Why Work With Proportions?
A sample proportion is often easier to interpret than a count: it describes the share of trials that produce a specified outcome. But the proportion is not continuous. With a fixed sample size, it can take only values spaced \(1/n\) apart. For probability calculations, the most reliable route is to translate the proportion event into an event about a binomial count, then approximate that count with a normal distribution.
The central practical issue is the boundary. For example, “at least 53%” in 400 trials means at least 212 successes—not a continuous interval beginning at 53%. A continuity correction places the normal boundary halfway between adjacent possible counts. This tutorial develops that translation and standardization carefully. The resulting procedure is aligned with the SOA Exam P (Probability) syllabus.
The normal approximation uses \(K\) with mean \(np\) and standard deviation \(\sqrt{np(1-p)}\). It is appropriate when the binomial distribution is not too concentrated near an endpoint; a common practical check is that both \(np\) and \(n(1-p)\) are at least 10. This check is a guide, not a guarantee of high accuracy. The count-boundary continuity correction, introduced in the earlier material on continuity corrections, is essential to the usual approximation.
From a Proportion Event to a Count Event
Since \(\hat p=K/n\) and \(n\) is positive, a proportion inequality can be multiplied by \(n\) without changing its direction. The resulting event must then be expressed using integer values of \(K\). For instance, \(K\le 51.7\) means \(K\le 51\), because \(K\) can only be an integer.
The bounds \(L\) and \(U\) are integers after the event has been translated. If the original proportion threshold does not correspond to an attainable proportion, the integer event still determines the correction. Strict and inclusive inequalities can produce different integer events, so do not apply a correction until that distinction is resolved.
Confirm that the trials are independent, have the same success probability \(p\), and that \(n\) is fixed. Then \(K\sim\operatorname{Bin}(n,p)\) and \(\hat p=K/n\).
Multiply proportion boundaries by \(n\), preserve strict or inclusive signs, and identify the attainable integer values of \(K\).
Move each integer boundary outward by 0.5: subtract 0.5 from a lower included count or add 0.5 to an upper included count.
Use \(Z=(K-np)/\sqrt{np(1-p)}\) at the corrected boundary or boundaries, then evaluate the corresponding standard normal probability.
Why the Correction Is Half a Count
The binomial count has probability at integer points, while a normal variable spreads probability continuously over intervals. To approximate the probability assigned to an integer count \(k\), associate that count with the interval from \(k-0.5\) to \(k+0.5\). The interval has midpoint \(k\) and extends halfway toward each neighboring integer.
For a cumulative event \(K\le U\), the integer bars for \(0,1,\ldots,U\) together reach from below the first count to \(U+0.5\). Thus the normal approximation uses \(P(N\le U+0.5)\), where \(N\) has mean \(np\) and variance \(np(1-p)\). The same reasoning puts the lower boundary for \(K\ge L\) at \(L-0.5\). For a range \(L\le K\le U\), the associated interval is \(L-0.5\le N\le U+0.5\). This explains both the direction and the size of the correction.
Equivalently, on the proportion scale the correction is \(0.5/n\). It is usually clearer to work on the count scale, because the possible values of \(K\) are integers and the correction is then visibly half a count.
Worked Applications
Worked Example: An Upper Tail for a Sample Proportion
A hypothetical quality-control process inspects 400 independently selected units. Each unit has probability \(0.50\) of meeting a specified condition, independently of the other units. Approximate the probability that the observed sample proportion meeting the condition is at least \(0.53\).
Method and model. Let \(K\) be the number meeting the condition. The assumptions give \(K\sim\operatorname{Bin}(400,0.50)\), and \(\hat p=K/400\). The normal approximation check gives \(np=200\) and \(n(1-p)=200\).
Translate the event. Since \(400(0.53)=212\), the event is \(\hat p\ge0.53\), or \(K\ge212\). This is an inclusive integer lower tail boundary.
Correct and standardize. The corrected lower boundary is \(211.5\). The count mean is \(400(0.50)=200\), and the standard deviation is \(\sqrt{400(0.50)(0.50)}=\sqrt{100}=10\). Therefore
The estimated probability is \(P(Z\ge1.15)=1-\Phi(1.15)\). Using \(\Phi(1.15)\approx0.8749\), this is approximately \(0.1251\), to four decimal places.
Check. The cutoff is above the mean, so the upper-tail probability must be less than 0.5. The result satisfies that check. The common mistake avoided here is using 212 as the normal boundary; that omits the half-count correction.
Worked Example: A Central Range of Sample Proportions
In a hypothetical independent survey, each of 400 respondents has probability \(0.50\) of giving a specified response. Approximate the probability that the sample proportion giving that response is between \(0.46\) and \(0.54\), inclusive.
Method and model. The response count satisfies \(K\sim\operatorname{Bin}(400,0.50)\). Both expected counts are 200, so the normal approximation is suitable for this calculation.
Translate the event. The lower threshold gives \(400(0.46)=184\), and the upper gives \(400(0.54)=216\). Thus the inclusive event is \(184\le K\le216\).
Correct and standardize. The corrected interval is \(183.5\le N\le216.5\). The mean is \(200\) and the standard deviation is \(10\), so the standardized boundaries are
The probability is \(\Phi(1.65)-\Phi(-1.65)\). Using \(\Phi(1.65)\approx0.9505\) and symmetry, the probability is approximately \(2\Phi(1.65)-1\approx0.9011\), to four decimal places.
Check. The interval is centered at the expected proportion \(0.50\), so the two standardized boundaries are symmetric about zero. The common mistake avoided is leaving the count boundaries at 184 and 216, which treats the discrete endpoints as if they were the edges of the continuous normal interval.
Worked Example: A Nonattainable Threshold and a Strict Inequality
In a hypothetical reliability test, 200 independent components are tested, and each independently passes with probability \(0.40\). Approximate the probability that the sample proportion passing is at least \(0.386\).
Method and model. Let \(K\) count passing components. Then \(K\sim\operatorname{Bin}(200,0.40)\), and \(\hat p=K/200\). The expected counts are \(np=80\) and \(n(1-p)=120\).
Translate the event. Multiplying the threshold by 200 gives \(77.2\). Since \(K\) is an integer, \(K/200\ge0.386\) is equivalent to \(K\ge78\), not \(K\ge77.2\). The attainable proportions immediately around the threshold are \(77/200=0.385\) and \(78/200=0.390\).
Correct and standardize. The corrected lower boundary is \(77.5\). The count variance is \(200(0.40)(0.60)=48\), so the standard deviation is \(\sqrt{48}\approx6.9282\). Hence
The required upper-tail probability is approximately \(P(Z\ge-0.3608)=\Phi(0.3608)\approx0.6409\), to four decimal places.
Check. The threshold is below the expected proportion of \(0.40\), so a probability above 0.5 is plausible. The common mistake avoided is treating \(77.2\) as a possible count or correcting the original proportion threshold directly without first finding the integer count event.
Common Boundary Errors
A continuity correction cannot repair an incorrectly translated event. First settle what integer counts are included; only then move the boundary by half a count. For example, with \(n=120\), the event \(\hat p\le0.45\) becomes \(K\le54\). Its corrected upper boundary is \(54.5\), not \(55.5\). The mean and standard deviation for \(p=0.50\) are \(60\) and \(\sqrt{30}\), giving
A frequent wrong calculation uses the uncorrected boundary \(54\), or rounds a normal CDF value too early and then performs inconsistent arithmetic. Keep intermediate values unrounded when possible, and round the final probability to four decimal places.
- Forgetting the sample size. Proportions change in increments of \(1/n\); multiply thresholds by \(n\) before identifying counts.
- Ignoring strictness. “Greater than” and “at least” can correspond to different counts when the threshold is attainable.
- Correcting the wrong boundary. For a lower-tail event use the upper edge \(U+0.5\); for an upper-tail event use the lower edge \(L-0.5\).
- Using a normal model without checking the binomial setting. The trials must meet the stated independence and common-probability assumptions, and a normal approximation can be poor when expected counts are small.
Check Your Understanding
For each problem, translate the proportion event into a count event before applying the continuity correction. Approximate probabilities are rounded to four decimal places.
- A hypothetical process has 100 independent trials, each with success probability \(0.30\). Approximate the probability that the sample proportion is at least \(0.35\). (Answer: 0.1631)
- For 400 independent trials with success probability \(0.25\), approximate the probability that the sample proportion is at most \(0.22\). (Answer: 0.0921)
- For 250 independent trials with success probability \(0.60\), approximate the probability that the sample proportion is between \(0.56\) and \(0.64\), inclusive. (Answer: 0.8248)
- A hypothetical test has 120 independent trials, each with success probability \(0.50\). Approximate the probability that the sample proportion is strictly greater than \(0.55\). (Answer: 0.1176)
- For 120 independent trials with success probability \(0.50\), approximate the probability that the sample proportion is at most \(0.45\). (Answer: 0.1577)