inspection process records the number \(X\) of flagged items among 400 items. Suppose \(X\sim\operatorname{Bin}(400,0.50)\), and use a normal approximation to estimate the probability that at most 190 items are flagged.

Method. The event is \(X\leq190\), an inclusive lower tail. First calculate the normal parameters:

$$ \mu=np=400(0.50)=200,\qquad \sigma^2=np(1-p)=400(0.50)(0.50)=100,\qquad \sigma=10. $$

The last included count is 190, so the corrected boundary is \(190+0.5=190.5\). Standardizing gives

$$ z=\frac{190.5-200}{10}=-0.95. $$

Therefore,

$$ P(X\leq190)\approx\Phi(-0.95)=0.171056\ldots\approx\boxed{0.1711}. $$

The estimate is a probability between 0 and 1, and its direction is sensible: 190 is below the mean of 200, so the lower-tail probability is less than 0.5. The calculation avoids the common mistake of using 190 itself as the continuous endpoint.

Worked Example: A Strict Upper Tail

A hypothetical batch review counts \(X\), the number of items requiring a second check, among 400 items. Assume \(X\sim\operatorname{Bin}(400,0.50)\). Estimate the probability that more than 210 items require a second check.

Method. Since \(X\) is integer-valued, \(X>210\) means \(X\geq211\). The smallest included count is 211, so the corrected normal region begins at \(210.5\). The parameters remain \(\mu=200\) and \(\sigma=10\), giving

$$ z=\frac{210.5-200}{10}=1.05. $$

This is an upper tail, so use the complement of the standard normal CDF:

$$ P(X>210)\approx P(Y\geq210.5) =1-\Phi(1.05) =1-0.8531409\ldots =0.1468591\ldots \approx\boxed{0.1469}. $$

The event lies above the mean, so an estimated probability below 0.5 is appropriate. The strict inequality matters: \(X>210\) starts at 211, not at 210.

Worked Example: A Two-Sided Count Interval

In a hypothetical audit, let \(X\) be the number of records with a specified feature among 400 independently selected records, with \(X\sim\operatorname{Bin}(400,0.50)\). Estimate the probability that the count is between 190 and 210, inclusive.

Method. The included integers run from 190 through 210. The corrected normal interval is therefore \([189.5,210.5]\). The mean is 200 and the standard deviation is 10, so the standardized endpoints are

$$ z_{\mathrm{low}}=\frac{189.5-200}{10}=-1.05,\qquad z_{\mathrm{high}}=\frac{210.5-200}{10}=1.05. $$

Subtract the lower-tail area from the upper endpoint’s cumulative probability:

$$ \begin{aligned} P(190\leq X\leq210) &\approx \Phi(1.05)-\Phi(-1.05)\\ &=0.8531409\ldots-0.1468591\ldots\\ &=0.7062818\ldots\\ &\approx\boxed{0.7063}. \end{aligned} $$

The final probability is rounded to four decimal places. The interval is centered on the mean, and its estimated probability is greater than 0.5, as expected. Retaining unrounded CDF values until the subtraction is important: subtracting separately rounded values can change the final fourth decimal place.

Two-Sided Events and Complements

A two-sided event can be expressed either as an interval or as the union of two tails. For a central interval \(a\leq X\leq b\), correct both endpoints: use \(a-0.5\) at the lower end and \(b+0.5\) at the upper end. For an event outside an interval, such as \(X\leq a\) or \(X\geq b\), the two tail regions are disjoint when \(a<b\), so their probabilities can be added.

$$ \begin{aligned} P(X\leq a\ \text{or}\ X\geq b) &\approx \Phi\!\left(\frac{a+0.5-\mu}{\sigma}\right) + 1-\Phi\!\left(\frac{b-0.5-\mu}{\sigma}\right), \qquad a<b. \end{aligned} $$

The formula assumes the two count tails do not overlap, which is ensured by \(a<b\). If an event is easier to express as a complement, identify exactly which integer values are excluded, correct the complement’s interval, and subtract its probability from 1. The complement changes the event being calculated; it does not remove the need for a continuity correction.

Worked Example: Two Disjoint Extreme Tails

For a hypothetical count \(X\sim\operatorname{Bin}(400,0.50)\), estimate the probability that the count is at most 189 or at least 211.

Method. The two events are disjoint. The lower tail \(X\leq189\) ends at 189.5, while the upper tail \(X\geq211\) begins at 210.5. With \(\mu=200\) and \(\sigma=10\), their z-scores are

$$ z_{\mathrm{low}}=\frac{189.5-200}{10}=-1.05,\qquad z_{\mathrm{high}}=\frac{210.5-200}{10}=1.05. $$

Add the two tail probabilities:

$$ \begin{aligned} P(X\leq189\ \text{or}\ X\geq211) &\approx\Phi(-1.05)+1-\Phi(1.05)\\ &=0.1468591\ldots+0.1468591\ldots\\ &=0.2937182\ldots\\ &\approx\boxed{0.2937}. \end{aligned} $$

The answer is twice the one-sided tail area from the preceding calculation because the boundaries are equally far from the mean. It is less than 1, and the two probabilities are added because the events cannot occur together.

Common Mistake: Using the Integer Cutoff

For the strict upper-tail example, a tempting but incorrect calculation is to use 210 as the normal boundary:

$$ 1-\Phi\!\left(\frac{210-200}{10}\right) =1-\Phi(1.00) =1-0.8413447\ldots =0.1587. $$

That calculation represents an upper normal tail beginning at 210, but the event \(X>210\) contains only integer counts beginning at 211. Its corrected boundary is 210.5, giving \(1-\Phi(1.05)\approx0.1469\). The uncorrected result is not the requested continuity-corrected estimate. The general safeguard is to translate the inequality into the first or last integer included before choosing a boundary.

A Reliable Tail-Calculation Sequence

For an exam calculation, write the event in integer form first, then proceed in order. This separates the interpretation of strictness from the normal arithmetic.

  1. Rewrite strict inequalities. For integer \(X\), \(X>k\) is \(X\geq k+1\), and \(X<k\) is \(X\leq k-1\).
  2. Find the included endpoint. Identify the first included integer for an upper tail, or the last included integer for a lower tail.
  3. Move half a unit outward. Extend the normal region to \(k+0.5\) for \(X\leq k\), or start it at \(k-0.5\) for \(X\geq k\).
  4. Standardize with the count’s parameters. Calculate \(z=(c-\mu)/\sigma\) at each corrected boundary \(c\).
  5. Match the probability operation to the event. Use \(\Phi(z)\) for a lower tail, \(1-\Phi(z)\) for an upper tail, a difference for an interval, and a sum for disjoint tails.
Key takeaway: Apply the continuity correction to the count boundary before standardizing. Inclusive lower tails end at \(k+0.5\); inclusive upper tails begin at \(k-0.5\). For strict inequalities, first determine which integer counts are actually included.

Check Your Understanding

For each problem, use the stated normal approximation and apply a continuity correction. Round probabilities to four decimal places.

  1. Suppose \(X\sim\operatorname{Bin}(400,0.50)\). Estimate \(P(X<190)\). (Answer: 0.1469)
  2. Suppose \(X\sim\operatorname{Bin}(400,0.50)\). Estimate \(P(X\geq210)\). (Answer: 0.1711)
  3. Suppose \(X\sim\operatorname{Bin}(400,0.50)\). Estimate \(P(194\leq X\leq206)\). The corrected z-score endpoints are \(-0.65\) and \(0.65\). (Answer: 0.4843)
  4. Suppose \(X\sim\operatorname{Bin}(400,0.50)\). Estimate \(P(X\leq189\text{ or }X\geq211)\). (Answer: 0.2937)
  5. Suppose \(X\sim\operatorname{Bin}(400,0.50)\). Estimate \(P(195\leq X\leq205)\). The corrected z-score endpoints are \(-0.55\) and \(0.55\). (Answer: 0.4177)

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