Central limit theorem – a worked example – Sho't left to data science

Remember our formal definition: The CLT states that, provided enough samples are taken, the sample distribution of the sample mean will be normally distributed, regardless of the population distribution.

In mathematical terms we say therefore that the sample mean is equal to the population mean:

$\mu_{\bar{x}} = \mu$

With enough samples this also happens – the sample standard deviation will be equal to the standard error:

$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

Which ultimately allows us to calculate the Z-score:

$Z = \frac{x - \mu}{\sigma}$

And using a Z-table, this allows us to find the probability of a value being <= 𝒙.

Here is an example of how this can be used:

Problem statement:

A group of pensioners receive average payouts of $110 per week with a standard deviation of $20. 25 pensioners are due to be paid today but you only have $3000 on hand. What is the probability you will have sufficient cash on hand?

Worked solution:

Let’s write down what we know as a start:

$\mu = 110$

$\sigma = 20$

$n = 25$

We can therefore conclude the following:

$\mu_{\bar{x}} = 110$

$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{25}} = \frac{20}{5} = 4$

The maximum average payout (𝒙) should not exceed $120 ($3000 / 25 pensioners). So our final equation works out as follows:

$Z = \frac{x - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{120 - 110}{4} = 25 = 99.38\%$

In other words the probability of your maximum average payout being <= $120 is 99.38% and you are therefore likely to have enough cash on hand for all 25 pensioners.

Sho't left to data science

Central limit theorem – a worked example

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