Central Limit Theorem - Steps & Examples

# Central Limit Theorem

Central limit theorem is a concept of probability. It states that when we take the distribution of the average of the sum of a big number of identically distributed and independent variables, the distribution will be normal approximately, invariant of the distribution underlying. Central limit theorem is mainly the reason of working of many statistical procedures.

In other words we can state the theorem of central limit as when some certain conditions are given with given enough of large number of iterates on independent variables, each of which has a well-defined variance along with well-defined expected value. Then when the arithmetic mean of these variables is taken, it will be distributed normally approximately, being regardless of the distribution underlying.

Let $Y_1$, $Y_2$, …, $Y_n$ be ‘n’ random independent variables. Each Y_i has a probability distribution P $(y_1, y_2, …., y_n)$ having mean $\mu_i$ and fixed variance $\sigma^2_i$. Then the variate of normal form  below

$Y_{norm}$ = $\frac{sum_{i = 1}^n (y_i) - \sum_{i = 1}^n (\mu_i)}{\sqrt {sum_{i = 1}^n (\sigma^2_i)}}$

will be having a cumulative distribution function which will be limiting and will be approaching a normal distribution.

If some more conditions are added on the distribution, then the probability density gets normal itself having mean equal to zero ($\mu$= 0) and variance equal to 1 ($\sigma^2$ = 1). If the conversion to the normal form is not been performed, then we have the variate,

Y = $\frac{1}{n}$ $\sum_{i = 1}^n (y_i)$ is distributed normally with $\mu_Y$ = $\sigma_y$ and $\sigma_Y$ = $\frac{\sigma_x}{\sqrt (n)}$.

## Steps

The common steps that are used to solve central limit theorem that are either involving “greater than”, “less than” or “between” are below:

1) The parts of the problem like the mean, population size, standard deviation, sample size and a number that is associated with “greater than”, “less than”, or two numbers associated with both values for range of “between”.

2) The graph is drawn with centre as mean and shading of the region. This is an optional step.

3) The given formula may be used to find z-score.

z = $\frac{(\bar{x}- \mu)}{(\frac{\sigma}{\sqrt n})}$

4) Look up in z table for the ‘z’ value you obtained in previous step.

Case 1: Now to solve a central limit theorem involving “greater than”, the following steps are continued:

5) Subtract the value of z-score obtained from 0.5.

Case 2: To solve a central limit theorem involving “less than”, we perform the following steps:

6) Add 0.5 to the value of z-score obtained.

Case 3: To solve a central limit theorem involving “between”, we perform the following steps:

In this case we continue after step 3.

Case 4: We use the previous step formula to find the z – values. We use x bar2 from step 1.

Case 5: Look up in z table for the ‘z’ value you obtained in 3rd and 4th step.

Case 6:. Add these two values together.

now again the final step is common in all three types of problems that is convert the decimal obtained in previous step in every case to percentage.

## Examples

Few examples on central limit theorem are given below:

Example 1: In a survey of a company, mean salary of employees is 29321 dollars with SD of 2120 dollars. Consider the sample of 100 employees and find the probability their mean salary will be less than 29000 dollars?

Solution:

Total number of employees (n) = 100

Mean ($\mu$) = 29321

standard deviation ($\sigma$) = 2120

Substitute all the values in z-formula i.e. z = $\frac{(\bar{x}- \mu)}{(\frac{\sigma}{\sqrt n})}$

z = $\frac{(29,000-29,321)}{(\frac{2,120}{\sqrt {100}})}$

= -$\frac{321}{212}$

= -1.51

Using z-table, we found -1.51 has an area of 93.45%.

Since we have to find result for "less than", so minus 93.45 from 100 to get required result.

=> 100 - 93.45 = 0.07

Hence the probability of employees having mean salary less than 29000 dollars is 0.07%.

Example 2: The distribution of y$_n$ converges to the standard normal distribution as n approches to infinity. Show that $\chi_n(t)$ approches to exp($\frac{−t^2}{2})$ as n approches to infinity for each t, where t belongs to R.

Solution:

We know that y$_n$ = $\frac{1}{\sqrt{n}}$ $\sum_{i=1}^n$ $\frac{X_i - \mu}{\sigma}$

By Taylor's theorem, for every t $\in$ R

Chracterstic function property: $\chi_n$(t) = $\chi^n(\frac{t}{\sqrt{n}})$

(Use results: $\chi$(0) = 1, $\chi'$(0) = 0 and $\chi''$(0) = -1)

=> $\chi(\frac{t}{\sqrt{n}})$ = 1 + $\frac{1}{2}$ $\chi'(k_n)$ $\frac{t^2}{n}$ where |k$_n$| $\leq$ $\frac{|t|}{n}$

As k$_n$ approches to 0 and therefore $\chi''$(k$_n$) = -1 as n approches to infinity.

So $\chi_n(t)$ = [1 +$\frac{1}{2}$ $\chi''(k_n)$ $\frac{t^2}{n}$]$^n$ approches to exp($\frac{−t^2}{2}$) as n $\in$ $\infty$.