Loav · VCE Mathematical Methods · Units 3 and 4
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A sample proportion estimates the fraction of a population with a particular characteristic. It changes from sample to sample even when the population stays unchanged. This guide distinguishes parameters from statistics, connects a proportion to a binomial count, calculates exact and approximate probabilities, and uses simulation to investigate sampling variation. You should know binomial distributions, normal probabilities and square roots.
The central idea is that one statistic has a distribution across repeated samples. A particular survey supplies one observed proportion. Imagining or simulating many surveys of the same size reveals the distribution of that proportion. This is different from a distribution of the individual responses within a single survey.
1. Population, sample and characteristic
The population is the full collection about which a conclusion is intended. The sample is the subset observed. Define the characteristic as a clear yes/no outcome. The population proportion p is the fraction with that characteristic in the population. In the model, p is fixed, although it may be unknown.
Let X count sampled individuals with the characteristic in a sample of size n, where n is a positive integer. Before sampling, X and the sample proportion are random variables. After sampling, their observed values are numbers.
Random sample proportion: P̂ = X/n.
Observed sample proportion: p̂ = x/n.
Population parameter: p.
A hat indicates an estimate. It does not make p̂ equal to p. If 58 out of 100 sampled people have a characteristic, the sample proportion is 0.58. The population proportion could differ. The purpose of statistical inference is to account for uncertainty when using the observed statistic to learn about that fixed population parameter.
2. Random selection and model assumptions matter
A useful sampling design gives a fair opportunity for population members to be included. A convenience sample or voluntary-response poll may systematically overrepresent particular groups. A larger biased sample can give a very precise estimate of the wrong population pattern. Increasing n reduces random sampling variation under a suitable model; it does not automatically repair selection bias, non-response or misleading questions.
For the binomial model, each sampled observation has the same success probability p and observations are independent. Independent sampling with replacement meets this structure. Sampling without replacement from a population very large compared with the sample can often be approximated this way. Sampling a large fraction of a small population does not preserve independence, so the binomial formulas should not be presented as exact there.
Specify the target population and the sampling assumptions before calculating. If a question explicitly supplies a binomial model, use it consistently. If you must choose a model from a story, explain what makes it suitable and identify important limitations.
3. The exact sample proportion remains discrete
If X ∼ Bi(n,p), then P̂ can take only 0, 1/n, 2/n, …, 1. A proportion of 0.337 is impossible when n = 100, because it would require 33.7 successes. Convert proportion events into integer count events before using exact binomial probabilities.
P(P̂ = r/n) = P(X = r).
P(a ≤ P̂ ≤ b) = P(na ≤ X ≤ nb), with X restricted to integers.
For an inclusive lower bound, round up to the first allowed integer; for an inclusive upper bound, round down. With strict inequalities, check whether the bound itself is an integer. For example, X < 60 includes only counts up to 59, while X < 60.4 includes counts up to 60.
Do the multiplication by n before rounding. Rounding the proportion threshold too early can change which counts are included. Listing the smallest and largest admissible count is a useful audit step.
4. Centre and spread of the sampling distribution
Under the binomial sampling model, the sample proportion is centred at the population proportion. Its standard deviation decreases with the square root of sample size.
E(P̂) = p
Var(P̂) = p(1 − p)/n
sd(P̂) = √[p(1 − p)/n]
The standard deviation of a statistic is also called its standard error. Here the formula uses the population value p because the model is being specified. In a later confidence-interval calculation, p is unknown and p̂ is used to estimate this spread.
For fixed p, multiplying n by four halves the standard deviation. Doubling n divides it by √2, not by two. For fixed n, the spread is greatest at p = 0.5 because p(1 − p) = 0.25 − (p − 0.5)2 ≤ 0.25. This gives a conservative way to plan sample size when p is unknown.
Worked examples
Worked example 1: exact and approximate interval probabilities
An original sampling model has population proportion p = 0.35. Independently select n = 200 observations and let P̂ be the sample proportion. Find its mean and standard deviation, then the probability that it lies from 0.30 to 0.40 inclusive.
- E(P̂) = 0.35. Its standard deviation is √[0.35(0.65)/200] ≈ 0.0337268.
- For the exact calculation, X = 200P̂ ∼ Bi(200,0.35). The interval becomes 60 ≤ X ≤ 80.
- P(60 ≤ X ≤ 80) = P(X ≤ 80) − P(X ≤ 59) ≈ 0.8807.
- For a direct normal approximation, use mean 0.35 and standard deviation 0.0337268. The area from 0.30 to 0.40 is approximately 0.8618.
The two answers differ because one sums discrete masses, including endpoint masses, while the other measures continuous area. Label the requested method. This guide uses the direct normal approximation as a continuous model; it is not an exact replacement for a binomial probability.
5. When a normal approximation is reasonable
When n is sufficiently large and p is not too near zero or one, the sampling distribution is approximately normal. A common course rule of thumb is that both np and n(1 − p) exceed five. These are expected counts of successes and failures, not probabilities. Passing the check supports the approximation but does not guarantee every tail probability is highly accurate.
P̂ is approximately N(p, p(1 − p)/n).
Z ≈ (P̂ − p)/√[p(1 − p)/n].
The normal model is continuous and unbounded, while the actual proportion is discrete and lies in [0,1]. The approximation is useful when most of its area stays within plausible values and the discrete spacing 1/n is small relative to the overall spread. If expected successes or failures are very small, use the exact binomial model when it is available.
Worked example 2: a strict proportion threshold
Suppose p = 0.60 and n = 400 under independent sampling. Estimate P(P̂ > 0.65) using the normal approximation and compare it with the exact count calculation.
- Expected successes and failures are 240 and 160, comfortably above five.
- The standard deviation is √[0.6(0.4)/400] ≈ 0.0244949.
- The threshold standardises to (0.65 − 0.60)/0.0244949 ≈ 2.04124. The upper-tail probability is approximately 0.0206.
- Exactly, P̂ > 0.65 means X > 260, so X ≥ 261. For X ∼ Bi(400,0.6), that probability is approximately 0.0176.
Using X ≥ 260 would answer “at least 0.65”, a different discrete event. In the continuous approximation, a strict or inclusive boundary has the same area. Preserve that distinction when comparing methods.
Worked example 3: choose sample size for a spread target
A survey planner wants the standard deviation of P̂ to be at most 0.02, regardless of the unknown population proportion. Find a conservative minimum sample size under the binomial model.
- The maximum possible p(1 − p) is 0.25, attained at p = 0.5.
- Require √(0.25/n) ≤ 0.02. Squaring gives 0.25/n ≤ 0.0004.
- Therefore n ≥ 625. At n = 625 the maximum standard deviation is exactly 0.02.
This controls model-based sampling spread, not all survey error. It also does not itself specify a confidence level or a confidence-interval margin of error; those require an additional normal cutoff.
6. Simulate repeated samples thoughtfully
A simulation makes the sampling distribution visible. Choose a population probability p and a sample size n. For each simulated sample, generate n independent random numbers uniformly between zero and one; count a success whenever a number is below p; divide the success count by n; and store the resulting proportion. Repeat this entire sample many times.
- Set p, n and a positive integer R for the number of repeated samples.
- For each repetition, reset the success count to zero.
- Generate n independent trials and update the success count.
- Store count/n once for that sample.
- Plot the R stored proportions and compare their centre and spread with p and √[p(1 − p)/n].
Resetting the count is essential. Otherwise the algorithm mixes results across samples and no longer simulates the stated statistic. Dividing by R instead of n is another common error: R controls how clearly the simulation reveals the distribution, whereas n controls the variation of each sample proportion.
Repeat the simulation for different n values while holding p fixed, then for different p values while holding n fixed. Larger n should narrow the cloud of proportions. Values of p near a boundary can produce a visibly skewed distribution, especially with small n. Finite simulations fluctuate, so their empirical mean and standard deviation need not exactly equal the theoretical values.
Practice and reasoned answers
1. A sample contains 47 successes among 80 observations. Identify x, n and p̂. Is p known?
x = 47, n = 80 and p̂ = 47/80 = 0.5875. The population parameter p is not supplied and is not known to equal 0.5875. The observed sample proportion is an estimate based on this particular sample.
2. For n = 60, convert P̂ < 0.35 into an integer count event.
X/60 < 0.35 gives X < 21. Since X is an integer, the event is X ≤ 20. The strict boundary excludes the count 21, whose sample proportion is exactly 0.35.
3. p = 0.25 and n increases from 100 to 400. How do the mean and standard deviation change?
The mean stays 0.25. The standard deviation changes from √(0.1875/100) ≈ 0.04330 to √(0.1875/400) ≈ 0.02165. Quadrupling sample size halves sampling spread; it does not move the centre.
4. p = 0.4, n = 100. Find P(0.35 ≤ P̂ ≤ 0.45) exactly and approximately.
Exactly, X ∼ Bi(100,0.4) and 35 ≤ X ≤ 45, giving approximately 0.7386. The direct normal model has mean 0.4 and standard deviation √0.0024 ≈ 0.04899, giving approximately 0.6926. Report which method was used rather than presenting the approximation as exact.
5. p = 0.02 and n = 100. Does the usual expected-count check support a normal approximation?
No. np = 2 while n(1 − p) = 98. The expected success count is too small under the rule of thumb. The binomial distribution is concentrated near zero and skewed, so an exact binomial calculation is preferable.
6. A simulation uses 1000 repeated samples of size 50. Which number should divide each sample’s success count?
Divide by 50. Each proportion describes one sample of 50 observations. The 1000 repetitions provide 1000 separate proportions for displaying the sampling distribution; they are not the denominator of each statistic.
7. p = 0.2, n = 250. Calculate P(P̂ < 0.24) exactly and with a direct normal approximation.
The exact event is X < 60, or X ≤ 59, for X ∼ Bi(250,0.2), giving approximately 0.9311. The normal standard deviation is √0.00064 ≈ 0.0252982; its area below 0.24 is approximately 0.9431. The different endpoint treatment helps explain the discrepancy.
Learning check
Identify the fixed parameter, the random statistic and its observed value. Convert a strict proportion inequality into a count event without rounding prematurely. Explain what n and R each control in a simulation. Once these distinctions are secure, use an observed sample proportion to construct and interpret an approximate confidence interval for the unknown population proportion.
Curriculum and study route
Loav Mathematical Methods Units 3 and 4. These module divisions are a teaching sequence. Use the current VCAA Mathematics study design, Mathematical Methods Units 3 and 4, Area of Study 4, and the VCAA Mathematical Methods FAQ for course requirements. Scope checked 15 October 2026. All worked scenarios, questions and explanations in this guide are original.
