How does the mean of a sample behave when we take many samples from a population?
The sampling distribution of the sample mean is approximately normal, centred on the population mean, with a standard deviation that shrinks as the sample size grows.
What a sampling distribution is, the mean and standard error of the sample mean, and how the Central Limit Theorem makes the sample mean approximately normal.
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What this dot point is asking
A single sample gives one estimate of the population mean, but if you took another sample you would get a slightly different . The sampling distribution of the sample mean describes how these sample means vary from sample to sample.
Mean and standard error
The sampling distribution has a smaller spread than the population, because averaging cancels out extreme individual values.
The standard error shrinks as grows, but only with - to halve the spread you must take four times as many observations.
The Central Limit Theorem
This is what makes statistical inference possible: even with a skewed population, you can use normal-distribution methods on the sample mean.
Common errors
Why it matters
The sampling distribution is the conceptual core of Topic 6. The standard error and the Central Limit Theorem are exactly what allow you to build confidence intervals for a population mean in the next dot points, and they explain why larger samples give more precise estimates.
Exam-style practice questions
Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2023 SACE Stage 22 marksDaily milk production per cow has a mean of 21.9 litres and a standard deviation of 3.26 litres. Let S90 be the total daily milk production of a random sample of 90 cows. Show that S90 has a mean of 1971 litres and a standard deviation of 30.93 litres (correct to four significant figures).Show worked answer →
For a sum of n independent observations, the mean adds and the variances add (so the standard deviation is multiplied by sqrt(n)).
Mean of S90 = n times population mean = 90 times 21.9 = 1971 litres.
Standard deviation of S90 = sqrt(n) times population sd = sqrt(90) times 3.26 = 9.4868 times 3.26 = 30.93 litres (to four significant figures).
Marks: one for the mean (multiply by 90), one for the standard deviation (multiply by sqrt(90), not by 90). The common error is multiplying the standard deviation by 90 instead of sqrt(90); only variances add directly, so the standard deviation scales with the square root of the sample size.
2017 SACE Stage 22 marksThe number of shares of a meme has a mean of 131.25 and a standard deviation of 221. Let S75 be the sum of the shares of a random sample of 75 memes. Show that S75 has a mean of 9844 shares and a standard deviation of 1914 shares.Show worked answer →
For the sum of n independent variables, multiply the mean by n and the standard deviation by sqrt(n).
Mean of S75 = 75 times 131.25 = 9843.75, which rounds to 9844 shares.
Standard deviation of S75 = sqrt(75) times 221 = 8.6603 times 221 = 1913.9, which rounds to 1914 shares.
Marks: one for the mean, one for the standard deviation using sqrt(75). As with any sum, the standard deviation grows with sqrt(n) because variances (not standard deviations) are additive. By the Central Limit Theorem, S75 is also approximately normal, which lets later parts compute probabilities for the total number of shares.