Distribution of the Sample Mean

A2 · S2 · 13 min

Take a random sample and work out its mean. Take another sample of the same size and you get a different mean. Before the sample is drawn, the sample mean is uncertain, so it is a random variable with its own distribution. This note finds that distribution: its mean, its variance, and its exact shape when the population is normal. These facts drive every confidence interval and every test for a mean on Paper 6, and questions that ask directly for $P(\bar{X} > k)$ appear on most papers.

The sample mean is a random variable

Suppose $X$ is some quantity measured on members of a population: the mass of a loaf, the score on a spinner, the lifetime of a bulb. A random sample of size $n$ gives $n$ observations $X_1, X_2, \dots, X_n$ . Because the sample is random, these are independent and each has the same distribution as $X$ . The sample mean is

\bar{X} = \frac{X_1 + X_2 + \cdots + X_n}{n}.

The capital letter matters. $\bar{X}$ is the random variable: the mean of a sample not yet taken. $\bar{x}$ is a number: the mean of the sample you actually took.

A tiny example makes this concrete. A spinner gives $1$ , $2$ or $3$ , each with probability $\tfrac{1}{3}$ . Spin it twice and take the mean. The nine equally likely outcomes give:

Spins	$(1,1)$	$(1,2), (2,1)$	$(1,3), (2,2), (3,1)$	$(2,3), (3,2)$	$(3,3)$
$\bar{x}$	$1$	$1.5$	$2$	$2.5$	$3$
$P(\bar{X} = \bar{x})$	$\tfrac{1}{9}$	$\tfrac{2}{9}$	$\tfrac{3}{9}$	$\tfrac{2}{9}$	$\tfrac{1}{9}$

A single spin has mean $2$ and variance $E(X^2) - 2^2 = \tfrac{14}{3} - 4 = \tfrac{2}{3}$ . The sample mean also has mean $2$ , but its variance is

E(\bar{X}^2) - 2^2 = \frac{1 + 2(2.25) + 3(4) + 2(6.25) + 9}{9} - 4 = \frac{39}{9} - 4 = \frac{1}{3}.

That is exactly half of $\tfrac{2}{3}$ , and the sample size was $2$ . The extreme values $1$ and $3$ need both spins to be extreme, so they are rarer for the mean than for a single spin, and the distribution of $\bar{X}$ is pulled in towards the centre. This is the whole story in miniature.

Mean and variance of $\bar{X}$

The general result follows from the rules for linear combinations of random variables. Let $X$ have mean $\mu$ and variance $\sigma^2$ . Then

E(\bar{X}) = \frac{1}{n}\big(E(X_1) + \cdots + E(X_n)\big) = \frac{1}{n}(n\mu) = \mu,

and, because the $X_i$ are independent,

\text{Var}(\bar{X}) = \frac{1}{n^2}\big(\text{Var}(X_1) + \cdots + \text{Var}(X_n)\big) = \frac{1}{n^2}(n\sigma^2) = \frac{\sigma^2}{n}.

Key result

For a random sample of size $n$ from a population with mean $\mu$ and variance $\sigma^2$ :

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

For the sample total $T = X_1 + X_2 + \cdots + X_n$ :

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

Two things to notice. First, the sample mean is centred on the population mean: on average it gets the right answer, which is why $\bar{x}$ is a sensible estimate of $\mu$ . Second, the spread shrinks as $n$ grows, but only with $\sqrt{n}$ . To halve the standard deviation of the mean you need four times as many observations.

The standard deviation $\dfrac{\sigma}{\sqrt{n}}$ is often called the standard error of the mean. The syllabus does not require the term, but you will meet it in textbooks.

Sum of a sample is not a multiple of one value

The total of $n$ independent observations has variance $n\sigma^2$ , not $n^2\sigma^2$ . The expression $nX$ means one observation multiplied by $n$ , with variance $n^2\sigma^2$ . Six eggs in a box are six different eggs, so their total is $X_1 + \cdots + X_6$ , not $6X$ .

When the population is normal

The mean and variance of $\bar{X}$ hold for any population. The shape of the distribution needs one more fact. A sum of independent normal variables is normal (see linear combinations of normal variables), and dividing by $n$ keeps it normal. So:

Key result

If $X \sim N(\mu, \sigma^2)$ , then for a random sample of any size $n$

\bar{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right) \quad \text{exactly}, \qquad T = \sum X_i \sim N(n\mu, n\sigma^2).

The graph shows a population $X \sim N(80, 10^2)$ (the widest curve) with the distributions of the sample mean for $n = 4$ , where $\bar{X} \sim N(80, 5^2)$ , and for $n = 16$ , where $\bar{X} \sim N(80, 2.5^2)$ (the tallest curve). All three are centred on $80$ ; the bigger the sample, the more tightly the mean clusters around $80$ .

y = exp(-(x - 80)^2 / 200) / (10 sqrt(2 pi)) y = exp(-(x - 80)^2 / 50) / (5 sqrt(2 pi)) y = exp(-(x - 80)^2 / 12.5) / (2.5 sqrt(2 pi))

If the population is not normal, $\bar{X}$ still has mean $\mu$ and variance $\dfrac{\sigma^2}{n}$ , but its shape is not exactly normal. For a large sample it is approximately normal anyway: that is the Central Limit Theorem, the next note. For a small sample from a non-normal population, Paper 6 cannot ask you for probabilities about $\bar{X}$ beyond listing outcomes as in the spinner example.

Probabilities for a sample mean

Define the variable and state the population distribution: " $X \sim N(800, 15^2)$ , where $X$ is the mass of a loaf in grams".
Write down the distribution of $\bar{X}$ , with the variance divided by $n$ : " $\bar{X} \sim N\left(800, \tfrac{15^2}{9}\right)$ ". Say why it is normal.
Standardise with the standard deviation of the mean: $z = \dfrac{\bar{x} - \mu}{\sigma/\sqrt{n}}$ .
Use the normal tables and a sketch to find the probability.

Routine: one sample mean

The masses of loaves from a bakery are normally distributed with mean $800$ g and standard deviation $15$ g.

(a) Find the probability that a single randomly chosen loaf has mass less than $790$ g.

(b) Find the probability that the mean mass of a random sample of $9$ loaves is less than $790$ g.

Solution

Let $X$ g be the mass of a loaf, $X \sim N(800, 15^2)$ .

(a)

P(X < 790) = P\left(Z < \frac{790 - 800}{15}\right) = P(Z < -0.667) = 1 - 0.7476 = 0.252

(b) The population is normal, so $\bar{X} \sim N\left(800, \dfrac{15^2}{9}\right) = N(800, 25)$ , with standard deviation $5$ .

P(\bar{X} < 790) = P\left(Z < \frac{790 - 800}{5}\right) = P(Z < -2) = 1 - 0.9772 = 0.0228

(c) The sample mean has a smaller variance than a single loaf. For the mean to be below $790$ g, the nine loaves would have to be light on average; one heavy loaf would pull the mean back up. So a low mean is much less likely than one light loaf.

A discrete population

The random variable $X$ has the following probability distribution.

$x$	$0$	$1$	$2$	$3$
$P(X = x)$	$0.1$	$0.3$	$0.4$	$0.2$

(a) Find $E(\bar{X})$ and $\text{Var}(\bar{X})$ , where $\bar{X}$ is the mean of a random sample of $10$ observations of $X$ .

(b) A random sample of $2$ observations is taken. Find $P(\bar{X} = 1)$ and $P(\bar{X} > 2.5)$ .

Solution

(a)

E(X) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) = 1.7

E(X^2) = 0 + 0.3 + 1.6 + 1.8 = 3.7, \qquad \text{Var}(X) = 3.7 - 1.7^2 = 0.81

E(\bar{X}) = 1.7, \qquad \text{Var}(\bar{X}) = \frac{0.81}{10} = 0.081

(b) $\bar{X} = 1$ means the two values add to $2$ : $(0, 2)$ , $(2, 0)$ or $(1, 1)$ .

P(\bar{X} = 1) = 0.1(0.4) + 0.4(0.1) + 0.3(0.3) = 0.04 + 0.04 + 0.09 = 0.17

$\bar{X} > 2.5$ needs a total greater than $5$ , which is only $(3, 3)$ .

P(\bar{X} > 2.5) = 0.2^2 = 0.04

Note that the population is not normal and the sample in (b) is tiny, so the only way to find these probabilities is to list the outcomes.

Finding a sample size

The time taken by a machine to complete a cycle is normally distributed with mean $50$ seconds and standard deviation $6$ seconds. Find the smallest sample size $n$ for which the probability that the mean time of the sample exceeds $52$ seconds is less than $0.01$ .

Solution

$\bar{X} \sim N\left(50, \dfrac{36}{n}\right)$ , so the standard deviation of $\bar{X}$ is $\dfrac{6}{\sqrt{n}}$ .

We need

P(\bar{X} > 52) < 0.01 \quad\Longleftrightarrow\quad \frac{52 - 50}{6/\sqrt{n}} > 2.326.

The value $2.326$ comes from the table of critical values: $\Phi(2.326) = 0.99$ .

\frac{2\sqrt{n}}{6} > 2.326 \;\Rightarrow\; \sqrt{n} > 6.978 \;\Rightarrow\; n > 48.69

The smallest sample size is $n = 49$ .

Check the direction: a larger $n$ makes the mean less spread out, so the probability of it straying above $52$ falls. The inequality $n > 48.69$ is consistent with that.

Working backwards to the mean

Rods are cut to lengths that are normally distributed with mean $\mu$ cm and standard deviation $0.4$ cm. For a random sample of $25$ rods, the probability that the sample mean exceeds $20.1$ cm is $0.05$ . Find $\mu$ .

Solution

$\bar{X} \sim N\left(\mu, \dfrac{0.4^2}{25}\right)$ , standard deviation $\dfrac{0.4}{5} = 0.08$ .

$P(\bar{X} > 20.1) = 0.05$ , so $20.1$ is $1.645$ standard deviations above the mean:

\frac{20.1 - \mu}{0.08} = 1.645 \;\Rightarrow\; \mu = 20.1 - 0.1316 = 19.97 \text{ (4 s.f.)}

Exam-hard: comparing two samples

The masses of apples of variety A are normally distributed with mean $150$ g and standard deviation $20$ g. The masses of apples of variety B are normally distributed with mean $140$ g and standard deviation $15$ g. A random sample of $8$ apples of variety A and a random sample of $10$ apples of variety B are chosen.

(a) Find the probability that the mean mass of the A apples is greater than the mean mass of the B apples.

(b) Find the probability that the total mass of the $8$ A apples is greater than the total mass of the $10$ B apples.

Solution

(a)

\bar{A} \sim N\left(150, \frac{400}{8}\right) = N(150, 50), \qquad \bar{B} \sim N\left(140, \frac{225}{10}\right) = N(140, 22.5)

The samples are independent, so

\bar{A} - \bar{B} \sim N(150 - 140,\ 50 + 22.5) = N(10, 72.5).

P(\bar{A} - \bar{B} > 0) = P\left(Z > \frac{0 - 10}{\sqrt{72.5}}\right) = P(Z > -1.174) = 0.880

(b) Totals: $T_A \sim N(8 \times 150,\ 8 \times 400) = N(1200, 3200)$ and $T_B \sim N(10 \times 140,\ 10 \times 225) = N(1400, 2250)$ .

T_A - T_B \sim N(-200,\ 5450)

P(T_A - T_B > 0) = P\left(Z > \frac{200}{\sqrt{5450}}\right) = P(Z > 2.709) = 1 - 0.99663 = 0.00337

The variances add in both parts even though the variables are subtracted. Part (b) is very different from part (a) because there are more B apples, so the B total is larger on average.

Common mistakes

Using $\sigma$ instead of $\dfrac{\sigma}{\sqrt{n}}$ . When the question is about a sample mean, standardise with the standard deviation of the mean. This is the single most common error in the topic.
Dividing the standard deviation by $n$ . The variance is divided by $n$ ; the standard deviation is divided by $\sqrt{n}$ . $N\left(800, \tfrac{15^2}{9}\right)$ has standard deviation $5$ , not $\tfrac{15}{9}$ .
Writing $N(800, 15)$ when the standard deviation is $15$ . The second parameter of $N(\mu, \sigma^2)$ is the variance. Write $N(800, 15^2)$ or $N(800, 225)$ .
Treating a total as $nX$ . The total of $n$ observations has variance $n\sigma^2$ , not $n^2\sigma^2$ .
Claiming $\bar{X}$ is normal for a small sample from a non-normal population. It is exactly normal only if $X$ is normal; otherwise you need a large sample and the Central Limit Theorem.
Subtracting variances. For $\bar{A} - \bar{B}$ the variances add.

Exam tip

Always write the distribution of $\bar{X}$ in full before standardising, for example " $\bar{X} \sim N\left(800, \tfrac{225}{9}\right)$ ". This line usually carries a mark of its own.
When the population is stated to be normal, say "since $X$ is normal, $\bar{X}$ is normal". Do not quote the Central Limit Theorem here: the result is exact, and examiners penalise quoting the CLT when it is not needed.
"Find the smallest $n$ " questions: set up the inequality, solve it, and round up to the next integer, whatever the decimal part.
Use the critical values printed under the normal table ( $1.645$ , $1.96$ , $2.326$ , $2.576$ ) rather than reading them inexactly from the main table.
Give probabilities to 3 significant figures, keeping 4 or more figures in intermediate values such as $z$ .

Summary

$\bar{X}$ , the mean of a random sample of size $n$ , is a random variable.
$E(\bar{X}) = \mu$ and $\text{Var}(\bar{X}) = \dfrac{\sigma^2}{n}$ for any population; the standard deviation of $\bar{X}$ is $\dfrac{\sigma}{\sqrt{n}}$ .
The sample total has mean $n\mu$ and variance $n\sigma^2$ .
If $X$ is normal, $\bar{X} \sim N\left(\mu, \dfrac{\sigma^2}{n}\right)$ exactly, for every $n$ .
Standardise a sample mean with $z = \dfrac{\bar{x} - \mu}{\sigma/\sqrt{n}}$ .
Larger samples give means that cluster more tightly around $\mu$ ; quadrupling $n$ halves the spread.
For differences of sample means or totals, the means subtract and the variances add.

Practice questions

Question

$X \sim N(24, 5^2)$ . A random sample of $10$ observations of $X$ is taken. State the distribution of $\bar{X}$ and find $P(\bar{X} > 25.5)$ .
The random variable $X$ takes the values $1$ , $2$ and $5$ with probabilities $0.5$ , $0.3$ and $0.2$ respectively. Find the mean and variance of the mean of a random sample of $40$ observations of $X$ .
The masses of eggs are normally distributed with mean $62$ g and standard deviation $4$ g. Eggs are packed in boxes of $6$ . Find the probability that the total mass of eggs in a randomly chosen box is less than $360$ g.
$X \sim N(\mu, 3^2)$ . For a random sample of $16$ observations, $P(\bar{X} < 48.2) = 0.1$ . Find $\mu$ .
IQ scores are normally distributed with mean $100$ and standard deviation $15$ . Find the smallest sample size for which the probability that the sample mean lies within $3$ of $100$ is at least $0.95$ .
The heights of a species of plant are normally distributed with mean $70$ cm and standard deviation $12$ cm. Find the probability that the mean height of a random sample of $9$ plants lies between $66$ cm and $75$ cm.
$X \sim N(30, 2^2)$ and $Y \sim N(29, 3^2)$ are independent. A random sample of $5$ observations of $X$ and a random sample of $4$ observations of $Y$ are taken. Find the probability that $\bar{X} > \bar{Y}$ .
$X \sim N(20, 16)$ . A random sample of $n$ observations of $X$ is taken, and $P(\bar{X} > 21) = 0.0668$ . (a) Find $n$ . (b) Find the probability that the sum of the $n$ observations exceeds $740$ .

Answers

Since $X$ is normal, $\bar{X} \sim N\left(24, \dfrac{25}{10}\right) = N(24, 2.5)$ . $P(\bar{X} > 25.5) = P\left(Z > \dfrac{1.5}{\sqrt{2.5}}\right) = P(Z > 0.949) = 1 - 0.8287 = 0.171$ .
$E(X) = 0.5 + 0.6 + 1.0 = 2.1$ ; $E(X^2) = 0.5 + 1.2 + 5 = 6.7$ ; $\text{Var}(X) = 6.7 - 2.1^2 = 2.29$ . $E(\bar{X}) = 2.1$ and $\text{Var}(\bar{X}) = \dfrac{2.29}{40} = 0.0573$ (3 s.f.).
$T = X_1 + \cdots + X_6 \sim N(6 \times 62,\ 6 \times 16) = N(372, 96)$ . $P(T < 360) = P\left(Z < \dfrac{360 - 372}{\sqrt{96}}\right) = P(Z < -1.225) = 1 - 0.8897 = 0.110$ .
$\bar{X} \sim N\left(\mu, \dfrac{9}{16}\right)$ , standard deviation $0.75$ . Since $P(\bar{X} < 48.2) = 0.1$ , $48.2$ is below the mean: $\dfrac{48.2 - \mu}{0.75} = -1.282$ , so $\mu = 48.2 + 0.9615 = 49.16$ (4 s.f.).
$\bar{X} \sim N\left(100, \dfrac{225}{n}\right)$ . $P(97 < \bar{X} < 103) \ge 0.95$ needs $\dfrac{3}{15/\sqrt{n}} \ge 1.96$ , so $\sqrt{n} \ge 9.8$ and $n \ge 96.04$ . The smallest sample size is $97$ .
$\bar{X} \sim N\left(70, \dfrac{144}{9}\right) = N(70, 16)$ , standard deviation $4$ . $P(66 < \bar{X} < 75) = P(-1 < Z < 1.25) = 0.8944 - (1 - 0.8413) = 0.8944 - 0.1587 = 0.736$ .
$\bar{X} \sim N\left(30, \dfrac{4}{5}\right)$ and $\bar{Y} \sim N\left(29, \dfrac{9}{4}\right)$ , so $\bar{X} - \bar{Y} \sim N(1,\ 0.8 + 2.25) = N(1, 3.05)$ . $P(\bar{X} - \bar{Y} > 0) = P\left(Z > \dfrac{-1}{\sqrt{3.05}}\right) = P(Z > -0.5726) = 0.717$ .
(a) $\bar{X} \sim N\left(20, \dfrac{16}{n}\right)$ . $P(\bar{X} > 21) = 0.0668$ means $\Phi(z) = 0.9332$ , so $z = 1.5$ . $\dfrac{21 - 20}{4/\sqrt{n}} = 1.5 \Rightarrow \sqrt{n} = 6 \Rightarrow n = 36$ . (b) $T \sim N(36 \times 20,\ 36 \times 16) = N(720, 576)$ , standard deviation $24$ . $P(T > 740) = P\left(Z > \dfrac{20}{24}\right) = P(Z > 0.833) = 1 - 0.7976 = 0.202$ . (Equivalently, $T > 740$ is the same event as $\bar{X} > \tfrac{740}{36} = 20.56$ .)

The sample mean is a random variable

Mean and variance of Xˉ\bar{X}Xˉ

When the population is normal

Practice questions

Console

Mean and variance of $\bar{X}$