Checking Normality of p-hat with np and n(1-p)

Two Expected Counts Determine Whether a Normal Model Is Reasonable

In When the 10 Percent Condition Applies, you learned to check whether sampling without replacement creates enough dependence to affect the usual standard deviation formula. That condition and the normality check answer different questions. Now we will focus on the Large Counts condition: whether the expected numbers of successes and failures are both large enough for a Normal model of \(\hat{p}\) to be appropriate.

Here, a success means having the categorical characteristic being studied; a failure means not having it. These labels do not imply that one outcome is desirable. If the population proportion of successes is \(p\), then the expected number of successes in a sample of size \(n\) is \(np\), and the expected number of failures is \(n(1-p)\).

When both inequalities hold, the condition supports using a Normal model for the sampling distribution of \(\hat{p}\), provided the sampling process is also appropriate. As covered in Normal Models for Sample Means and Proportions, the model is centered at \(p\), with standard deviation \(\sqrt{p(1-p)/n}\) under the relevant sampling conditions. This tutorial is about checking the shape condition, not calculating probabilities from that model.

A Reliable Way to Check Both Counts

Use the population proportion \(p\), not the sample proportion \(\hat{p}\). The check describes expected counts under the population proportion or model being considered. For example, if \(p=0.30\) and \(n=50\), the expected number of successes is \(50(0.30)=15\), while the expected number of failures is \(50(0.70)=35\). These are expected counts, not a claim that every sample will contain exactly 15 successes and 35 failures.

A quick arithmetic check is that the two expected counts must add to the sample size:

If the counts do not add to \(n\), revisit the complement \(1-p\) or the multiplication. Then compare each count with 10 separately. Passing one part does not make up for failing the other: the condition requires both.

Worked Example: Both Expected Counts Are Large Enough

Worked Example: Expected Successes Fall Short

Worked Example: Expected Failures Fall Short

Worked Example: Meeting the Boundary Exactly

Finding a Sample Size That Meets the Condition

The same inequalities can help plan a sample. For a known value of \(p\), both \(np\geq10\) and \(n(1-p)\geq10\) must hold. Dividing by the positive proportions gives \(n\geq10/p\) and \(n\geq10/(1-p)\). Since the sample size must be a whole number, round up to the smallest integer that satisfies both requirements. This planning technique does not replace checking the sample’s other conditions.

What Passing the Check Does—and Does Not—Mean

The Large Counts condition is a practical check for whether the sampling distribution of \(\hat{p}\) is reasonably modeled by a Normal distribution. It is not a claim that the original categorical observations themselves have a Normal distribution. Each observation is still a success or a failure; the Normal model concerns the distribution of sample proportions across repeated samples.

Meeting this condition does not establish that the sample is random, that the population proportion is known without uncertainty, or that the sampling method avoids bias. If sampling is without replacement from a finite population, use the 10% condition from the earlier tutorial to assess whether the usual independence approximation is reasonable. A suitable normality check cannot fix a biased or nonrandom sample.

If either expected count is below 10, the standard AP decision is that the Large Counts condition is not met. That does not prove that every possible Normal approximation would be inaccurate by the same amount; it means the usual condition does not support using that model here. The next tutorial examines how the shape of the sampling distribution changes when \(p\) is near 0 or 1.

Common Mistakes and AP Exam Communication

• Checking only one count. A large \(np\) does not compensate for a small \(n(1-p)\), or vice versa. A full check reports both values and compares each with 10.
• Using \(\hat{p}\) instead of \(p\). For this check, use the population proportion or the proportion specified by the model. Do not substitute an observed sample proportion without a reason supported by the problem.
• Forgetting the complement. The failure proportion is \(1-p\), not \(p\). Check that the two expected counts add to \(n\) as a simple arithmetic safeguard.
• Rejecting a count equal to 10. The requirement is at least 10, so equality passes. A count of 9.9 does not pass.
• Rounding a count before deciding. Keep the unrounded product for the comparison. For example, \(9.94\) is below 10 even though it might be casually rounded to 10.
• Calling a Normal model guaranteed. Say that the condition is met and supports using a Normal model, provided the other sampling conditions are appropriate. The check does not remove the need to justify the sampling process.

Key Takeaway

The Normal model for \(\hat{p}\) is supported by the Large Counts condition only when the expected numbers of successes and failures both reach the threshold. Check the two sides separately, use \(p\) and \(1-p\), and keep this shape check distinct from the 10% condition and from the question of whether the sample was selected appropriately.