A population has σ = 20. What sample size gives a standard error of 4? [2 marks]

SQA AH style: standard error-style practice: A population has mean μ = 70 and standard deviation σ = 12. Samples of size 36 are taken. State the mean and standard error of the sample mean X, and find P(X > 73).

Mean of the sample mean: E(X) = μ = 70; standard error: σsqrt(n) = 12sqrt(36) = 126 = 2 (1 mark). By the central limit theorem X ≈ N(70, 2^2). Standardise: Z = 73 - 702 = 1.5 (1 mark). P(X > 73) = P(Z > 1.5) = 1 - 0.9332 = 0.0668 (1 mark). Markers reward the standard error, the standardisation using it, and the tail probability.

ScotlandStatisticsSyllabus dot point

How does the sample mean vary from sample to sample, and why is it approximately normal?

Describe the sampling distribution of the sample mean, calculate its mean and standard error, and state and apply the central limit theorem to find probabilities for a sample mean.

A focused answer to the SQA Advanced Higher Statistics sampling distributions content: the sampling distribution of the sample mean, its expected value and standard error, the central limit theorem, and how to find probabilities for a sample mean by standardising.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

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

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What this dot point is asking
The sampling distribution of the sample mean
The central limit theorem
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What this dot point is asking

A statistic such as the sample mean is itself random: a fresh sample gives a fresh value. The SQA wants you to describe how the sample mean varies, to compute its centre and spread (the standard error), and to use the central limit theorem to treat it as approximately normal so you can find probabilities. This result is the engine behind every confidence interval and hypothesis test in the course.

The sampling distribution of the sample mean

Imagine taking every possible sample of size $n$ and recording each sample mean; the distribution of those means is the sampling distribution of $\bar{X}$ .

The $\sqrt{n}$ in the denominator is why precision improves only slowly: to halve the standard error you must quadruple the sample size.

Find the standard error and a probability

To find a probability for a sample mean, we first compute the standard error (the standard deviation of the sampling distribution), then use the central limit theorem to treat $\bar{X}$ as approximately normal, and finally standardise in the usual way.

Step 1: Identify the population parameters and compute the standard error

The population has mean $\mu = 50$ and standard deviation $\sigma = 8$ ; samples of size $n = 64$ are drawn. The standard error is $\sigma$ divided by $\sqrt{n}$ : a larger sample produces a smaller standard error, meaning the sample means are clustered more tightly around $\mu$ .

\text{SE} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{8}{\sqrt{64}} = \dfrac{8}{8} = 1

By the central limit theorem, $\bar{X} \approx N(50,\ 1^2) = N(50, 1)$ .

Step 2: Standardise to convert the event to a $z$ -score

Subtract the mean of $\bar{X}$ and divide by its standard error, exactly as with any normal variable.

P(\bar{X} < 52) = P\!\left(Z < \dfrac{52 - 50}{1}\right) = P(Z < 2)

Step 3: Look up the probability from the table

$P(Z < 2) = 0.9772$ .

Final answer: $P(\bar{X} < 52) = 0.9772$ .

The central limit theorem

The central limit theorem is the most important result in the course because it makes the normal distribution apply to almost any sample mean.

Two consequences matter for the exam. First, the result holds regardless of the parent distribution's shape, so even a skewed or oddly shaped population gives an approximately normal sample mean once $n$ is large (a sample size of about $30$ is the usual rule of thumb). Second, if the parent population is already normal, then $\bar{X}$ is exactly normal for every $n$ , however small.

How quickly the approximation kicks in depends on the parent shape: a nearly symmetric population needs only a small $n$ , whereas a strongly skewed one needs a larger $n$ before the sample mean looks normal. This is why the rule of thumb is a guide, not a guarantee, and why you should note the parent shape when you justify using the CLT. The total of $n$ observations behaves the same way, since $\sum X = n\bar{X}$ is also approximately normal, with mean $n\mu$ and variance $n\sigma^2$ .

Try this

Q1. A population has $\sigma = 20$ . What sample size gives a standard error of $4$ ? [2 marks]

Cue. $\dfrac{20}{\sqrt{n}} = 4 \Rightarrow \sqrt{n} = 5 \Rightarrow n = 25$ .

Q2. State the distribution of the sample mean of size $9$ from a normal population $N(100, 36)$ . [2 marks]

Cue. A normal parent gives an exactly normal sample mean: $\bar{X} \sim N\!\left(100, \dfrac{36}{9}\right) = N(100, 4)$ , standard error $2$ .

Exam-style practice questions

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

AH style: standard error3 marksA population has mean

\mu = 70

and standard deviation

\sigma = 12

. Samples of size

36

are taken. State the mean and standard error of the sample mean

\bar{X}

, and find

P(\bar{X} > 73)

Show worked answer →

Mean of the sample mean: $E(\bar{X}) = \mu = 70$ ; standard error: $\dfrac{\sigma}{\sqrt{n}} = \dfrac{12}{\sqrt{36}} = \dfrac{12}{6} = 2$ (1 mark).

By the central limit theorem $\bar{X} \approx N(70, 2^2)$ . Standardise: $Z = \dfrac{73 - 70}{2} = 1.5$ (1 mark).

$P(\bar{X} > 73) = P(Z > 1.5) = 1 - 0.9332 = 0.0668$ (1 mark). Markers reward the standard error, the standardisation using it, and the tail probability.

AH style: CLT statement3 marksState the central limit theorem and explain why it lets you treat the mean of a large sample from a skewed population as approximately normal.

Show worked answer →

The central limit theorem states that for a sample of size $n$ from any population with mean $\mu$ and finite standard deviation $\sigma$ , the distribution of the sample mean $\bar{X}$ approaches $N\!\left(\mu, \dfrac{\sigma^2}{n}\right)$ as $n$ becomes large (2 marks).

Because the result holds whatever the shape of the parent population, a large enough sample mean is approximately normal even when the population itself is skewed; the skew of the parent is "averaged out" as $n$ grows (1 mark). Markers reward a correct statement including the mean and variance and the point that the parent distribution need not be normal.

Related dot points

Sources & how we know this

SQA Advanced Higher Statistics Course Specification (C803 77) — SQA (2023)
SQA Advanced Higher Statistics Data Booklet — SQA (2019)