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 $\sigma_Y$ = $\frac{\sigma_x}{\sqrt (n)}$.

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

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

In this case we continue after step 3.

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.

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%.

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 $\in$ $\infty$.