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Topic 3: Statistical inference

Understand the distribution of the sample mean, apply the central limit theorem to describe its shape, mean and standard deviation, and use these to compute probabilities for sample means drawn from a population

A focused answer to the QCE Specialist Mathematics Unit 4 dot point on the sampling distribution of the mean. Covers the mean and standard error of the sample mean, the central limit theorem, standardising to compute probabilities, and how sample size affects spread, with a verified worked example and the standard-error mistake QCAA markers watch for.

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

QCAA wants you to understand that the sample mean Xˉ\bar{X} is itself a random variable with its own distribution, to state and apply the central limit theorem, and to use the resulting normal model to compute probabilities about sample means. This is the foundation of statistical inference, assessed in IA3 and the external assessment, and it precedes confidence-interval work.

The answer

The sample mean is random

If you take a random sample of size nn from a population and compute its mean Xˉ\bar{X}, a different sample gives a different mean. So Xˉ\bar{X} is a random variable. Its distribution is called the sampling distribution of the mean, and it has its own mean and standard deviation.

Mean and standard error

If the population has mean μ\mu and standard deviation σ\sigma, then for a sample of size nn:

E(Xˉ)=μ,SD(Xˉ)=σn.E(\bar{X}) = \mu, \qquad \text{SD}(\bar{X}) = \frac{\sigma}{\sqrt{n}}.

The standard deviation of the sample mean, σn\dfrac{\sigma}{\sqrt{n}}, is called the standard error. It is smaller than the population standard deviation, and it shrinks as nn grows: larger samples give more reliable estimates of μ\mu. Crucially the divisor is n\sqrt{n}, not nn.

The central limit theorem

The central limit theorem states that for a sufficiently large sample size nn, the distribution of the sample mean Xˉ\bar{X} is approximately normal,

XˉN ⁣(μ,σ2n),\bar{X} \sim N\!\left(\mu, \frac{\sigma^2}{n}\right),

regardless of the shape of the original population. If the population is already normal, Xˉ\bar{X} is exactly normal for any nn. A common rule of thumb takes n30n \geq 30 as large enough for the approximation when the population is not too skewed.

Standardising to find probabilities

To compute a probability for Xˉ\bar{X}, standardise using the standard error:

Z=Xˉμσ/n.Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}.

This ZZ has the standard normal distribution N(0,1)N(0,1), so probabilities follow from the normal model. The only change from a single-observation calculation is dividing by σn\dfrac{\sigma}{\sqrt{n}} rather than σ\sigma.

Effect of sample size

Because the standard error is σn\dfrac{\sigma}{\sqrt{n}}, quadrupling the sample size halves the spread of Xˉ\bar{X}. This is why larger samples produce tighter estimates and narrower confidence intervals: the sampling distribution concentrates around μ\mu.

Working backwards to find a sample size

A favourite extended-response task gives a probability statement about Xˉ\bar{X} and asks for the sample size. The method reverses standardising: convert the probability to a critical zz-value, then solve (valueμ)σ/n=z\dfrac{\text{(value} - \mu)}{\sigma/\sqrt{n}} = z for nn. Because nn must be a whole number of observations, round the result, and state explicitly that a sample size is an integer. Reading the correct zz from the stated tail probability (lower tail, upper tail, or central region) is where care is needed, and a sketch of the normal curve with the area shaded prevents sign errors.

Normal population versus the central limit theorem

It is worth distinguishing two reasons Xˉ\bar{X} might be normal. If the population itself is normal, then Xˉ\bar{X} is exactly normal for every sample size, however small, because a sum of normal variables is normal. If the population is not normal, Xˉ\bar{X} is only approximately normal, and only for a large enough nn, by the central limit theorem. An exam question that asks you to justify normality is testing exactly this distinction: cite the normal population if one is given, and cite the central limit theorem only when the population shape is unknown or non-normal and nn is large.

Single observation versus the mean of a sample

The most consequential modelling decision is whether a question concerns one randomly chosen value or the average of a sample. A single observation XX has standard deviation σ\sigma; the sample mean Xˉ\bar{X} has the much smaller standard error σn\dfrac{\sigma}{\sqrt{n}}. Using the wrong one is the difference between a correct and an incorrect probability. Read the wording carefully: phrases like "the mean of the sample" or "the average" signal Xˉ\bar{X}, while "a randomly selected" individual signals XX.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

QCAA 20245 marksPaper 2 (complex familiar). Heights of Year 12 students are normal with mean 168.6168.6 cm and standard deviation 12.712.7 cm. A random sample of 2020 is taken. (a) Explain why the sample mean is normally distributed. (b) Determine P(Xˉ>170)P(\bar{X} > 170). (c) If there is a 75%75\% probability that Xˉ\bar{X} lies within ±h\pm h of the mean, determine P(Xˉ168.6+h)P(\bar{X} \geq 168.6 + h). (d) Hence determine hh.
Show worked answer →

Standard error =σn=12.7202.840= \dfrac{\sigma}{\sqrt{n}} = \dfrac{12.7}{\sqrt{20}} \approx 2.840 cm; mean μ=168.6\mu = 168.6.

(a) Because the population is itself normal, Xˉ\bar{X} is normal for any sample size, not relying on a large nn.

(b) z=170168.62.8400.493z = \dfrac{170 - 168.6}{2.840} \approx 0.493, so P(Xˉ>170)=P(Z>0.493)0.31.P(\bar{X} > 170) = P(Z > 0.493) \approx 0.31.

(c) The central 75%75\% leaves 25%25\% in the two tails, so each tail is 12.5%12.5\%; P(Xˉ168.6+h)=0.125.P(\bar{X} \geq 168.6 + h) = 0.125.

(d) The upper critical value for a 0.1250.125 tail is z1.150z \approx 1.150, so h=1.150×2.8403.27h = 1.150 \times 2.840 \approx 3.27 cm.

Markers reward the normality reason, the standardised probability, the tail logic, and the value of hh.

QCAA 20237 marksPaper 2 (complex unfamiliar). University travel times are normal with mean 25.225.2 min and standard deviation 4.74.7 min. A random sample of 120120 gives Xˉ\bar{X}. (a) Determine P(24.5Xˉ25.9)P(24.5 \leq \bar{X} \leq 25.9). (b) Given P(Xˉk)=0.8P(\bar{X} \leq k) = 0.8, determine kk. (c) A second sample has P(Xˉ24.6)=0.05P(\bar{X} \leq 24.6) = 0.05; determine its size.
Show worked answer →

First sample: standard error =4.71200.4291= \dfrac{4.7}{\sqrt{120}} \approx 0.4291 min; mean 25.225.2.

(a) zz-values 24.525.20.42911.631\dfrac{24.5 - 25.2}{0.4291} \approx -1.631 and 25.925.20.42911.631\dfrac{25.9 - 25.2}{0.4291} \approx 1.631, so P(1.631Z1.631)0.897.P(-1.631 \leq Z \leq 1.631) \approx 0.897.

(b) For a lower area 0.80.8, z0.8416z \approx 0.8416, so k=25.2+0.8416×0.429125.56k = 25.2 + 0.8416 \times 0.4291 \approx 25.56 min.

(c) P(Xˉ24.6)=0.05P(\bar{X} \leq 24.6) = 0.05 puts 24.624.6 at z=1.645z = -1.645: 24.6=25.21.6454.7n24.6 = 25.2 - 1.645\cdot\dfrac{4.7}{\sqrt{n}}, so n=1.645×4.70.612.886\sqrt{n} = \dfrac{1.645 \times 4.7}{0.6} \approx 12.886, giving n166.n \approx 166.

Markers reward the standard error, both standardisations, and solving for nn with rounding to a whole sample size.

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