Finding Probabilities for Sample Proportions

From a Sample Proportion to a Probability

A sample proportion varies from sample to sample. Even when the population proportion is fixed, one random sample may contain a little more than the population share of successes and another a little less. The central limit theorem lets us estimate probabilities about that sample-to-sample variation with a normal distribution, provided the sample is large enough for the approximation to be reasonable.

The key is to work with the sampling distribution of the sample proportion, not with the variation among individual outcomes. For a probability question, begin with the population success probability $p$ and sample size $n$. Then check the success and failure counts, calculate the standard error, and translate the requested event into a standard normal probability.

Model and approximation: Suppose $X_1,\ldots,X_n$ are independent Bernoulli outcomes with the same probability $p$ of success, where each $X_i$ is 1 for a success and 0 for a failure. The sample proportion is $\hat p=(X_1+\cdots+X_n)/n$. If both $np$ and $n(1-p)$ are sufficiently large, commonly at least 10 each, then $$ \hat p \approx N\left(p,\frac{p(1-p)}{n}\right). $$ Thus its approximate mean is $p$, and its standard error is $SE=\sqrt{p(1-p)/n}$. The count rule is a practical guideline, not a guarantee of perfect accuracy.

Why the Standard Error and Normal Approximation Have This Form

Each $X_i$ has mean $p$ and variance $p(1-p)$: it equals 1 with probability $p$, and 0 with probability $1-p$. Because the outcomes are independent, the variance of their sum is the sum of their variances. Dividing the sum by $n$ to form $\hat p$ divides its variance by $n^2$.

E(\hat p)=\frac{np}{n}=p,\qquad \operatorname{Var}(\hat p)=\frac{np(1-p)}{n^2} =\frac{p(1-p)}{n}.

The central limit theorem says that a suitably standardized sum of independent outcomes is approximately standard normal when there are enough successes and failures. Applying that result to the sample proportion gives

Z=\frac{\hat p-p}{\sqrt{p(1-p)/n}}\approx N(0,1).

This equation is the working tool for probability questions. A proposed cutoff for $\hat p$ is converted to a $z$-score by subtracting the mean $p$ and dividing by the standard error. The requested probability is then read as a lower tail, upper tail, or area between two $z$-scores.

A Reliable Calculation Sequence

Specify the model.
State the sample size $n$, the population success probability $p$, and what counts as a success. The approximation assumes independent outcomes with the same success probability.

Check the counts.
Calculate $np$ and $n(1-p)$. If either expected count is small, the usual normal approximation may be unreliable.

Calculate the standard error.
Use $SE=\sqrt{p(1-p)/n}$. This measures the typical sample-to-sample spread of $\hat p$, in proportion units.

Standardize the cutoff or cutoffs.
For a cutoff $c$, use $z=(c-p)/SE$. Keep track of whether the event asks for values below, above, or between the cutoffs.

Find and interpret the normal area.
Use the standard normal cumulative distribution function $\Phi(z)=P(Z\le z)$. State the resulting probability as an approximation and interpret it in the setting.

Worked Probability Calculations

Worked Example: Probability of a Proportion Above a Cutoff

Suppose a hypothetical process produces an item meeting a specified standard with probability $p=0.40$. A random sample of $n=100$ items is taken, and $\hat p$ is the proportion meeting the standard. Estimate the probability that at least 45% meet it.

Check the model and counts. Treat each item as an independent outcome with the same success probability. The expected success count is $np=100(0.40)=40$, and the expected failure count is $n(1-p)=100(0.60)=60$. Both meet the usual count guideline.

Find the standard error.

SE=\sqrt{\frac{p(1-p)}{n}} =\sqrt{\frac{(0.40)(0.60)}{100}} =\sqrt{0.0024} \approx 0.04899.

The cutoff $0.45$ is above the mean $0.40$. Its standardized value is

z=\frac{0.45-0.40}{0.04899}\approx 1.021.

“At least 45%” is an upper-tail event, so

P(\hat p\ge 0.45)\approx P(Z\ge 1.021) =1-\Phi(1.021) \approx 1-0.8464 =0.1536.

Using the unrounded standardized cutoff gives approximately $0.1537$; either calculation gives about $0.154$. Under this model, the estimated probability that at least 45% of the 100 items meet the standard is about 15.4%.

Worked Example: Probability Between Two Proportion Cutoffs

In a hypothetical survey setting, suppose the probability that a randomly selected person gives a particular response is $p=0.55$. For a random sample of $n=80$, estimate the probability that the sample proportion is between $0.50$ and $0.65$.

Check the model and counts. Assume independent responses with the same probability $0.55$. The expected success count is $80(0.55)=44$, and the expected failure count is $80(0.45)=36$, so both counts meet the guideline.

The standard error is

SE=\sqrt{\frac{(0.55)(0.45)}{80}} =\sqrt{0.00309375} \approx 0.05562.

Standardize each endpoint:

z_{\text{lower}}=\frac{0.50-0.55}{0.05562}\approx -0.899,\qquad z_{\text{upper}}=\frac{0.65-0.55}{0.05562}\approx 1.798.

The probability between the cutoffs is the larger cumulative area minus the smaller one:

P(0.50<\hat p<0.65) \approx \Phi(1.798)-\Phi(-0.899) \approx 0.9639-0.1843 =0.7796.

The estimated probability is about $0.780$, or 78.0%. In context, if this hypothetical sampling process were repeated many times under the stated assumptions, roughly 78% of samples would have a proportion between 0.50 and 0.65 by this normal approximation.

Worked Example: A Lower-Tail Probability

Suppose a hypothetical population has a success probability $p=0.30$. From independent trials, take a sample of $n=150$. Estimate the probability that the sample proportion is at most $0.25$.

Check the model and counts. The expected number of successes is $150(0.30)=45$, and the expected number of failures is $150(0.70)=105$. Both are large enough for the usual normal approximation guideline.

The standard error and standardized cutoff are

SE=\sqrt{\frac{(0.30)(0.70)}{150}} =\sqrt{0.0014} \approx 0.03742, $$ $$ z=\frac{0.25-0.30}{0.03742}\approx -1.336.

The event is a lower-tail event. Therefore,

P(\hat p\le 0.25)\approx P(Z\le -1.336) =\Phi(-1.336)\approx 0.0908.

The estimated probability is about $0.091$, or 9.1%. The negative $z$-score is sensible because $0.25$ is below the sampling distribution’s mean of $0.30$.

When the Count Check Fails

The success-failure check is part of the method, not a formality. For example, suppose a hypothetical outcome has $p=0.08$ and a sample has $n=100$. The expected success count is $100(0.08)=8$, while the expected failure count is $100(0.92)=92$. The success count is below the common guideline of 10. Even though the failure count is large, the usual normal approximation is not well supported by this check.

In that situation, do not present a normal-tail calculation as a dependable estimate merely because the standard-error formula can be evaluated. The distribution of $\hat p$ may not be close enough to normal, especially near its lower boundary of zero. The count rule does not prove that an approximation will be accurate when satisfied, but failing it is a clear reason for caution.

Common Mistakes and Why the Method Matters

Using the wrong standard error. In this probability setting, $p$ is the specified population probability, so use $\sqrt{p(1-p)/n}$. Do not substitute a cutoff such as $0.45$ for $p$; the cutoff is not the assumed mean.
Forgetting to check both counts. A large sample size by itself does not guarantee a good approximation. Check $np$ and $n(1-p)$, since one may be small when $p$ is near 0 or 1.
Reversing the tail. A cutoff above $p$ will have a positive $z$-score. “At least” asks for the area to its right, $1-\Phi(z)$, not $\Phi(z)$.
Confusing individual variability with sampling variability. The variance of one Bernoulli outcome is $p(1-p)$; the variance of the sample proportion is $p(1-p)/n$. The standard error describes the spread of sample proportions, not the spread of individual outcomes.
Treating the approximation as exact. The sample proportion takes discrete values, whereas the normal model is continuous. A calculated normal area is an approximation, even when the count check is satisfied.

These calculations turn the central limit theorem into a practical tool: they quantify how unusual a sample proportion would be under a stated model. The result is useful only when the sampling assumptions and count check are taken seriously, and the probability must be described as an approximation rather than a guaranteed exact value.

Check Your Understanding

For each question, identify the relevant check or calculation before interpreting the result.

For $p=0.20$ and $n=60$, calculate the expected success and failure counts. Does the usual count guideline support a normal approximation?
For a model with $p=0.50$ and $n=100$, write the standard error of $\hat p$ and explain what it measures.
A standardized cutoff is $z=1.25$. Which expression gives the approximate probability that $\hat p$ is at least that cutoff from its mean: $\Phi(1.25)$ or $1-\Phi(1.25)$?
Why is the standard error for a sample proportion smaller than the standard deviation of one Bernoulli outcome when $n>1$?
A proposed model has $p=0.04$ and $n=100$. Which count check raises concern, and what should you say about a normal approximation?