Why Check Conditions Before Using a Normal Approximation?
A large number of trials does not, by itself, make a binomial count suitable for a normal approximation. If success is rare, even a substantial number of trials may produce only a few expected successes. The count then remains concentrated near its lower boundary, zero, and a normal model may be a poor stand-in. A useful condition check therefore examines the trial model and both expected counts, not just the number of trials.
This tutorial focuses on deciding whether a binomial count passes common screening conditions. Passing a screen supports considering a normal approximation; it does not guarantee a specified level of accuracy for every probability calculation. The conditions are aligned with the SOA Exam P (Probability) syllabus.
The usual normal model considered for a binomial count has mean \(np\) and variance \(npq\). In the course notation for a normal random variable, the candidate approximation is \(N(np,npq)\), where the second parameter is the variance. The condition check asks whether the binomial count is in a setting where this model is a reasonable approximation to consider. It is not a replacement for verifying that the count is actually binomial.
The Two Expected-Count Conditions
A common introductory guideline is to require at least 5 expected successes and at least 5 expected failures. A more conservative guideline, often used when a stronger screen is wanted, requires at least 10 of each. In symbols, the two checks are
where \(c\) is the chosen guideline, commonly 5 or 10. Both inequalities matter. A large value of \(np\) does not compensate for a small value of \(n(1-p)\), and vice versa. The expected counts are not required to be whole numbers: they are averages over repeated sets of trials, not predictions of the exact outcome in one set.
These cutoffs are rules of thumb, not universal mathematical guarantees. Different courses or applications may use different thresholds, so state the criterion being applied. If a problem specifies one, use that one. If it does not, report both expected counts and identify the guideline used. Passing the more conservative check is stronger evidence than passing only the less conservative one, but it still does not promise exact accuracy for every tail or boundary event.
| Check | Calculation | What it screens for |
|---|---|---|
| Expected successes | \(np\) | Whether success is too rare, given the number of trials |
| Expected failures | \(n(1-p)\) | Whether failure is too rare, given the number of trials |
| Both checks | Compare each with the selected cutoff | Whether the count passes the chosen expected-count guideline |
Why Both Expected Counts Appear
The two checks can be understood by comparing the binomial mean with its natural boundaries. A binomial count can only lie between 0 and \(n\). Its mean is \(np\), so the distance from the mean to the lower boundary 0 is \(np\). The distance from the mean to the upper boundary \(n\) is \(n-np=n(1-p)\). The standard deviation is \(\sqrt{np(1-p)}\), when \(0<p<1\).
Dividing each boundary distance by the standard deviation gives
Thus the expected counts describe how many successes and failures the model expects, and together they give information about how close the mean is to each possible boundary relative to the standard deviation. If one expected count is very small, the distribution can be strongly constrained near that boundary. Requiring both counts to exceed a cutoff is a practical way to screen out that situation.
Apply the Check in a Fixed Order
A reliable decision begins with the probability model, not with a numerical threshold. First verify that the problem describes a fixed number of independent trials with a common success probability. Only then calculate \(np\) and \(n(1-p)\). A large \(n\) cannot make an incorrect binomial model correct.
Check for a fixed number of trials, two outcomes classified as success or failure, independence, and the same success probability on every trial.
Identify \(n\) and \(p\), then set \(q=1-p\). Keep the event called a success consistent throughout the calculation.
Find \(np\) and \(nq\). Do not check only the more convenient or larger value.
Apply the specified cutoff, or state whether you are using 5 or 10 when the problem does not specify one. Report a failed check plainly.
Say whether the model passes the chosen screen and whether a normal approximation is reasonable to consider. Do not describe the screen as a guarantee of accuracy.
Worked Decisions
Worked Example: A Moderate Success Probability
A hypothetical sensor independently flags a defective component with probability \(0.12\) for each component inspected. A fixed batch contains 150 components. Let \(X\) be the number flagged. Check whether a normal approximation passes the guideline requiring at least 10 expected successes and 10 expected failures.
Method and model check. There are \(n=150\) fixed trials. The problem states that outcomes are independent and that each has the same flag probability \(p=0.12\). Thus \(X\sim\operatorname{Bin}(150,0.12)\), and
Expected successes. The expected number of flagged components is
Expected failures. The expected number not flagged is
Both values meet the cutoff: \(18\geq10\) and \(132\geq10\). The count passes this expected-count screen, so a normal approximation is reasonable to consider. As a parameter check, its binomial mean is \(18\), and its variance is \(npq=150(0.12)(0.88)=15.84\). This screen alone does not establish the accuracy of any particular approximated probability.
Common mistake avoided: It would be incomplete to say that the approximation is appropriate just because 150 is a large trial count. Here both expected counts were actually calculated and compared with the stated cutoff.
Worked Example: Many Trials but Rare Successes
A hypothetical monitoring system independently detects a particular event on each of 60 days, with the same probability \(p=0.04\) each day. Let \(X\) count the detected days. Does the count pass the guideline requiring at least 5 expected successes and 5 expected failures? Does it pass the more conservative guideline of 10 for each?
Method and model check. The trial count is fixed at \(n=60\); the daily outcomes are stated to be independent and to have a common probability. Therefore \(X\sim\operatorname{Bin}(60,0.04)\). The failure probability is
The expected successes and failures are
For the cutoff of 5, the success condition fails because \(2.4<5\), although the failure condition passes because \(57.6\geq5\). The count therefore does not pass the two-part guideline. It also fails the cutoff of 10 because \(2.4<10\), even though \(57.6\geq10\). The large imbalance between the expected counts is important: the number of trials alone does not establish suitability.
Common mistake avoided: Checking only expected failures would give \(57.6\geq10\) and could lead to a false “pass.” The condition requires both inequalities, so the small expected number of successes determines the result.
Worked Example: Finding the Trial Count Needed for a Chosen Cutoff
A hypothetical quality test has independent trials with a constant success probability \(p=0.035\). Find the smallest integer trial count that meets the conservative guideline of at least 10 expected successes and 10 expected failures.
Method. Let \(n\) be the number of trials and \(q=1-p=0.965\). The two requirements are
Solving the success requirement gives
so the integer \(n\) must be at least 286. Solving the failure requirement gives
so that condition alone requires an integer \(n\) of at least 11. Both conditions must hold; the stricter requirement is \(n\geq286\). Check the boundary value \(n=286\):
Both expected counts are at least 10. At \(n=285\), however, \(np=285(0.035)=9.975<10\), so 285 trials do not suffice. The smallest qualifying integer is therefore 286.
Common mistake avoided: Rounding \(285.714\ldots\) to 285 would violate the condition. A minimum trial count must be rounded up, then checked in both inequalities.
Worked Example: Expected Counts Do Not Fix a Non-Binomial Design
A hypothetical laboratory has 400 labeled specimens, of which exactly 80 are positive. It selects 50 specimens at random without replacement and counts the positives. A technician proposes applying the binomial condition check using \(p=80/400\). Is the count exactly binomial, and what do the proposed expected counts indicate?
Method and model check. The proposed success probability for any one draw is \(p=80/400=0.20\), with \(q=0.80\). If one mechanically used a binomial model with \(n=50\), its expected counts would be
Those values meet the 10-and-10 guideline. However, the draws are made without replacement from a finite group with a fixed number of positive specimens. The result of one draw changes the composition of the specimens left for the next draw, so the outcomes are not independent. The count is not exactly binomial, and passing the numerical expected-count check cannot make it binomial.
The correct conclusion is that the count does not meet the exact independent-trial model required for a binomial approximation check. Its sampling design must be modeled appropriately before deciding on an approximation. The expected counts above are only what the proposed binomial calculation would produce; they do not validate that model.
Common mistake avoided: Correct arithmetic is not sufficient. Always verify the independence and common-probability assumptions before treating \(n\) and \(p\) as binomial parameters.
Common Exam Traps
- Using \(n\) as the only test. A large sample or trial count may still have too few expected successes when \(p\) is small, or too few expected failures when \(p\) is close to 1.
- Checking one expected count only. The standard screen has two requirements: \(np\) and \(n(1-p)\) must both meet the selected cutoff.
- Confusing observed and expected counts. These conditions use the model quantities \(np\) and \(n(1-p)\), not the number of successes or failures that happened to be observed.
- Treating a guideline as a guarantee. Passing a cutoff supports considering a normal approximation; it does not prove accuracy for all events or replace the problem’s stated assumptions.
- Ignoring how the trials were selected. Sampling without replacement, changing probabilities, or dependence can invalidate the binomial model even when the arithmetic for expected counts looks favorable.
- Rounding down a minimum \(n\). When solving \(np\geq c\) for an integer number of trials, round the resulting lower bound up and verify both inequalities at the proposed integer.
Check Your Understanding
For each item, check the model before applying the expected-count guideline. Unless stated otherwise, use the 10-and-10 guideline.
- A fixed set of 90 independent trials has common success probability \(0.15\). Find the expected successes and failures, then decide whether the count passes the 10-and-10 screen. (Answer: \(13.5\) expected successes and \(76.5\) expected failures; passes.)
- A process has 75 independent trials with success probability \(0.06\). Does it pass the 5-and-5 guideline? Does it pass the 10-and-10 guideline? (Answer: \(4.5\) expected successes and \(70.5\) expected failures; fails the 5-and-5 guideline and the 10-and-10 guideline.)
- For independent trials with common success probability \(0.025\), find the smallest integer \(n\) meeting both expected-count requirements of 10. (Answer: \(n=400\); the expected counts are \(10\) and \(390\).)
- A fixed number of 200 independent trials each has success probability \(0.50\). Find both expected counts and state whether the 10-and-10 screen is met. (Answer: \(100\) expected successes and \(100\) expected failures; passes.)
- A sample of 80 objects is drawn without replacement from a batch of 500 containing exactly 30 defective objects. A proposed calculation uses \(p=30/500\). Give the proposed binomial expected counts and explain whether the count is exactly binomial. (Answer: \(4.8\) expected defective objects and \(75.2\) expected nondefective objects under the proposed binomial calculation; it is not exactly binomial because sampling without replacement does not give independent trials.)
- A stated binomial model has \(n=100\) and \(p=0.09\). Compare the count with both the 5-and-5 and 10-and-10 guidelines. (Answer: \(9\) expected successes and \(91\) expected failures; it passes the 5-and-5 guideline but fails the 10-and-10 guideline.)