18.1.1 Central Limit Theorem

Central Limit Theorem

We already know that if $X \sim N(\mu, \sigma^2) \Rightarrow \bar{X} \sim N(\mu, \frac{\sigma^2}{n})$ .

The central limit theorem states that for any distribution with mean \mu and variance \sigma^2, the sample mean is distributed as follows:

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

For non-normal populations with large sample size n, this relationship is only approximate. For normal populations, it is exact.

Note: The central limit theorem only applies to the approximate case for non-normal or unknown distributions.

Worked Examples

lightbulbExample

Example Alex obtained the actual waist measurements, in inches, of a random sample of 50 pairs of jeans, each of which was labeled as having a 32-inch waist. The results are summarized by:

n = 50, \quad \Sigma W = 1615.0, \quad \Sigma W^2 = 52214.50

Test: At the 0.1% significance level, whether this sample provides evidence that the mean waist measurement of jeans labeled as having 32-inch waists is in fact greater than 32 inches. State your hypotheses clearly.

Hypotheses:

M = \text{mean waist measurement of the jeans}

H_0: \mu = 32 \quad \text{(This is the parameter} \ \mu \ \text{we will use in the distribution)}

H_1: \mu > 32

By the central limit theorem, since n is large, the sample mean $\bar{X}$ is distributed approximately as:

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

where $\sigma^2$ is estimated by $s^2$ .

Calculating $s^2$ :

\bar{w^2}-\bar{w}^2 = 1044.29 - 1043.29 = 1

Thus:

s^2 = \frac{50}{49} \times 1 = \frac{50}{49}

\therefore W \sim N\left( 32, \frac{(\frac {50}{49})}{50} \right) = N\left( 32, \frac{1}{49} \right)

The test statistic is:

\bar{W} = \frac{1615.0}{50} = 32.3

P(W \geq 32.3) = 0.01786

Since:

0.01786 > 0.001

We accept $H_0$ .

There is insufficient evidence to suggest that the mean waist size is greater than 32 inches.

infoNote

Q5. (June 2018, Q4) The discrete random variable Y has a probability distribution given by:

y	0	1	2	3
P(Y = y')	0.4	0.2	0.3	0.1

Entire population data

$\bar{Y}$ denotes the mean of 50 random independent observations of $Y$ .

(i) Find the approximate distribution of $\bar{Y}$ , giving the value(s) of any parameter(s).

(ii) State the possible values taken by $\bar{Y}$ in the range from $1.4$ to $1.5$ inclusive.

Solution:

(i) Since n = 50 ≥ 25, by the central limit theorem, we know:

\bar{Y} \sim N\left( \mu, \frac{\text{Var}(Y)}{n} \right)

by Central limit Theorem

We calculate:

M = 0 \times 0.4 + 1 \times 0.2 + 2 \times 0.3 + 3 \times 0.1 = 1.1

E(Y^2) = 0^2 \times 0.4 + 1^2 \times 0.2 + 2^2 \times 0.3 + 3^2 \times 0.1 = 2.3

\text{Var}(Y) = 2.3 - 1.1^2 = 1.09

Thus:

\bar{Y} \sim N\left( 1.1, \frac{1.09}{50} \right)

(ii) Possible values for $\bar{Y}$ in the range 1.4 to 1.5 are:

1.4, \, 1.42, \, 1.44, \, 1.46, \, 1.48, \, 1.5

Goes up in $\frac {1}{50}$ thus as we divide by 50.

Central Limit Theorem Simplified Revision Notes for A-Level Edexcel Further Maths Further Statistics 1

18.1.1 Central Limit Theorem

Central Limit Theorem

Worked Examples

500K+ Students Use These Powerful Tools to Master Central Limit Theorem For their A-Level Exams.

Other Revision Notes related to Central Limit Theorem you should explore