[NORMAL DISTRIBUTION FUNCTION ]

μ is mean / mode / median as in the above distribution, all the three are same. σ is Standard Deviation. If μ = 0 and σ =1, the distribution is called standard normal distribution. The normal distribution is a sub-class of elliptical distribution. It is symmetric about its mean. It also has maximum entropy for a given mean and variance. The total area under the curve in standard normal functions is 1. To see the graphing, click (works in Firefox,Chrome or IE9). Normally, the x-axis of distribution curve is represented by z where z = x /  σ  i.e. x-axis is measured in the unit of Standard Deviation(SD) in stead of absolute value of x. So it is f(z) vs. z. For more on normal distribution, click . In the normal probability distribution, the point of inflexion in the curve are ±1SD from the mean.These are the points at which the curve changes its direction from concave to convex. The quartile deviation Q or PE (probable error) is given by the formula PE=0.6745σ . One PE laid off in the + and - directions from the M includes the middle 50% of the cases.

Carl Fredrich Gauss, in 1809, discovered the normal distribution as a tool to rationalize the Method of Least Squares. An American mathematician Adrain published in the same year, two derivatives of normal distribution independently but his work came to light only in 1871 when Abbe pointed it out.

[STANDARD NORMAL DISTRIBUTION FUNCTION ]

Q-Q Plot

It is a plot of percentiles or quantiles of a Standard Normal Distribution against the corresponding percentile of the observed data. If the observations follow approximately a normal distribution, the resulting plot should be roughly a straight line with positive slope.

Meaning of Normal Probability Distribution

►The meaning of normal probability distribution may be explained through 2 approaches- (1) by way of binomial distribution (2) by deriving an equation  fit the known properties of the curve.

♐ Binomial expression ( p + q )ⁿ approaches the normal form when p=q=1/2 and n becomes infinitely large. The tendency for the distribution of many human traits to take the bell shaped form suggests- but does not prove-that the occurence of physical and mental traits may be determined by the operation of a very large number of genetic factors which combine by chance.One can see here the relationship to Mendel's theory. Examination of normal probability distribution equation

shows that when x (deviation of a score from Mean which is at the center of the curve) = 0, the term e^{-(x2 / 2σ2)} equals 1 and  y₀ =N /(σ √2π) . Thus y0 is the maximum ordinate and stands at the center of the curve. When y = 0, the equation becomes 1/ e^{(x2 / 2σ2)} = 0 and x → ∞ as y → 0 .Hence it is clear that the equation has a maximum at the mean and the curve stretches to infinity the + and - direction of the mean. Since x in the formula is squared, either + or - values have the same effect and the curve is bilaterally symmetrical.

★ T S Kelly proposed a differential equation dy/dx = Cxy as a starting point of the derivation of normal equation  The equation has an origin at 0 and the slope of the curve is 0 at both x=0 and y=0. These are important characteristics of the normal probability curve. On integration, log y = (-Cx²/2) + k where C and k are constants or y = ke^{-(Cx2 / 2)}   . To evaluate k, we go to the Moment of the Distribution. The zero moment is N or k√(2π/c) ; 1st moment is 0 and 2nd moment is N*σ² or (k/c)√(2π/c) . Solving for c and k, we get c=1/σ²  and k= N/ σ√(2π ) . On substitution, this gives the integral equation.

Y-value

for

Normal Probability Distribution of Discrete Nature

Mean value=m=∫xf(x) dx =∫x*(1/√2π)*(1/σ)* e^-z*z/2  dx  where z=(x-μ)/σ and x= σz+μ and dx=σdz

m=<f(x)>=∫(σz+μ)*(1/√2π)*e^-z*z/2  dz =σ*(1/√2π)* ∫ze^-z*z/2  dz +μ/√(2π)*∫e^-z*z/2  dz . The red portion is an odd function of z and hence with limit from -infinity to +infinity, it is zero.

∫e^-z*z/2  dz can be evaluated by taking y=-z² /2 and dy=zdz

So =μ/√(2π)*∫e^-z*z/2  dz =(2μ/2√π) ∫ e^-y y^-1/2 dy with limit from 0 to infinity since this is an even function and hence limit -infinity to + infinity=2* limit zero to infinity.

integral is (μ/√π) ∫ e^-y y^1/2-1 dy . The green portion integral is a gama function  ∫ e^-x x^n-1 dx where n=1/2

Γ(1/2)= √π  Hence the entire blue integral μ/√(2π)*∫e^-z*z/2  dz =μ

So mean value of f(x) is μ

∫ f(x)dx =σ ∫e^-z*z/2  dz=(2σ/√2)  ∫ e^-y y^-1/2 dy=σ(√2)  Γ(1/2)= σ(√2)√π=σ√(2π)

  Which is area under the curve . If we normalize the integral by multiplying 1/σ√(2π) , the area becomes 1 .

To find standard Deviation of the distribution

variance= <X² > - <X>² =  ∫ x²f(x)dx - [∫xf(x) dx]²

 ∫ x²f(x)dx =∫(σz+μ)² σ e^-z*z/2  dz = σ³ ∫z²e^-z*z/2  dz +μ²  ∫σe^-z*z/2  dz +2σ²μ∫ze^-z*z/2  dz (the limit from -infinity to +infinity)

The Brown integral is an odd function and hence zero.

The red part is μ²σ ∫e^-z*z/2  dz =μ²σ√(2π)..... On normalization , it is μ²

The Blue part is 2√2σ³ Γ(3/2)=√(2π)σ³   w= √(2π)σ*σ² = .....On normalisation , it becomes σ²   .

Hence <X² >- <X>² =Blue part+<X>²    + 0 -<X>² =Blue part = standard deviation² = σ² ;