Central Limit Theorem (Edexcel A Level Further Maths: Further Statistics 1)

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It's easier to explain this with an example
Let $X ~ B (10, \frac{1}{2})$ where $X$ is the number of heads in 10 flips of a fair coin
- $E (X) = n p = 5$ heads
- $Var (X) = n p (1 - p) = 2.5$
Imagine doing this experiment each day for 7 days
- $X_{1} ~ B (10, \frac{1}{2})$ is the number of heads on day 1
- $X_{2} ~ B (10, \frac{1}{2})$ is the number of heads on day 2
- ...
- $X_{7} ~ B (10, \frac{1}{2})$ is the number of heads on day 7
$X_{1}$ to $X_{7}$ are independent random variables, each with identical $B (10, \frac{1}{2})$ distributions
- Identical distributions don't make the number of heads on each day identical
$X ~ B (10, \frac{1}{2})$ is called the population distribution
- With population mean $μ = 5$
- And population variance $σ^{2} = 2.5$
$X_{1}, . . ., X_{7}$ is called the random sample of size 7 taken from the population distribution $X$
- Where each $X_{i}$ is an independent identical $X$ distribution

Let $X_{1}, X_{2}, . . ., X_{n}$ , be a random sample of size $n$ taken from the population distribution $X$
The sample mean is given by $X = \frac{X_{1} + X_{2} + . . . + X_{n}}{n}$
- It is not a fixed number
- Different samples of size $n$ have different sample means
From the example before
- Each day there's a different number of heads
  - Here's one sample: 4, 6, 5, 5, 3, 5, 6
  - This particular sample mean is 4.857...
  - This is close to the population mean of $μ = 5$ heads
- A second sample of 7 days would give a different sample mean
- As would a third sample of 7 days
  - Generating lots of samples like this will give a distribution of sample means

The Central Limit Theorem (CLT) says that
- If $X_{1}, X_{2}, . . ., X_{n}$ is a random sample of $n$ independent distributions
- Taken from any population distribution $X$
- With population mean $μ$ and population variance $σ^{2}$
- Then, provided $n$ is large
- The sample mean has the approximate normal distribution $X \approx ~ N (μ, \frac{σ^{2}}{n})$
  - $\approx ~$ means approximately modelled by
This works for any population distribution
- For example, $X_{1}$ to $X_{50}$ could be a random sample of size 50 taken from a $Po (λ)$ distribution ( $μ = λ$ and $σ^{2} = λ$ )
  - The CLT says $X \approx ~ N (μ, \frac{σ^{2}}{n})$
  - $n = 50$ is a large sample
- The sample mean has an approximate normal distribution, despite the population distribution being Poisson
In the special case when the population distribution is itself normal
- Then the CLT is exact
  - $X ~ N (μ, \frac{σ^{2}}{n})$
  - No approximation symbol

To use $X \approx ~ N (μ, \frac{σ^{2}}{n})$ you need to find $μ$ and $σ^{2}$ from the question
- The population mean and variance
If the population, $X$ , is a discrete random variable, then
- $μ = E (X)$
- $σ^{2} = Var (X)$
Use known formulae for standard discrete models
- For example, if $X ~ Geo (p)$
  - Then $μ = \frac{1}{p}$ and $σ^{2} = \frac{1 - p}{p^{2}}$
  - So $\frac{σ^{2}}{n}$ is $\frac{1 - p}{n p}$
Don't forget to divide the population variance by $n$
Don't confuse $n$ for sample size here with the $" n "$ from a binomial distribution
- Use a different letter, for example
  - $X ~ B (m, p)$
  - So $\frac{σ^{2}}{n}$ is $\frac{m p (1 - p)}{n}$

Key phrases to look out for are
- "Estimate the probability that the mean number of ... is greater than ..."
- "A random sample $X_{1}, . . ., X_{50}$ is taken"
Look out for samples being large

Let $X$ represent the value shown when a four-sided spinner is spun, given by the distribution below.

$x$	0	2	4	8
$P (X = x)$	0.5	0.2	0.1	0.2

The spinner will be spun 60 times and the values shown on each spin will be recorded.

Estimate the probability that the mean of the values recorded is less than 2.8.

clt-1 clt-2

to absolutely everything:

the (exam) results speak for themselves:

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