Normal distribution, also known as Gaussian distribution, of a continuous random variable `X` has its probability density function presented in the form:

The mean and standard deviation of this distribution are `mu` and `sigma` respectively, they are 2 parameters of the distribution. So, this distribution is symbolized by `N(mu,sigma^2)`.

The form of this probability density function (19) has the shape of a bell (Fig. 1)

Fig. 1 The bell curve of probability density function of normal distribution

We can divide this curve to 2 branches at the mean, and these branches are symmetrical about the mean.

These are some important values concerning normal distribution.

•   `P(mu-sigma<=X<=mu+sigma) = 0,6827`
•   `P(mu-2sigma<=X<=mu+2sigma) = 0,9545`
•   `P(mu-3sigma<=X<=mu+3sigma) = 0,9973`

These results have many applications in practice. For example, if we know that 95,45% of data located in the interval `+-2sigma`, we can reject the data outside this interval in some cases

To facilitate the manual calculation of normal distribution, we transpose from `X` to `Z` by:

(19) transforms to :

(21) is also a probability density function of normal distribution with mean 0 and standard deviation 1, and known as standard normal distribution. This is an even function and ordinate is its symmetrical axis (Fig. 2).

Fig. 2 The bell curve of probability density function of standard normal distribution.

On Fig. 2, the blue area represent the value of cumulative distribution function `F(z)`:

Because of the symmetry, we get :

`F(z)+F(-z)=1`(23)

Note : In formula (20), `z` is known as "z-score". We can consider `z` as relative distance between a value and its mean, in which standard deviation is unit. This quantity is frequently used in statistics, especially in comparison.

Because normal distribution is often used in statistical calculation, tables of this distribution are constructed to help us in manual calculation. These tables give us the relation between value `a` and its cumulative distribution function `F(a)`:

There are mainly two types of tables :

• given `a`, determine `F(a)` (distribution table)
• given `F(a)`, determine `a` (percentage point table)

Distribution table

Percentage point table

When `n` in binomial distribution (number of trials or number of elements in sample) is large, probability of this distribution approach normal distribution and the difference between these two distribution can be ignored. This approximation facilitate the calculation of probability, because when `n` is large, it is difficult to do it by using formula (7).

The mean and standard deviation of approximate normal distribution are:

`mu=np`(25)

`sigma^2=np(1-p)`(26)

We have to note that this equivalence exists in the conditions `np>=5`  or  `n(1-p)>=5`

Example : 48% of shirts produced by company X is classified as class A. Supermarket S bought 800 shirts from this company. What is the probability that not less than 50% of these shirts belong to class A.

By principle, we have to apply binomial distribution with `n=800` ; `p=0,48` and we have to calculate `P(X>= 400)`. This is a difficult and tedious task, requires a lot of time.

Using equivalent normal distribution with:

•   `mu=np=800xx0,48=384`
•   `sigma^2= np(1-p)=800xx0,48xx0,52=199,68`   or   `sigma=14,13`

Put :   `Z=(X-mu)/sigma`

then :   `P(X>=400)=P(Z>=(400-384)/(14,13))=P(Z>=1,13)`

  `P(X>=400)=P(Z >=1,13)=1-P(Z<=1,13)=1-F(1,13)=1-0,87=0,13`

Note : From the relation between binomial and hypergeometric distribution, we can conclude that normal distribution is the approximative one to hypergeometric distribution when the number of elements in sample `n` is much less than the number of elements of population `N`.

Similar to the case of binomial distribution, Poisson distribution is equivalent to normal distribution when the number of elements of sample is large. Two parameters of equivalent normal distribution are:

•   `mu=lambda`,
•   `sigma^2=lambda`.

Note that this equivalence only exist when `lambda>5`

Example : Each gram of product A contains 1040 cells of microorganism M. Taken a sample, What is the probability that the microorganism M content in this sample is less than 1000 g^{- 1} .

Microorganism M contents of this product have Poisson distribution. So we have to calculate:

  `P(X< 1000)=sum_(x=1)^999 (e^(-1040)1040^x)/(x!)`

It is very difficult to calculate this formula. So we use approximate normal distribution with:

•   `mu=lambda=1040`
•   `sigma^2=lambda=1040`   or   `sigma=32,25`

Put :   `Z=(X-mu)/sigma`

So :   `P(X< 1000)=P(Z<= (1000-1040)/(32,25))=P(Z<= -1,24)=0,11`