The distribution of the sample mean | CAIE 9709 Probability & Statistics 2 Notes

Probability & Statistics 2

Mathematics9709

The centre stays; the spread becomes smaller

E(\bar X)=\mu

Symbols

$\bar X$ = sample mean (same as X)
$\mu$ = population mean (same as X)

\operatorname{Var}(\bar X)=\frac{\sigma^2}{n}

Symbols

$\sigma^2$ = population variance (squared units)
$n$ = sample size (none)

\operatorname{sd}(\bar X)=\frac{\sigma}{\sqrt n}

Same centre, less spread

centre stays at μ = 50

n = 1 → SE = 10

n = 4 → SE = 5

n = 16 → SE = 2.5

A larger sample does not move the centre; it reduces the standard error.

The sample mean is centred at the population mean.
Its standard deviation, called the standard error, is $\sigma/\sqrt n$ .
To halve the standard error, multiply the sample size by four.

Decide why the sample mean is normal

If the population variable X is normal, $\bar X$ is exactly normal for every sample size.
If X is not normal, a large random sample makes $\bar X$ approximately normal by the Central Limit Theorem.

Definition

1 mark

What is meant by Central Limit Theorem?

Model answer: For a sufficiently large random sample, the distribution of the sample mean is approximately normal, whatever the shape of the population distribution.

Key idea

The Central Limit Theorem changes the shape used for the sample mean. It does not change its mean

\mu

or variance

\sigma^2/n

Build the distribution before finding a tail

Worked example

A non-normal population needs the CLT

A current paper gives $X\sim\mathrm B(8,3/4)$ and a random sample of 100 values.

E(X)=8(3/4)=6

\operatorname{Var}(X)=8(3/4)(1/4)=1.5

\bar X\mathrel{\dot\sim}\mathrm N(6,1.5/100)

z=\frac{6.2-6}{\sqrt{1.5/100}}=1.633

Use the upper-tail probability $1-\Phi(1.633)$ .

Answer

P(\bar X>6.2)=0.0512

The CLT is needed because the population distribution is binomial, not normal. This question explicitly says a continuity correction is not expected. (9709/61/M/J/25 Q2)

Examiner note

Calculator route: use Normal CDF with lower bound 6.2, a very large upper bound, mean 6 and standard deviation

\sqrt{1.5/100}

. Still write the distribution and standardisation for method marks.

Your turnIndependent sample-mean check [4 marks]

A population has mean 40 and variance 25. For a random sample of size 64, use the CLT to find $P(\bar X>41)$ .

Show worked answer

\bar X\mathrel{\dot\sim}\mathrm N(40,25/64)

z=\frac{41-40}{5/8}=1.6

Use the upper-tail probability $1-\Phi(1.6)$ .

Answer

P(\bar X>41)=0.0548

Common mistake

\mathrm N(\mu,\sigma^2)

, the second parameter is the variance. Divide the population variance by n, then take a square root only when standardising.

The centre stays; the spread becomes smaller

E(\bar X)=\mu

Symbols

$\bar X$ = sample mean (same as X)
$\mu$ = population mean (same as X)

\operatorname{Var}(\bar X)=\frac{\sigma^2}{n}

Symbols

$\sigma^2$ = population variance (squared units)
$n$ = sample size (none)

\operatorname{sd}(\bar X)=\frac{\sigma}{\sqrt n}

Same centre, less spread

centre stays at μ = 50

n = 1 → SE = 10

n = 4 → SE = 5

n = 16 → SE = 2.5

A larger sample does not move the centre; it reduces the standard error.

The sample mean is centred at the population mean.
Its standard deviation, called the standard error, is $\sigma/\sqrt n$ .
To halve the standard error, multiply the sample size by four.

Decide why the sample mean is normal

If the population variable X is normal, $\bar X$ is exactly normal for every sample size.
If X is not normal, a large random sample makes $\bar X$ approximately normal by the Central Limit Theorem.

Definition

1 mark

What is meant by Central Limit Theorem?

Model answer: For a sufficiently large random sample, the distribution of the sample mean is approximately normal, whatever the shape of the population distribution.

Key idea

The Central Limit Theorem changes the shape used for the sample mean. It does not change its mean

\mu

or variance

\sigma^2/n

Build the distribution before finding a tail

Worked example

A non-normal population needs the CLT

A current paper gives $X\sim\mathrm B(8,3/4)$ and a random sample of 100 values.

E(X)=8(3/4)=6

\operatorname{Var}(X)=8(3/4)(1/4)=1.5

\bar X\mathrel{\dot\sim}\mathrm N(6,1.5/100)

z=\frac{6.2-6}{\sqrt{1.5/100}}=1.633

Use the upper-tail probability $1-\Phi(1.633)$ .

Answer

P(\bar X>6.2)=0.0512

The CLT is needed because the population distribution is binomial, not normal. This question explicitly says a continuity correction is not expected. (9709/61/M/J/25 Q2)

Examiner note

Calculator route: use Normal CDF with lower bound 6.2, a very large upper bound, mean 6 and standard deviation

\sqrt{1.5/100}

. Still write the distribution and standardisation for method marks.

Your turnIndependent sample-mean check [4 marks]

A population has mean 40 and variance 25. For a random sample of size 64, use the CLT to find $P(\bar X>41)$ .

Show worked answer

\bar X\mathrel{\dot\sim}\mathrm N(40,25/64)

z=\frac{41-40}{5/8}=1.6

Use the upper-tail probability $1-\Phi(1.6)$ .

Answer

P(\bar X>41)=0.0548

Common mistake

\mathrm N(\mu,\sigma^2)

, the second parameter is the variance. Divide the population variance by n, then take a square root only when standardising.