Mathematics 9709 · Probability & Statistics 2

The distribution of the sample mean

Average a bunch of measurements and the flukes cancel out. The sample mean $\bar X$ is itself random, but it has a beautiful distribution: $\bar X\sim N\!\big(\mu,\ \tfrac{\sigma^2}{n}\big)$ — same centre as the population, $\sqrt n$ times tighter, and normal even when the population isn't.

Same centre, √n times tighter

Different samples give different means, so $\bar X$ is a random variable with its own distribution. Two facts hold for any population: averaging keeps the centre, and averaging shrinks the spread by $\sqrt n$ (the variance by $n$ ).

$E(\bar X)=\mu,\qquad \operatorname{Var}(\bar X)=\dfrac{\sigma^2}{n},\qquad z=\dfrac{\bar x-\mu}{\sigma/\sqrt n}$

The whole chapter is the /n: Var(X̄) = σ²/n, so you standardise by σ/√n, NOT σ. Same centre μ, spread shrinks by √n.

The sampling distribution of X̄ keeps the same centre μ but gets taller and narrower as n grows (its spread is σ/√n). Average more, and X̄ huddles around the true mean.

Worked example

Start simple: build it from a die

Roll a fair die, $X\in\{1,\dots,6\}$ : $E(X)=3.5$ , $\operatorname{Var}(X)=\tfrac{35}{12}\approx2.917$ . Take the mean of two rolls, $\bar X$ .

Centre is unchanged: $E(\bar X)=3.5$ .
Spread halves: $\operatorname{Var}(\bar X)=\tfrac{\sigma^2}{2}=\tfrac{35}{24}\approx1.458$ .

Answer

E(\bar X)=3.5,\ \operatorname{Var}(\bar X)=\tfrac{35}{24}

No normal curve yet — just the centre holding and the spread halving.

If the population is normal, $\bar X$ is exactly normal for any $n$ . If it isn't, the Central Limit Theorem makes $\bar X$ approximately normal once $n$ is large. In this course, $n>50$ is usually large enough, whatever the population's shape.

Worked example

State the distribution, then one tail

$X$ has mean 6 and variance 8; $\bar X$ is the mean of $80$ observations. State its distribution and find $P(\bar X<6.4)$ .

$\bar X\approx N\!\big(6,\tfrac{8}{80}\big)=N(6,0.1)$ by the CLT ( $n$ large).
$z=\tfrac{6.4-6}{\sqrt{0.1}}=1.265$ , so $\Phi(1.265)$ .

Answer

P(\bar X<6.4)=\Phi(1.265)=0.897

Worked example

Normal population (exact, small n)

Women drink 500 ml on average, SD 50 ml. For 25 women, 13 litres is enough if the mean is under $520$ ml. The parent is treated as normal, so $\bar X\sim N(500,\tfrac{50^2}{25})=N(500,100)$ exactly, and $z=\tfrac{520-500}{10}=2$ .

Answer

P(\bar X<520)=\Phi(2)=0.977

Worked example

Answer time: normal parent, n = 16

Call-answer times are $N(20,5^2)$ seconds. For 16 random calls, find $P(\bar X<18)$ .

Normal parent → $\bar X\sim N\!\big(20,\tfrac{25}{16}\big)$ , so $\sigma/\sqrt n=\tfrac{5}{4}=1.25$ .
$z=\tfrac{18-20}{1.25}=-1.6$ , so $\Phi(-1.6)$ .

Answer

P(\bar X<18)=0.0548

For a normal population, X̄ is normal for ANY n (no CLT, no “n large” needed); the σ/√n is the only change.

Totals: same idea, a different framing

A question about a total of $n$ items is the same job: $\sum X=n\bar X$ , so $\sum X\sim N(n\mu,\ n\sigma^2)$ . Solve it either way — they agree.

Worked example

Sum vs mean: a bag of 6 pears

Pears are $N(45,52)$ g, packed in bags of 6. Find the percentage of bags whose total mass exceeds 300 g.

Total route: $\sum X\sim N(6\cdot45,\ 6\cdot52)=N(270,312)$ ; $z=\tfrac{300-270}{\sqrt{312}}=1.698$ .
Mean route: 300 g over 6 is a mean of 50 g; $\bar X\sim N(45,\tfrac{52}{6})$ gives the same $z=\tfrac{50-45}{\sqrt{52/6}}=1.698$ .
$P=1-\Phi(1.698)=0.0447$ .

Answer

\approx4.47\%

Total of n items: ΣX ~ N(nμ, nσ²) — multiply BOTH mean and variance by n. A “total > T” question can be rewritten as a “mean > T/n” question.

n = 1·X̄ ~ N(50, 10²/n)·SE = σ/√n = 10.00

The CLT in one picture: a skewed parent (left) becomes a symmetric bell for X̄ (right) once you average enough — the sampling distribution forgets the parent's shape.

Worked example

CLT on a non-normal population

$f(x)=x/2$ on $[0,2]$ (a skewed parent); sample of 50, find $P(\bar X<\tfrac32)$ . Here $E(X)=\tfrac43$ , $\operatorname{Var}(X)=\tfrac29$ .

$n=50$ is large, so by the CLT $\bar X\approx N\!\big(\tfrac43,\tfrac{2/9}{50}\big)$ .
$z=\tfrac{3/2-4/3}{\sqrt{2/450}}=2.5$ , so $\Phi(2.5)$ .

Answer

P(\bar X<\tfrac32)=\Phi(2.5)=0.994

Say “ $n$ is large, so by the CLT $\bar X$ is approximately normal” — that sentence scores a mark.

Worked example

Inverse: reverse from a probability

A normal population has mean 15, variance 9; $\bar X$ is the mean of 35. (a) Find $P(\bar X<16.2)$ . (b) Find $k$ with $P(\bar X<k)=0.75$ .

Standard error: $\sigma/\sqrt n=\sqrt{9/35}=0.5071$ ; $\bar X\sim N(15,\tfrac{9}{35})$ .
(a) $z=\tfrac{16.2-15}{0.5071}=2.366$ , so $\Phi(2.366)=0.991$ .
(b) $\Phi^{-1}(0.75)=0.674$ , so $k=15+0.674\times0.5071=15.34$ .

Answer

P(\bar X<16.2)=0.991,\quad k=15.34

Population	Is X̄ normal?
Normal	exactly, for any n
not Normal, n large (>50)	approximately, by the CLT
not Normal, n small	no — can't assume normality

Your turn— tap to reveal the worked answer (Flour: a total just below the line)

Ciara needs 5 kg of flour, so buys 10 bags labelled 500 g; in fact each bag has mean 510 g, variance 120 g². Find $P(\text{total}<5000)$ .

$\sum X\sim N(10\cdot510,\ 10\cdot120)=N(5100,1200)$ , $z=\tfrac{5000-5100}{\sqrt{1200}}=-2.887$ .

Answer

\Phi(-2.887)\approx0.00194

Examiner note

Standardise with the standard error

\sigma/\sqrt n

, not

\sigma

; cite the CLT (“

n

large”) only when the population isn't already Normal. For a total, scale mean and variance by

n

. (9709/62/O/N/24 Q4)

Why σ/√n, not σ/n

$\operatorname{Var}(\bar X)=\operatorname{Var}\!\big(\tfrac1n\textstyle\sum X_i\big)=\tfrac{1}{n^2}\cdot n\sigma^2=\tfrac{\sigma^2}{n}$ (variances add; the $\tfrac1n$ comes out squared). The SD is its root, $\sigma/\sqrt n$ .

Common mistake

Using

\sigma^2

instead of

\sigma^2/n

; dividing by

n

instead of

\sqrt n

when standardising; for a total, forgetting to scale the variance by

n

; and claiming normality for a non-normal population with small

n

(the CLT needs

n

large).

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The distribution of the sample mean

Same centre, √n times tighter

$E(\bar X)=\mu,\qquad \operatorname{Var}(\bar X)=\dfrac{\sigma^2}{n},\qquad z=\dfrac{\bar x-\mu}{\sigma/\sqrt n}$

The whole chapter is the /n: Var(X̄) = σ²/n, so you standardise by σ/√n, NOT σ. Same centre μ, spread shrinks by √n.

The sampling distribution of X̄ keeps the same centre μ but gets taller and narrower as n grows (its spread is σ/√n). Average more, and X̄ huddles around the true mean.

Worked example

Start simple: build it from a die

Roll a fair die, $X\in\{1,\dots,6\}$ : $E(X)=3.5$ , $\operatorname{Var}(X)=\tfrac{35}{12}\approx2.917$ . Take the mean of two rolls, $\bar X$ .

Centre is unchanged: $E(\bar X)=3.5$ .
Spread halves: $\operatorname{Var}(\bar X)=\tfrac{\sigma^2}{2}=\tfrac{35}{24}\approx1.458$ .

Answer

E(\bar X)=3.5,\ \operatorname{Var}(\bar X)=\tfrac{35}{24}

No normal curve yet — just the centre holding and the spread halving.

Worked example

State the distribution, then one tail

$X$ has mean 6 and variance 8; $\bar X$ is the mean of $80$ observations. State its distribution and find $P(\bar X<6.4)$ .

$\bar X\approx N\!\big(6,\tfrac{8}{80}\big)=N(6,0.1)$ by the CLT ( $n$ large).
$z=\tfrac{6.4-6}{\sqrt{0.1}}=1.265$ , so $\Phi(1.265)$ .

Answer

P(\bar X<6.4)=\Phi(1.265)=0.897

Worked example

Normal population (exact, small n)

Answer

P(\bar X<520)=\Phi(2)=0.977

Worked example

Answer time: normal parent, n = 16

Call-answer times are $N(20,5^2)$ seconds. For 16 random calls, find $P(\bar X<18)$ .

Normal parent → $\bar X\sim N\!\big(20,\tfrac{25}{16}\big)$ , so $\sigma/\sqrt n=\tfrac{5}{4}=1.25$ .
$z=\tfrac{18-20}{1.25}=-1.6$ , so $\Phi(-1.6)$ .

Answer

P(\bar X<18)=0.0548

For a normal population, X̄ is normal for ANY n (no CLT, no “n large” needed); the σ/√n is the only change.

Totals: same idea, a different framing

A question about a total of $n$ items is the same job: $\sum X=n\bar X$ , so $\sum X\sim N(n\mu,\ n\sigma^2)$ . Solve it either way — they agree.

Worked example

Sum vs mean: a bag of 6 pears

Pears are $N(45,52)$ g, packed in bags of 6. Find the percentage of bags whose total mass exceeds 300 g.

Total route: $\sum X\sim N(6\cdot45,\ 6\cdot52)=N(270,312)$ ; $z=\tfrac{300-270}{\sqrt{312}}=1.698$ .
Mean route: 300 g over 6 is a mean of 50 g; $\bar X\sim N(45,\tfrac{52}{6})$ gives the same $z=\tfrac{50-45}{\sqrt{52/6}}=1.698$ .
$P=1-\Phi(1.698)=0.0447$ .

Answer

\approx4.47\%

Total of n items: ΣX ~ N(nμ, nσ²) — multiply BOTH mean and variance by n. A “total > T” question can be rewritten as a “mean > T/n” question.

n = 1·X̄ ~ N(50, 10²/n)·SE = σ/√n = 10.00

The CLT in one picture: a skewed parent (left) becomes a symmetric bell for X̄ (right) once you average enough — the sampling distribution forgets the parent's shape.

Worked example

CLT on a non-normal population

$f(x)=x/2$ on $[0,2]$ (a skewed parent); sample of 50, find $P(\bar X<\tfrac32)$ . Here $E(X)=\tfrac43$ , $\operatorname{Var}(X)=\tfrac29$ .

$n=50$ is large, so by the CLT $\bar X\approx N\!\big(\tfrac43,\tfrac{2/9}{50}\big)$ .
$z=\tfrac{3/2-4/3}{\sqrt{2/450}}=2.5$ , so $\Phi(2.5)$ .

Answer

P(\bar X<\tfrac32)=\Phi(2.5)=0.994

Say “ $n$ is large, so by the CLT $\bar X$ is approximately normal” — that sentence scores a mark.

Worked example

Inverse: reverse from a probability

A normal population has mean 15, variance 9; $\bar X$ is the mean of 35. (a) Find $P(\bar X<16.2)$ . (b) Find $k$ with $P(\bar X<k)=0.75$ .

Standard error: $\sigma/\sqrt n=\sqrt{9/35}=0.5071$ ; $\bar X\sim N(15,\tfrac{9}{35})$ .
(a) $z=\tfrac{16.2-15}{0.5071}=2.366$ , so $\Phi(2.366)=0.991$ .
(b) $\Phi^{-1}(0.75)=0.674$ , so $k=15+0.674\times0.5071=15.34$ .

Answer

P(\bar X<16.2)=0.991,\quad k=15.34

Population	Is X̄ normal?
Normal	exactly, for any n
not Normal, n large (>50)	approximately, by the CLT
not Normal, n small	no — can't assume normality

Your turn— tap to reveal the worked answer (Flour: a total just below the line)

Ciara needs 5 kg of flour, so buys 10 bags labelled 500 g; in fact each bag has mean 510 g, variance 120 g². Find $P(\text{total}<5000)$ .

$\sum X\sim N(10\cdot510,\ 10\cdot120)=N(5100,1200)$ , $z=\tfrac{5000-5100}{\sqrt{1200}}=-2.887$ .

Answer

\Phi(-2.887)\approx0.00194

Examiner note

Standardise with the standard error

\sigma/\sqrt n

, not

\sigma

; cite the CLT (“

n

large”) only when the population isn't already Normal. For a total, scale mean and variance by

n

. (9709/62/O/N/24 Q4)

Why σ/√n, not σ/n

Common mistake

Using

\sigma^2

instead of

\sigma^2/n

; dividing by

n

instead of

\sqrt n

when standardising; for a total, forgetting to scale the variance by

n

; and claiming normality for a non-normal population with small

n

(the CLT needs

n

large).