Each condition has a job

Draw a simple random sample of size $n$ from a population in which a proportion $p$ have some characteristic, and let $\hat{p}$ be the proportion in the sample. Across all such samples,

$$\mu_{\hat{p}} = p, \qquad \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}.$$

The first equation is the unbiasedness of $\hat{p}$ from the previous topic. The second is the one that gets rebuilt wrong. Three details decide it:

1. $n$ sits in the denominator inside the radical, so the standard deviation falls like $\frac{1}{\sqrt{n}}$, not like $\frac{1}{n}$. Quadrupling the sample halves the spread.
2. The ingredients are the population proportion $p$ and its complement, not the sample proportion, whenever $p$ is known or hypothesized.
3. The whole expression has no units and is a spread of proportions, so it is always a small number when $n$ is large.

With $p = 0.60$ and $n = 100$, $\sigma_{\hat{p}} = \sqrt{\frac{0.60(0.40)}{100}} = \sqrt{0.0024} \approx 0.049$. At $n = 400$ it is $0.0245$: four times the data, half the spread.

The results above hold under conditions, and each one is doing a distinct piece of work:

1. Random. The data come from a simple random sample or a randomized experiment. This is what makes $\hat{p}$ unbiased and what licenses generalizing beyond the data at all. Nothing else on this list can repair its absence.
2. 10% condition. The sample is no more than 10% of the population, $n \le 0.10 N$. Sampling without replacement makes observations slightly dependent, and this keeps that dependence small enough for $\sqrt{\frac{p(1-p)}{n}}$ to be accurate.
3. Large counts. $np \ge 10$ and $n(1-p) \ge 10$. This is what makes the shape approximately normal.

Checking these means naming the study's actual numbers, not the condition. "Large counts: $np = 100(0.60) = 60 \ge 10$ and $n(1-p) = 100(0.40) = 40 \ge 10$" is a check; "large counts, OK" is not. Two substitutions are especially common and both are wrong: using $n \ge 30$ for a proportion, which is a rule of thumb for means and says nothing about the counts of successes and failures, and citing the 10% condition as though it established randomness.

The failure modes differ, which is why the conditions are not interchangeable. A non-random sample gives a distribution centered in the wrong place. A sample that is too large a share of its population gives a standard deviation that is too big. Counts that are too small give a skewed shape, so normal-based answers land in the wrong place even though the center and spread are right.

When the large-counts condition holds, the sampling distribution of $\hat{p}$ is approximately normal. Combined with the center and spread above, that means a probability question about $\hat{p}$ becomes a z-score calculation:

$$z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}.$$

For $p = 0.60$ and $n = 100$, the probability that a sample proportion exceeds 0.65 is

$$z = \frac{0.65 - 0.60}{0.049} \approx 1.02, \qquad P(\hat{p} > 0.65) \approx 0.154.$$

Why the counts and not $n$ alone: $\hat{p}$ is bounded between 0 and 1, so when $p$ sits near an edge, the distribution runs out of room on that side and skews. With $p = 0.02$ and $n = 100$, $np = 2$, and the distribution of $\hat{p}$ piles up near zero with a long right tail that no normal curve describes. The same $n = 100$ with $p = 0.60$ is comfortably symmetric. The condition is about the counts because the counts are what measure the distance to the boundary.

Two spreads are in play and only one of them shrinks. The population has its own variability, fixed and unaffected by sampling. The statistic $\hat{p}$ has the spread $\sqrt{\frac{p(1-p)}{n}}$, and that is what falls as $n$ grows.

Reading it back in context keeps the referent attached: "in repeated random samples of 100 voters from this population, the sample proportion supporting the measure typically falls about 0.049 from 0.60". The phrase in repeated samples is what identifies the object being described, and an interpretation without it usually ends up describing the voters instead of the statistic.

One more consequence worth carrying forward: because the spread depends on $n$ only through $\sqrt{n}$, precision is expensive. Halving the standard deviation costs four times the sample; cutting it to a third costs nine times. Every margin of error in this unit inherits that arithmetic.