Central Limit Theorem

A2 · S2 · 13 min

If the population is normal, the sample mean is normal. But most real populations are not normal: waiting times are skewed, counts are discrete, incomes have long tails. The Central Limit Theorem says that this hardly matters once the sample is large, because the mean of a large random sample is approximately normally distributed whatever the shape of the population. It is the reason confidence intervals and tests for a mean work for almost any data, and Paper 6 regularly asks you to use it, to say when it is needed, and to explain why it is not needed.

What the theorem says

You already know from the distribution of the sample mean that, for any population with mean $\mu$ and variance $\sigma^2$ ,

E(\bar{X}) = \mu, \qquad \text{Var}(\bar{X}) = \frac{\sigma^2}{n}.

Those two facts say where $\bar{X}$ is centred and how spread out it is. They say nothing about its shape. The Central Limit Theorem supplies the shape.

Central Limit Theorem

For a random sample of size $n$ from any population with mean $\mu$ and variance $\sigma^2$ , if $n$ is large then, approximately,

\bar{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right).

Equivalently, the sample total $X_1 + X_2 + \cdots + X_n$ is approximately $N(n\mu, n\sigma^2)$ .

The approximation improves as $n$ increases. If the population is itself normal, the result is exact for every $n$ and the theorem is not needed.

The syllabus asks only for an informal understanding: you must know what the theorem says, when to use it and how to use it, but not prove it. There is no exact cut-off for "large". A common guide is $n \ge 30$ ; for a population that is nearly symmetrical a smaller sample is enough, and for a very skewed population a larger one is safer. In exam questions the sample sizes are chosen so that it is clear which case you are in: typically $n = 40, 50, 60, 100$ or more.

Why averaging produces a bell shape

Think about throwing one die. Each score from $1$ to $6$ is equally likely, so the distribution is flat. Now throw two dice and average them. A mean of $1$ needs both dice to show $1$ , but a mean of $3.5$ can happen in six ways. The distribution of the mean is a triangle, peaked in the middle. With more dice, extreme means need every die to be extreme, which becomes vanishingly unlikely, while middling means can be made in a huge number of ways. The distribution heaps up in the centre and tails off symmetrically on both sides: a bell.

The same happens for a skewed population. The graph shows a strongly skewed population (the curve that starts at its highest point and decays, an exponential distribution with mean $1$ ) together with the exact distribution of the sample mean for $n = 4$ and for $n = 30$ . At $n = 4$ the skew is still visible. At $n = 30$ the distribution of $\bar{X}$ is narrow, centred on $1$ , and very nearly symmetrical.

y = exp(-x) + 0*sqrt(x) y = 4^4 x^3 exp(-4 x) / fact(3) + 0*sqrt(x) y = 30^30 x^29 exp(-30 x) / fact(29) + 0*sqrt(x)

The theorem is about the distribution of the mean, not about individual values. Take a sample of $50$ waiting times from a skewed population: the $50$ values themselves are still skewed. It is only the means of many such samples that would form a bell.

Which distribution to use for $\bar{X}$

Every question about a sample mean begins with the same decision.

Population	Sample size	Distribution of $\bar{X}$	What to write
Normal	Any	$N\left(\mu, \dfrac{\sigma^2}{n}\right)$ exactly	" $X$ is normal, so $\bar{X}$ is normal"
Not normal, or unknown	Large	Approximately $N\left(\mu, \dfrac{\sigma^2}{n}\right)$	" $n$ is large, so by the Central Limit Theorem $\bar{X}$ is approximately normal"
Not normal, or unknown	Small	Mean $\mu$ , variance $\dfrac{\sigma^2}{n}$ , shape unknown	Normal probabilities cannot be justified

The population can be discrete (Poisson, binomial, any table of probabilities) or continuous (any pdf). Only its mean and variance are needed.

Using the Central Limit Theorem

Find the population mean $\mu$ and variance $\sigma^2$ . For a Poisson population both equal $\lambda$ ; for a binomial $B(m, p)$ they are $mp$ and $mp(1 - p)$ ; for a pdf, integrate.
Check that the population is not known to be normal and that $n$ is large.
Write "by the Central Limit Theorem, $\bar{X} \sim N\left(\mu, \dfrac{\sigma^2}{n}\right)$ approximately", with the numbers in.
Standardise with $\dfrac{\sigma}{\sqrt{n}}$ and use the normal tables.

Routine: a skewed population

The time a customer spends in a shop has mean $12$ minutes and standard deviation $8$ minutes. The distribution of times is skewed. Find the probability that the mean time spent by a random sample of $64$ customers exceeds $14$ minutes.

Solution

The population is not normal, but $n = 64$ is large, so by the Central Limit Theorem

\bar{X} \sim N\left(12, \frac{8^2}{64}\right) = N(12, 1) \text{ approximately.}

P(\bar{X} > 14) = P\left(Z > \frac{14 - 12}{1}\right) = P(Z > 2) = 1 - 0.9772 = 0.0228

A Poisson population

The number of emails a person receives in a day has the distribution $\text{Po}(6)$ .

(a) Find the probability that the mean number of emails per day, over a random sample of $50$ days, is more than $6.5$ .

(b) Explain why the Central Limit Theorem was needed in your answer.

Solution

(a) For $\text{Po}(6)$ , $\mu = 6$ and $\sigma^2 = 6$ . Since $n = 50$ is large, by the Central Limit Theorem

\bar{X} \sim N\left(6, \frac{6}{50}\right) = N(6, 0.12) \text{ approximately.}

P(\bar{X} > 6.5) = P\left(Z > \frac{6.5 - 6}{\sqrt{0.12}}\right) = P(Z > 1.443) = 1 - 0.9255 = 0.0745

(b) The number of emails per day has a Poisson distribution, which is not normal, so the sample mean is not exactly normal. The Central Limit Theorem says it is approximately normal because the sample is large.

A population defined by a pdf

The random variable $X$ has probability density function

f(x) = \begin{cases} \tfrac{3}{8}x^2 & 0 \le x \le 2, \\ 0 & \text{otherwise.} \end{cases}

A random sample of $60$ observations of $X$ is taken. Find the probability that the sample mean is less than $1.45$ .

Solution

First find the population mean and variance.

E(X) = \int_0^2 \tfrac{3}{8}x^3\,dx = \tfrac{3}{8}\left[\tfrac{x^4}{4}\right]_0^2 = \tfrac{3}{8}(4) = 1.5

E(X^2) = \int_0^2 \tfrac{3}{8}x^4\,dx = \tfrac{3}{8}\left[\tfrac{x^5}{5}\right]_0^2 = \tfrac{3}{8} \times \tfrac{32}{5} = 2.4, \qquad \text{Var}(X) = 2.4 - 1.5^2 = 0.15

$X$ is not normal, but $n = 60$ is large, so by the Central Limit Theorem

\bar{X} \sim N\left(1.5, \frac{0.15}{60}\right) = N(1.5, 0.0025) \text{ approximately},

with standard deviation $0.05$ .

P(\bar{X} < 1.45) = P\left(Z < \frac{1.45 - 1.5}{0.05}\right) = P(Z < -1) = 1 - 0.8413 = 0.159

The total of a large sample

Using the shop population from the first example (mean $12$ minutes, standard deviation $8$ minutes), find the probability that the total time spent in the shop by $64$ randomly chosen customers is less than $700$ minutes.

Solution

By the Central Limit Theorem, the total $T$ is approximately normal with

E(T) = 64 \times 12 = 768, \qquad \text{Var}(T) = 64 \times 8^2 = 4096, \qquad \text{standard deviation } 64.

P(T < 700) = P\left(Z < \frac{700 - 768}{64}\right) = P(Z < -1.0625) = 1 - 0.8559 = 0.144

The same answer comes from the mean: $T < 700$ is the same event as $\bar{X} < \tfrac{700}{64} = 10.9375$ , and $\dfrac{10.9375 - 12}{1} = -1.0625$ .

Exam-hard: when the theorem is and is not available

The lifetimes of a certain type of battery have mean $30$ hours and standard deviation $9$ hours. The distribution of lifetimes is not known.

(a) Explain why you cannot find the probability that a single battery lasts more than $32$ hours.

(b) Batteries are sold in packs of $50$ , and each pack may be regarded as a random sample. Find the probability that the mean lifetime of the batteries in a pack is more than $32$ hours.

(c) A shop sells $20$ packs. Find the probability that at least $2$ of these packs have a mean lifetime of more than $32$ hours.

Solution

(a) A probability for a single battery needs the distribution of lifetimes, and only its mean and standard deviation are known. The Central Limit Theorem says nothing about one observation.

(b) $n = 50$ is large, so by the Central Limit Theorem

\bar{X} \sim N\left(30, \frac{81}{50}\right) \text{ approximately}, \qquad \text{standard deviation } \frac{9}{\sqrt{50}} = 1.2728.

P(\bar{X} > 32) = P\left(Z > \frac{2}{1.2728}\right) = P(Z > 1.571) = 1 - 0.9419 = 0.0581

P(Y \ge 2) = 1 - 0.9419^{20} - 20(0.0581)(0.9419)^{19} = 1 - 0.3021 - 0.3726 = 0.325

Common mistakes

Saying "by the Central Limit Theorem, $X$ is normal". The theorem is about $\bar{X}$ (or the sample total), never about $X$ itself. Individual observations keep the population's shape.
Quoting the theorem when the population is normal. If $X \sim N(\mu, \sigma^2)$ , then $\bar{X}$ is exactly normal for every $n$ . Citing the Central Limit Theorem here is wrong and can lose a mark.
Using it for a small sample. With $n = 6$ from an unknown or skewed population, you cannot justify a normal distribution for $\bar{X}$ .
Forgetting to divide the variance by $n$ . The theorem gives $N\left(\mu, \dfrac{\sigma^2}{n}\right)$ , not $N(\mu, \sigma^2)$ .
Thinking the theorem needs a continuous population. Poisson, binomial and other discrete populations are fine; only $\mu$ and $\sigma^2$ are used.
Writing "the sample is large so the population is normal". Sample size cannot change the population.

Exam tip

The phrase examiners want is close to "since $n$ is large, by the Central Limit Theorem $\bar{X}$ is approximately normally distributed". Mention both the large sample and the mean.
"Explain whether it was necessary to use the Central Limit Theorem" is a common one-mark part. If the population was given as normal: "No, because the population is normal, so $\bar{X}$ is normal whatever the sample size." If not: "Yes, because the distribution of $X$ is not normal (or not known), so the theorem is needed to say $\bar{X}$ is approximately normal."
For a Poisson or binomial population, write down $\mu$ and $\sigma^2$ explicitly before using them; the mark for the distribution of $\bar{X}$ depends on both.
Paper 6 questions do not apply a continuity correction to a sample mean found by the Central Limit Theorem; the correction belongs to approximating a single binomial or Poisson count.
Probabilities to 3 significant figures; keep $z$ to at least 3 decimal places.

Summary

For any population with mean $\mu$ and variance $\sigma^2$ , the mean of a large random sample is approximately $N\left(\mu, \dfrac{\sigma^2}{n}\right)$ .
The sample total is approximately $N(n\mu, n\sigma^2)$ .
"Large" is informal; $n \ge 30$ is a common guide.
If the population is normal, $\bar{X}$ is exactly normal for all $n$ : no theorem needed.
Small samples from non-normal or unknown populations cannot be treated as normal.
The theorem describes $\bar{X}$ , not individual observations.
It works for discrete and continuous populations alike; find $\mu$ and $\sigma^2$ first.

Practice questions

Question

A population has mean $70$ and standard deviation $12$ . Find the probability that the mean of a random sample of $100$ observations lies between $68$ and $71$ .
For the population in question 1, find the least sample size $n$ for which $P(\bar{X} < 72) \ge 0.99$ .
The random variable $X$ has pdf $f(x) = \tfrac{1}{2}$ for $0 \le x \le 2$ , so $E(X) = 1$ and $\text{Var}(X) = \tfrac{1}{3}$ . Find the probability that the mean of a random sample of $48$ observations exceeds $1.1$ .
The number of flaws in a roll of cloth has the distribution $\text{Po}(3.2)$ . Find the probability that the mean number of flaws in a random sample of $100$ rolls is less than $3$ .
Each batch of $10$ components contains $X$ defective components, where $X \sim B(10, 0.3)$ . Find the probability that the mean number of defectives per batch, in a random sample of $70$ batches, exceeds $3.2$ .
In each case, state whether $\bar{X}$ is exactly normal, approximately normal by the Central Limit Theorem, or neither, giving a reason. (a) $X \sim N(40, 9)$ , $n = 5$ . (b) $X$ has an unknown distribution, $n = 6$ . (c) $X \sim \text{Po}(2)$ , $n = 80$ .
The random variable $X$ has pdf $f(x) = 2x$ for $0 \le x \le 1$ . A random sample of $72$ observations of $X$ is taken. Find the probability that the sample mean is greater than $0.69$ .
Boxes delivered to a warehouse have masses with mean $25$ kg and standard deviation $6$ kg; the distribution of masses is not known. A container can safely carry $1050$ kg. (a) Find the probability that the total mass of $40$ randomly chosen boxes exceeds $1050$ kg. (b) Find the largest number of boxes that can be loaded if the probability that their total mass exceeds $1050$ kg must be less than $0.01$ . Assume that the number of boxes is large enough for the Central Limit Theorem to apply.

Answers

By the Central Limit Theorem, $\bar{X} \sim N\left(70, \dfrac{144}{100}\right) = N(70, 1.44)$ approximately, standard deviation $1.2$ . $P(68 < \bar{X} < 71) = P(-1.667 < Z < 0.833) = 0.7976 - (1 - 0.9522) = 0.7976 - 0.0478 = 0.750$ .
Need $\dfrac{72 - 70}{12/\sqrt{n}} \ge 2.326$ , so $\sqrt{n} \ge \dfrac{2.326 \times 12}{2} = 13.956$ and $n \ge 194.8$ . The least sample size is $195$ .
By the Central Limit Theorem, $\bar{X} \sim N\left(1, \dfrac{1/3}{48}\right) = N\left(1, \dfrac{1}{144}\right)$ , standard deviation $\dfrac{1}{12}$ . $P(\bar{X} > 1.1) = P(Z > 1.2) = 1 - 0.8849 = 0.115$ .
$\mu = \sigma^2 = 3.2$ . By the Central Limit Theorem, $\bar{X} \sim N\left(3.2, \dfrac{3.2}{100}\right) = N(3.2, 0.032)$ , standard deviation $0.17889$ . $P(\bar{X} < 3) = P\left(Z < \dfrac{-0.2}{0.17889}\right) = P(Z < -1.118) = 1 - 0.8682 = 0.132$ .
$\mu = 10 \times 0.3 = 3$ , $\sigma^2 = 10 \times 0.3 \times 0.7 = 2.1$ . By the Central Limit Theorem, $\bar{X} \sim N\left(3, \dfrac{2.1}{70}\right) = N(3, 0.03)$ , standard deviation $0.17321$ . $P(\bar{X} > 3.2) = P(Z > 1.155) = 1 - 0.8760 = 0.124$ .
(a) Exactly normal, $N(40, 1.8)$ : the population is normal, so the sample mean is normal for any $n$ . (b) Neither: the population is not known to be normal and the sample is too small for the Central Limit Theorem. (c) Approximately normal, $N(2, 0.025)$ , by the Central Limit Theorem, since $n = 80$ is large.
$E(X) = \displaystyle\int_0^1 2x^2\,dx = \tfrac{2}{3}$ ; $E(X^2) = \displaystyle\int_0^1 2x^3\,dx = \tfrac{1}{2}$ ; $\text{Var}(X) = \tfrac{1}{2} - \tfrac{4}{9} = \tfrac{1}{18}$ . By the Central Limit Theorem, $\bar{X} \sim N\left(\tfrac{2}{3}, \dfrac{1/18}{72}\right) = N\left(\tfrac{2}{3}, \tfrac{1}{1296}\right)$ , standard deviation $\tfrac{1}{36}$ . $P(\bar{X} > 0.69) = P\left(Z > (0.69 - \tfrac{2}{3}) \times 36\right) = P(Z > 0.84) = 1 - 0.7995 = 0.200$ .
(a) By the Central Limit Theorem the total $T \sim N(40 \times 25,\ 40 \times 36) = N(1000, 1440)$ approximately, standard deviation $37.947$ . $P(T > 1050) = P(Z > 1.318) = 1 - 0.9062 = 0.0938$ . (b) For $n$ boxes, $T \sim N(25n, 36n)$ approximately. We need $\dfrac{1050 - 25n}{6\sqrt{n}} > 2.326$ . $n = 38$ : $\dfrac{1050 - 950}{6\sqrt{38}} = 2.704 > 2.326$ , so this works. $n = 39$ : $\dfrac{1050 - 975}{6\sqrt{39}} = 2.002 < 2.326$ , so this fails. The largest number of boxes is $38$ . (The expression decreases as $n$ increases, so no larger $n$ can work.)

What the theorem says

Why averaging produces a bell shape

Which distribution to use for Xˉ\bar{X}Xˉ

Practice questions

Console

Which distribution to use for $\bar{X}$