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Year 12 Specialist (Unit 3 & 4) Statistical inference

Distribution of the sample mean (CLT)

20 practice questions 0 video lessons Theory + worked examples

Master the distribution of the sample mean and the Central Limit Theorem in Year 12 Specialist Mathematics for Queensland (QCAA). The sample mean \(\bar X\) is a random variable with mean \(\mu\) and standard deviation \(\dfrac{\sigma}{\sqrt n}\), and for a large sample it is approximately normal whatever the parent distribution.

You will learn to find the standard deviation of \(\bar X\), state the model \(\bar X\sim N\!\left(\mu,\dfrac{\sigma^2}{n}\right)\) when \(n\ge 30\), standardise with \(z=\dfrac{\bar x-\mu}{s/\sqrt n}\) and find probabilities for the sample mean — the foundation of statistical inference in Unit 4.

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Theory

The Central Limit Theorem (CLT) says that for a large sample (\(n\ge 30\)) the sample mean \(\bar X\) is approximately normal, \(\bar X\sim N\!\left(\mu,\dfrac{\sigma^2}{n}\right)\), whatever the shape of the parent distribution. In Year 12 Specialist Mathematics (QCAA, Queensland) you find the sample-mean standard deviation \(\dfrac{\sigma}{\sqrt n}\), standardise with \(z=\dfrac{\bar x-\mu}{s/\sqrt n}\), and use the normal distribution to find probabilities for \(\bar X\).

Take a random sample of size \(n\) from a population and compute its mean. A different sample gives a different mean, so the sample mean \(\bar X\) is itself a random variable, with its own distribution called the sampling distribution of the sample mean.

If the parent random variable \(X\) has mean \(\mu\) and standard deviation \(\sigma\), then \(\bar X\) has the same mean \(\mu\), but a smaller standard deviation \(\dfrac{\sigma}{\sqrt n}\) (often called the standard error). Larger samples make \(\bar X\) cluster more tightly around \(\mu\).

The Central Limit Theorem is the key result: for a large sample \(n\ge 30\), the sampling distribution of \(\bar X\) is approximately normal, \(\bar X\sim N\!\left(\mu,\dfrac{\sigma^2}{n}\right)\), regardless of the shape of the parent distribution — even a strongly skewed parent gives a bell-shaped sampling distribution.

To find a probability you standardise: \(z=\dfrac{\bar x-\mu}{\sigma/\sqrt n}\) is approximately \(N(0,1)\). When \(\sigma\) is unknown, the sample standard deviation \(s\) is used in its place, and \(\dfrac{\bar X-\mu}{s/\sqrt n}\) is approximately standard normal for large \(n\).

Central Limit Theorem: skewed parent versus normal sampling distribution On the left a strongly right-skewed parent distribution of X. Repeated sampling of means produces, on the right, an approximately normal bell-shaped sampling distribution of the sample mean X-bar, narrower and centred at the same mean mu. μ Parent: X (skewed) CLT μ Sample mean X̄
The CLT in one picture: however skewed the parent \(X\) is, the sampling distribution of \(\bar X\) is approximately normal and centred at \(\mu\) once \(n\) is large.
Standardised normal curve with a shaded right tail A standard normal bell curve. The area to the right of the standardised value z is shaded; it equals the probability that the sample mean exceeds the given value. μ z area
Standardising turns a question about \(\bar X\) into a standard-normal area: \(P(\bar X>x)=P\!\left(Z>\dfrac{x-\mu}{\sigma/\sqrt n}\right)\).

For a sample of size \(n\) from a population with mean \(\mu\) and standard deviation \(\sigma\), the sample mean has:

\[ E(\bar X)=\mu,\qquad \text{sd}(\bar X)=\dfrac{\sigma}{\sqrt n} \]
sd(X¯)=σn

The Central Limit Theorem: for large \(n\) (\(n\ge 30\)), whatever the parent distribution,

\[ \bar X\;\sim\;N\!\left(\mu,\;\dfrac{\sigma^2}{n}\right) \]
X¯∼N(μ,σ2n)

To find a probability, standardise the sample mean (using \(s\) for \(\sigma\) when \(\sigma\) is unknown):

\[ Z=\dfrac{\bar X-\mu}{\sigma/\sqrt n}\;\approx\;N(0,1),\qquad Z=\dfrac{\bar X-\mu}{s/\sqrt n} \]
Z=X¯−μs/n
The standard deviation is \(\dfrac{\sigma}{\sqrt n}\), not \(\sigma\). Dividing by \(\sqrt n\) is what makes the sampling distribution narrower than the parent. The variance in \(N\!\left(\mu,\dfrac{\sigma^2}{n}\right)\) is \(\dfrac{\sigma^2}{n}\), so take its square root before standardising.

How to solve a sample-mean problem

  1. Check the condition. Confirm the sample is large, \(n\ge 30\) (or that the parent is already normal). Only then does the CLT make \(\bar X\) approximately normal.
  2. State the distribution. Write \(\bar X\sim N\!\left(\mu,\dfrac{\sigma^2}{n}\right)\); the standard deviation of \(\bar X\) is \(\dfrac{\sigma}{\sqrt n}\) (use \(s\) if \(\sigma\) is unknown).
  3. Standardise. Convert the value \(\bar x\) to \(z=\dfrac{\bar x-\mu}{\sigma/\sqrt n}\).
  4. Find the probability. Read the required area under the standard normal curve, subtracting for a "between" interval and using \(1-\Phi(z)\) for an upper tail.
Example 1 — Standard deviation of \(\bar X\)
The masses of oranges have standard deviation \(\sigma=20\) g. For random samples of \(n=100\) oranges, find the standard deviation of the sample mean \(\bar X\).
Solution

The sample-mean standard deviation is \(\sigma\) divided by \(\sqrt n\):

\(\text{sd}(\bar X)\)\(=\)\(\dfrac{\sigma}{\sqrt n}\)
\(=\)\(\dfrac{20}{\sqrt{100}}\)
\(=\)\(\dfrac{20}{10}\)
\(=\)\(2\)

The standard deviation of \(\bar X\) is \(2\) g.

Example 2 — State the distribution
A population has mean \(\mu=250\) and standard deviation \(\sigma=18\). For samples of size \(n=81\), give the approximate distribution of \(\bar X\) from the Central Limit Theorem.
Solution

Since \(n=81\ge 30\), the CLT applies; find the variance \(\dfrac{\sigma^2}{n}\):

\(\text{Var}(\bar X)\)\(=\)\(\dfrac{\sigma^2}{n}\)
\(=\)\(\dfrac{18^2}{81}\)
\(=\)\(\dfrac{324}{81}\)
\(=\)\(4\)

Write the normal model with mean \(\mu\) and this variance:

\(\bar X\)\(\sim\)\(N(250,\,4)\)

\(\bar X\sim N(250,4)\); the standard deviation of \(\bar X\) is \(2\).

Example 3 — Probability for a skewed parent
Service times at a call centre are strongly right-skewed with mean \(\mu=20\) minutes and standard deviation \(\sigma=6\) minutes. For a random sample of \(n=36\) calls, find \(P(\bar X>21.5)\).
Solution

The parent is skewed, but \(n=36\ge 30\), so the CLT applies. Find the sample-mean sd:

\(\dfrac{\sigma}{\sqrt n}\)\(=\)\(\dfrac{6}{\sqrt{36}}\)
\(=\)\(\dfrac{6}{6}\)
\(=\)\(1\)

Standardise the value \(21.5\):

\(z\)\(=\)\(\dfrac{21.5-20}{1}\)
\(=\)\(1.5\)

Read the upper tail of the standard normal:

\(P(\bar X>21.5)\)\(=\)\(P(Z>1.5)\)
\(=\)\(1-0.9332\)
\(=\)\(0.0668\)

\(P(\bar X>21.5)\approx 0.0668\).

Standardised normal curve with a shaded right tail A standard normal bell curve. The area to the right of the standardised value z is shaded; it equals the probability that the sample mean exceeds the given value. μ z area
Example 4 — Between two values (using \(s\))
The masses of cereal boxes have mean \(\mu=500\) g and sample standard deviation \(s=40\) g. For a random sample of \(n=64\) boxes, find \(P(490<\bar X<505)\).
Solution

Find the sample-mean sd using \(s\) in place of \(\sigma\):

\(\dfrac{s}{\sqrt n}\)\(=\)\(\dfrac{40}{\sqrt{64}}\)
\(=\)\(\dfrac{40}{8}\)
\(=\)\(5\)

Standardise both endpoints:

\(z_1\)\(=\)\(\dfrac{490-500}{5}=-2\)
\(z_2\)\(=\)\(\dfrac{505-500}{5}=1\)

Subtract the standard-normal areas:

\(P(490<\bar X<505)\)\(=\)\(\Phi(1)-\Phi(-2)\)
\(=\)\(0.8413-0.0228\)
\(=\)\(0.8185\)

\(P(490<\bar X<505)\approx 0.8185\).

Common pitfalls

Using \(\sigma\) instead of \(\dfrac{\sigma}{\sqrt n}\). The sample mean is far less variable than a single observation. Always divide by \(\sqrt n\): for \(\sigma=6\) and \(n=36\), \(\text{sd}(\bar X)=\dfrac{6}{6}=1\), not \(6\).
Confusing variance with standard deviation. In \(N\!\left(\mu,\dfrac{\sigma^2}{n}\right)\) the second entry is the variance. Take its square root, \(\dfrac{\sigma}{\sqrt n}\), before you standardise.
Applying the CLT when \(n\) is small. For a skewed parent you need \(n\ge 30\). If \(n\) is small and the parent is not normal, \(\bar X\) need not be approximately normal.
Thinking the parent must be normal. The whole power of the CLT is that the parent can be any shape — skewed, exponential, discrete — and \(\bar X\) is still approximately normal once \(n\) is large.

Frequently asked questions

What does the Central Limit Theorem say?

For a large sample (\(n\ge 30\)), the sample mean \(\bar X\) is approximately normal, \(\bar X\sim N\!\left(\mu,\dfrac{\sigma^2}{n}\right)\), no matter what shape the parent distribution has.

What is the standard deviation of the sample mean?

It is \(\dfrac{\sigma}{\sqrt n}\), the population standard deviation divided by the square root of the sample size. It is smaller than \(\sigma\), and it shrinks as \(n\) grows.

How large does the sample need to be?

As a rule, \(n\ge 30\) is large enough for the CLT to make \(\bar X\) approximately normal for most parent distributions. If the parent is already normal, \(\bar X\) is exactly normal for any \(n\).

Do I use \(\sigma\) or the sample standard deviation \(s\)?

Use \(\sigma\) if it is known. If it is not, use the sample standard deviation \(s\) in its place: for large \(n\), \(\dfrac{\bar X-\mu}{s/\sqrt n}\) is approximately standard normal.

How do I find a probability for \(\bar X\)?

Standardise: \(z=\dfrac{\bar x-\mu}{\sigma/\sqrt n}\), then read the required area under the standard normal curve, just as for any normal probability.

Does the parent distribution have to be normal?

No. The CLT works for any parent shape — skewed, exponential or discrete — provided the sample is large. That is exactly why the theorem is so useful.

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