The Central Limit Theorem
Lesson 23 built the sampling distribution and left two questions open: exactly how wide is it, and what shape does it take? This lesson answers both, and the second answer is the surprising one. The population can be any shape at all , flat, skewed, lumpy, and once the samples are big enough the distribution of their means is approximately normal anyway. That is the central limit theorem, and it is why the normal curve turns up everywhere.
Commit to an answer before reading on. Most people get the shape wrong, and that is the useful part.
A fair die is perfectly flat: $1$ to $6$ , each with probability $\tfrac{1}{6}$ , nothing like a bell. Roll $30$ dice, take the average, and repeat that many times. Sketch the shape those averages make, and say roughly how spread out they are compared with a single die.
- Check $n$ , then write the distribution down. If $n \geq 30$ , you may say $\bar{X}$ is approximately normal, and you can state its mean and variance immediately from $\mu$ , $\sigma^2$ and $n$ .
- Use $\sigma/\sqrt{n}$ , never $\sigma$ . The spread of the sample mean is not the spread of the population. Reaching for $\sigma$ out of habit is the single most expensive error in this topic.
- Use $E(\bar{X}) = \mu$ and $\operatorname{Var}(\bar{X}) = \dfrac{\sigma^2}{n}$ for the sampling distribution of the mean.
- Examine how the sample size changes that variance , including with digital tools.
- State the central limit theorem and apply it when $n \geq 30$ .
- Explain its significance: it holds whatever shape the population has .
Take samples of size $n$ from a population with mean $\mu$ and variance $\sigma^2$ . The sampling distribution of the mean has
$$E(\bar{X}) = \mu \qquad \text{and} \qquad \operatorname{Var}(\bar{X}) = \frac{\sigma^2}{n}$$and therefore a standard deviation of $\dfrac{\sigma}{\sqrt{n}}$ .
The first formula is Lesson 23's observation made general: the distribution is centred exactly on $\mu$ , for every $n$ , including $n = 1$ . The second is the new content, and it is the one that does the work.
Seeing $n$ do it. A single fair die has $\mu = 3.5$ and $\sigma^2 = \tfrac{35}{12} \approx 2.917$ , so $\sigma \approx 1.708$ . Averaging $n$ of them:
| $n$ | $\operatorname{Var}(\bar{X}) = \sigma^2/n$ | $\text{sd}(\bar{X}) = \sigma/\sqrt{n}$ | effect on the spread |
|---|---|---|---|
| $1$ | $2.9167$ | $1.7078$ | the population itself |
| $4$ | $0.7292$ | $0.8539$ | 4 times the sample, half the spread |
| $16$ | $0.1823$ | $0.4270$ | 16 times the sample, a quarter of the spread |
| $64$ | $0.0456$ | $0.2135$ | 64 times the sample, an eighth of the spread |
Each row quadruples $n$ and halves the standard deviation. That is the $\sqrt{n}$ law in one column.
B2 write =2.9167/A2 for the variance and in C2 write =SQRT(B2) for the standard deviation, then fill both down and chart column C against column A. The curve drops steeply to about $n = 30$ and then flattens: going from $n = 5$ to $n = 30$ buys a great deal, and going from $n = 70$ to $n = 100$ buys very little. The check that catches a broken model: column C at $n = 4$ must be exactly half its value at $n = 1$ . If it is a quarter, you have divided the standard deviation by $n$ instead of taking the square root.
For samples of size n from a population with mean μ and variance σ², the sampling distribution of the mean has E(X̄) = μ and Var(X̄) = σ²/n, so its standard deviation is σ/√n. Variance divides by n; the standard deviation divides by √n, so quadrupling the sample halves the spread.
Pause, copy both formulas, and write beneath them "variance $\div n$ , standard deviation $\div \sqrt{n}$ " with the $n = 1, 4, 16, 64$ row of the table beside it.
Quick check: A population has $\sigma = 12$ . For samples of size $n = 36$ , the standard deviation of the sample mean is:
Card 05 gave the centre and the spread. It said nothing about shape , and shape is what you need before you can compute a probability. That is what the theorem supplies.
Notice how little it asks of you. You need $\mu$ , you need $\sigma^2$ , and you need $n \geq 30$ . You are not asked what the population looks like , and that omission is the entire point.
The significance. Irrespective of the population's distribution, for a sufficiently large sample the sampling distribution of the mean is approximately normal. The population may be flat, skewed, spiky, bimodal, or a shape nobody has a name for. Average enough of it and the means go bell-shaped regardless.
Here is that happening, computed exactly rather than simulated. The population is a single fair die, as flat as a distribution gets:
- $n = 1$ is the population itself: six equal bars, no peak, no tails.
- $n = 2$ is already a triangle. Averaging two dice makes $3.5$ far more reachable than $1$ , because $1$ needs two ones and $3.5$ has many routes.
- $n = 5$ is recognisably bell-shaped.
- $n = 10$ is very close to a normal curve.
Every panel is centred on $\mu = 3.5$ , and the printed standard deviations fall exactly as $\sigma/\sqrt{n}$ demands: $1.708$ , $1.208$ , $0.764$ , $0.540$ .
Central limit theorem: for a population with mean μ and variance σ², provided n is large enough (typically n ≥ 30), the sampling distribution of the mean is approximately normal with mean μ and variance σ²/n. It holds irrespective of the shape of the population, which is what makes it useful.
Pause, copy the statement in full including the words "approximately" and "irrespective of the population distribution", and sketch the four dice panels beside it.
True or false: The central limit theorem can only be applied when the population itself is known to be normally distributed.
Worked examples · 3 in a row, reveal as you go
A population has mean $\mu = 64$ and standard deviation $\sigma = 15$ . Samples of size $n = 25$ are drawn. State the mean and variance of the sampling distribution of the mean, and its standard deviation. (3 marks)
In each case, decide whether the central limit theorem lets you treat $\bar{X}$ as approximately normal, and justify it. (a) A strongly right-skewed population, $n = 45$ . (b) The same population, $n = 8$ . (c) A normally distributed population, $n = 8$ . (3 marks)
A population has $\sigma = 20$ . (a) Find the standard deviation of $\bar{X}$ for $n = 25$ . (b) What sample size would halve that? (c) Comment on the cost. (3 marks)
Complete: By the central limit theorem, for $n$ large enough the sampling distribution of the mean is approximately , with mean $\mu$ and variance $\sigma^2$ divided by .
Misconceptions to fix · the 3 traps that cost marks
True or false: A population has $\sigma = 20$ . Taking samples of size $n = 100$ makes the individual values in the population less spread out.
Activities · practice with the ideas
A population has $\mu = 50$ and $\sigma = 8$ . For $n = 16$ , state $E(\bar{X})$ , $\operatorname{Var}(\bar{X})$ and $\text{sd}(\bar{X})$ .
A population has $\sigma = 30$ . What sample size gives $\text{sd}(\bar{X}) = 5$ ?
State the central limit theorem in full, from memory, including what it says about the shape of the population.
Build the spreadsheet from card 05: sample sizes $1$ to $100$ in column A, $=2.9167/A2$ in B, $=SQRT(B2)$ in C, then chart C against A. Describe the shape of the curve, and state where the gain from a bigger sample starts to fade.
A friend says "our population is very skewed, so we cannot use the normal distribution at all". Give the shortest correct reply.
Which does NOT belong? Consequences of the central limit theorem for $n \geq 30$ :
Earlier you sketched the shape made by the averages of $30$ dice, and guessed their spread.
The shape is a bell , almost exactly normal, even though one die is perfectly flat. The spread is $\dfrac{1.708}{\sqrt{30}} \approx 0.312$ , so almost every one of those averages sits between roughly $2.6$ and $4.4$ , while a single die still ranges over the whole of $1$ to $6$ . Averaging did two things at once: it pulled the spread in by a factor of $\sqrt{30}$ , and it changed the shape entirely.
That is the theorem's real content. You do not have to know what the population looks like, and most of the time you never will. Lesson 25 turns this into the thing it was always for: computing the probability that a sample mean lands within given bounds.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A population has mean $\mu = 120$ and standard deviation $\sigma = 18$ . Samples of size $n = 36$ are drawn. (a) State $E(\bar{X})$ , $\operatorname{Var}(\bar{X})$ and $\text{sd}(\bar{X})$ . (b) State whether the central limit theorem allows you to treat $\bar{X}$ as approximately normal, and why. (3 marks)
Q2. Explain the significance of the central limit theorem for populations that are not normally distributed. Include what the theorem requires of the population's shape, and state what happens differently when the population is normal. (3 marks)
Q3. A population has $\sigma = 24$ . (a) Find $\text{sd}(\bar{X})$ for $n = 9$ . (b) Find the sample size needed for $\text{sd}(\bar{X}) = 2$ . (c) A researcher plans to double their sample from $n = 100$ to $n = 200$ and expects the spread of $\bar{X}$ to halve. Explain why they are wrong, and state what sample size would actually halve it. (4 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $E(\bar{X}) = 50$ . $\sigma^2 = 64$ , so $\operatorname{Var}(\bar{X}) = 64 \div 16 = 4$ and $\text{sd}(\bar{X}) = 2$ . Check: $8 \div \sqrt{16} = 8 \div 4 = 2$ .
2. $\dfrac{30}{\sqrt{n}} = 5$ gives $\sqrt{n} = 6$ , so $n = 36$ .
3. For a population with mean $\mu$ and variance $\sigma^2$ , provided the sample size is large enough, typically $n \geq 30$ , the sampling distribution of the mean is approximately normal with mean $\mu$ and variance $\sigma^2/n$ . It requires nothing about the shape of the population: it holds irrespective of that shape.
4. The curve falls steeply at first and then flattens. Most of the gain is used up by about $n = 30$ ; beyond roughly $n = 70$ each extra observation buys very little, because the standard deviation falls only as $1/\sqrt{n}$ . The check is that column C at $n = 4$ must be exactly half its value at $n = 1$ .
5. "That is true of the population, but the central limit theorem is about the distribution of the sample mean, not the data. With $n \geq 30$ , $\bar{X}$ is approximately normal whatever the population's shape."
Q1 (3 marks): (a) $E(\bar{X}) = \mu = 120$ [1]. $\sigma^2 = 18^2 = 324$ , so $\operatorname{Var}(\bar{X}) = 324 \div 36 = 9$ and $\text{sd}(\bar{X}) = 3$ [1]. Check: $18 \div \sqrt{36} = 3$ . (b) Yes. $n = 36 \geq 30$ , so by the central limit theorem $\bar{X}$ is approximately $N(120, 9)$ . The justification is the sample size alone; nothing is needed about the population's shape [1].
Q2 (3 marks): Its significance is that it applies irrespective of the population distribution: for a sufficiently large sample the sampling distribution of the mean is approximately normal even when the population is skewed, flat, bimodal or any other shape [1]. This matters because in practice the population's distribution is unknown, so any method requiring it to be normal would be unusable; the theorem removes that requirement and replaces it with a condition on $n$ [1]. When the population is normal, the sampling distribution of the mean is exactly normal for every $n$ , including small ones, and the $n \geq 30$ condition is not needed at all [1].
Q3 (4 marks): (a) $\text{sd}(\bar{X}) = \dfrac{24}{\sqrt{9}} = \dfrac{24}{3} = 8$ [1]. (b) $\dfrac{24}{\sqrt{n}} = 2$ gives $\sqrt{n} = 12$ , so $n = 144$ [1]. (c) The standard deviation of $\bar{X}$ is $\sigma/\sqrt{n}$ , so it falls with the square root of the sample size, not with the sample size. Doubling $n$ from $100$ to $200$ divides the spread by $\sqrt{2} \approx 1.41$ , a reduction of about $29\%$ , not $50\%$ [1]. To halve it the sample must be quadrupled, to $n = 400$ [1].
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