Translating Integer Events Into Normal Intervals

Why the Event Must Be Translated First

A count can take only integer values, but the normal random variable used to approximate it is continuous. Thus, before finding a normal probability, first determine exactly which integer values satisfy the stated event. Only then should you place the half-unit boundaries around those integers. This two-stage method prevents a common error: carrying the event’s original real-valued endpoints directly into a normal calculation.

Suppose $K$ is an integer-valued count and its distribution is approximated by a normal random variable $Y\sim N(\mu,\sigma^2)$, where $\mu$ is the approximating mean and $\sigma^2$ is the approximating variance. The standard deviation is $\sigma=\sqrt{\sigma^2}$. The normal variable $Y$ is continuous; it represents the count approximately, but it does not take only integer values.

Definition: To continuity-correct an event for an integer-valued variable $K$, first identify the smallest and largest permitted integers, say $L$ and $U$. Replace the event $L\le K\le U$ by the normal interval $L-0.5\le Y\le U+0.5$. The endpoints are half a unit beyond the extreme included integers.

This conversion applies when the count values are consecutive integers from $L$ through $U$. A one-sided event is handled in the same way: for example, $K\le U$ becomes $Y\le U+0.5$, and $K\ge L$ becomes $Y\ge L-0.5$. If no integer satisfies the event, its probability is zero; there is no nonempty interval to approximate.

Translate the Inequalities Into Integers

For a real number $x$, $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$, and $\lceil x\rceil$ is the least integer greater than or equal to $x$. These functions make the treatment of strict and inclusive endpoints precise.

Rule: For an integer-valued $K$, the permitted integers are given by $$ \begin{aligned} a\le K\le b &\quad\Longleftrightarrow\quad \lceil a\rceil\le K\le\lfloor b\rfloor,\\ a<K\le b &\quad\Longleftrightarrow\quad \lfloor a\rfloor+1\le K\le\lfloor b\rfloor,\\ a\le K<b &\quad\Longleftrightarrow\quad \lceil a\rceil\le K\le\lceil b\rceil-1,\\ a<K<b &\quad\Longleftrightarrow\quad \lfloor a\rfloor+1\le K\le\lceil b\rceil-1. \end{aligned} $$ After finding the integer bounds, apply the half-unit correction to those bounds, not to $a$ and $b$.

The strictness matters when an endpoint is itself an integer. For instance, $K<26$ excludes 26, so its largest permitted value is 25. In contrast, $K\le26$ includes 26. If an endpoint is not an integer, strict and inclusive versions at that endpoint can permit exactly the same integers: because $K$ cannot equal 18.2, $K\ge18.2$ and $K>18.2$ both require $K\ge19$.

Why the Half-Unit Boundaries Work

Each integer $k$ is represented by the interval from $k-0.5$ to $k+0.5$. These intervals meet at half-integers, so the integers $L,L+1,\ldots,U$ together correspond to one continuous span from $L-0.5$ to $U+0.5$. This is the basis of the correction; the choice of whether to include a boundary does not affect a continuous normal probability.

\bigcup_{k=L}^{U}[k-0.5,k+0.5) =[L-0.5,U+0.5)

The half-open intervals on the left are nonoverlapping and cover the full span on the right. The event for $K$ includes exactly the integer values $L$ through $U$; the corresponding span for $Y$ extends half a unit beyond each extreme. Since a continuous normal variable has probability zero at any single endpoint, the probability of the corrected interval can be written with either strict or inclusive endpoint notation.

Once the corrected endpoints are set, standardize them using the normal approximation. If $Y\sim N(\mu,\sigma^2)$ and $\Phi$ is the standard normal cumulative distribution function, then

P(L\le K\le U)\ \approx\ \Phi\left(\frac{U+0.5-\mu}{\sigma}\right) - \Phi\left(\frac{L-0.5-\mu}{\sigma}\right).

This probability calculation uses a normal approximation already specified for $K$. Translating the event correctly does not, by itself, establish that a normal approximation is adequate. The examples below take the stated normal model as the supplied approximation and focus on translating the event.

Worked Applications

Worked Example: A Bounded Event With Mixed Endpoint Types

In a hypothetical portfolio, let $K$ be the number of reported claims in a period. For this calculation, suppose $K$ is approximated by $Y\sim N(22,4)$, so the approximating mean is 22 and the standard deviation is $\sqrt{4}=2$. Approximate $P(18.2\le K<26)$.

Method and integer bounds. The lower endpoint is included, so the least permitted integer is $\lceil18.2\rceil=19$. The upper endpoint is strict and is the integer 26, so the greatest permitted integer is 25. The event is therefore $19\le K\le25$.

Apply the correction. The corresponding normal interval is $18.5\le Y\le25.5$. Standardizing its endpoints gives

z_{\mathrm{lower}}=\frac{18.5-22}{2}=-1.75,\qquad z_{\mathrm{upper}}=\frac{25.5-22}{2}=1.75.

Thus the probability is approximately $\Phi(1.75)-\Phi(-1.75)=0.9599408-0.0400592=0.9198816$, or 0.9199 to four decimal places.

Check and common mistake avoided. The included integers are exactly 19 through 25, and the corrected boundaries lie half a unit outside them. Using 18.2 and 26 as the normal endpoints would fail to reflect which integer counts are included.

Worked Example: A Strict Lower Bound and an Inclusive Upper Bound

A hypothetical service team records the number $K$ of equipment alerts received in a shift. Suppose the normal approximation is $Y\sim N(7.5,4)$, with mean 7.5 and standard deviation 2. Find the approximate probability of $4.6<K\le10.2$.

Method and integer bounds. The strict inequality $K>4.6$ requires $K\ge\lfloor4.6\rfloor+1=5$. The inclusive inequality $K\le10.2$ allows $K\le\lfloor10.2\rfloor=10$. Thus the event is $5\le K\le10$.

Apply the correction and standardize. The corrected interval is $4.5\le Y\le10.5$. Its standardized endpoints are

z_{\mathrm{lower}}=\frac{4.5-7.5}{2}=-1.5,\qquad z_{\mathrm{upper}}=\frac{10.5-7.5}{2}=1.5.

The probability is approximately $\Phi(1.5)-\Phi(-1.5)=0.9331928-0.0668072=0.8663856$, or 0.8664 to four decimal places.

Check and common mistake avoided. The permitted counts are 5, 6, 7, 8, 9, and 10. The strict lower bound does not eliminate 5, because 5 is greater than 4.6. Applying a strict-bound rule as though 4.6 were an integer would be an unnecessary extra adjustment.

Worked Example: An Integer Endpoint in a Half-Open Interval

In a hypothetical quality audit, let $K$ be a count of inspected items meeting a criterion. Suppose $Y\sim N(32.5,9)$ approximates $K$, with mean 32.5 and standard deviation 3. Approximate $P(30\le K<36)$.

Method and integer bounds. The inclusive lower endpoint permits 30. The strict upper endpoint excludes 36, so the largest permitted integer is 35. Hence the event is $30\le K\le35$.

Apply the correction and standardize. The normal interval is $29.5\le Y\le35.5$. The standardized endpoints are

z_{\mathrm{lower}}=\frac{29.5-32.5}{3}=-1,\qquad z_{\mathrm{upper}}=\frac{35.5-32.5}{3}=1.

Therefore, the approximate probability is $\Phi(1)-\Phi(-1)=0.8413447-0.1586553=0.6826894$, or 0.6827 to four decimal places.

Check and common mistake avoided. There are six included integer values, from 30 to 35, and the corrected interval is centered at 32.5. Using 36 as the upper integer would incorrectly include the count excluded by the strict inequality.

A Reliable Calculation Sequence

For a count event with one or two bounds, keep the logical translation separate from the probability calculation. This makes it easier to find an endpoint error before it affects the final answer.

Read the inequality signs.
Mark which endpoints are included and which are excluded. Pay particular attention when an endpoint is an integer.

Find the extreme permitted integers.
Use the appropriate floor or ceiling rule at each endpoint. If the lower integer exceeds the upper integer, the event is empty.

Move out by one-half.
For included integers from $L$ to $U$, use the normal interval from $L-0.5$ to $U+0.5$.

Standardize and evaluate.
Subtract the approximating mean from each corrected endpoint, divide by the standard deviation, and take the difference of the corresponding standard normal probabilities.

Common Mistakes and a Final Check

The most tempting shortcut is to treat the original inequality as though $K$ were continuous. For example, $18.2\le K<26$ does not translate to $18.2\le Y\le26$. The count values satisfying the event are 19 through 25, so the normal interval is $18.5\le Y\le25.5$. The calculation with the original endpoints uses different boundaries and therefore approximates a different event.

Do not correct before identifying the integers. First find the smallest and largest allowed counts; only then subtract or add 0.5.
Do not ignore strictness at an integer endpoint. The events $K<26$ and $K\le26$ have different largest allowed integers.
Do not assume strictness changes a noninteger endpoint. For integer $K$, $K>4.6$ and $K\ge4.6$ both require $K\ge5$.
Check that the integer range is nonempty. A real interval can contain no integers, in which case its event probability is zero.
Keep the normal parameters clear. In $N(\mu,\sigma^2)$, the second parameter is the variance; divide by $\sigma$, not by $\sigma^2$, when standardizing.

Key takeaway: For a discrete count, the inequality signs determine the permitted integers; those integers determine the half-unit normal boundaries. Translate first, correct second, and calculate last. This event-to-interval method is aligned with the SOA Exam P (Probability) syllabus.

Check Your Understanding

For each problem, identify the integer event before applying the continuity correction. The stated normal model is the approximation to use.

Suppose $K$ is approximated by $Y\sim N(12,4)$. Approximate $P(K\le14.3)$. (Answer: corrected interval $Y\le14.5$; probability 0.8944.)
Suppose $K$ is approximated by $Y\sim N(12.5,6.25)$. Approximate $P(8<K\le17)$. (Answer: integers 9 through 17, corrected interval $8.5\le Y\le17.5$; probability 0.9225.)
Suppose $K$ is approximated by $Y\sim N(6,1)$. Approximate $P(3.1\le K<8.8)$. (Answer: integers 4 through 8, corrected interval $3.5\le Y\le8.5$; probability 0.9876.)
Suppose $K$ is approximated by $Y\sim N(44,4)$. Approximate $P(40\le K\le47)$. (Answer: corrected interval $39.5\le Y\le47.5$; probability 0.9477.)
For integer-valued $K$, determine $P(10.2<K<10.9)$ without a normal calculation. (Answer: no integer satisfies the event, so the probability is 0.)

What You'll Learn