Normal Distribution

A random variable X is said to follow a Normal Distribution with parameters \mu and \sigma if it has a probability density function given by:

f(x;\mu,\sigma) =  \frac{1}{\sqrt{2\pi}\sigma}  e^{-\frac{1}{2\sigma^2} \left( x - \mu \right)^2} ; -\infty < x<\infty , -\infty < \mu <\infty , \sigma > 0

A random variable X is said to follow a Normal Distribution with parameters \mu and \sigma if it has a distribution function given by:

F(x) = \int_{-\infty}^{x}   \frac{1}{\sqrt{2\pi}\sigma}  e^{-\frac{1}{2\sigma^2} \left( y - \mu \right)^2} dy

First Four Moments:

 E(X) =   \int_{-\infty}^{\infty} x f(x;\mu,\sigma)dx \\
= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} (\mu + \sigma y) e^{-\frac{y^2}{2}}dy \left[ \textit{Putting, } \sigma y = x-\mu  \right] \\
= \mu + 0 \left[ \textit{Since, } ye^{-\frac{y^2}{2}}  \textit{is an odd function} \right] \\
\mu_2 = V(X) = E(X-\mu)^2 =  \int_{-\infty}^{\infty} (x - \mu)^2 f(x;\mu,\sigma)dx \\
= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} ( \sigma y)^2 e^{-\frac{y^2}{2}}dy \left[ \textit{Putting, } \sigma y = x-\mu  \right] \\
=  \frac{\sigma^2}{\sqrt{2\pi}} \int_{-\infty}^{\infty} y^2 e^{-\frac{y^2}{2}}dy  \\
=  2 \frac{\sigma^2}{\sqrt{2\pi}} \int_{0}^{\infty} y^2 e^{-\frac{y^2}{2}}dy  \left[ \textit{Since, } y^2e^{-\frac{y^2}{2}}  \textit{is an even function} \right] \\
= \frac{2\sigma^2}{\sqrt{2\pi}}  \int_{0}^{\infty} \sqrt{2 p} e^{-p}dp \left[ \textit{Putting, } 2p = y^2   \right] \\
= \frac{2 \sigma^2}{\sqrt{\pi}} \Gamma(3/2) = \sigma^2 \\
\mu_3 = E(X-\mu)^3 = 0 \\
\mu_4 = E(X-\mu)^4 =  \frac{4 \sigma^4}{\sqrt{\pi}} \Gamma(5/2) = 3 \sigma^4

Skewness and Kurtosis

\gamma_1 = \frac{\mu_3}{\mu_2^{3/2}} = 0 , \\
\gamma_2 = \frac{\mu_4}{\mu_2^2} -3 = 3-3 =0

Some Interesting Properties of Normal Distribution

If a continuous random variable X follows N(\mu,\sigma) , then

  1. The distribution is symmetric about \mu .
  2. The normal probability curve has point of inflection at \mu \pm \sigma .
  3. Owing to the fact that a change in the mean value changes the location of the probability curve of a normal distribution and a change in the variance value changes the shape of the probability curve, \mu \textit{and } \sigma^2 are called the location and scale parameters respectively.
  4. The distribution is symmetric about \mu and hence the mean, median and mode of the distribution are the same. Also all odd order moments about the mean are zero.
  5. The mean deviation about mean is \sqrt{\frac{2}{\pi}} \sigma
  6. A normal distribution with mean 0 and variance 1 is known as standard/unit normal distribution. And the pdf and pdf of the standard normal distribution is usually denoted by \phi(.) \textit{and} \Phi(.) respectively. It satisfies the following relations,
1. \phi(x) = \phi(-x) , \forall x\in \mathbb{R} \quad \\
2. \Phi(x) = 1 - \Phi(x) , \forall  x\in \mathbb{R} \\
3. \Phi(0) = 0.5 \qquad \quad ...........
4. \quad 1-\Phi(x) \\
= \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \left\{ \frac{1}{x} - \frac{1}{x^3} + \frac{1.3}{x^5} - \frac{1.3.5}{x^7} + \dots + (-1)^k \frac{1.3\dots(2k-1)}{x^{2k+1}}\right\}

Graphical Solution of LPP



\[ z=x_1+x_2\] subject to, \[x_1+x_2\leq 2000\] \[x_1+x_2\leq 1500\] \[x_2\leq 600\] and \(x_1,x_2\geq 0\)

To solve the given LPP graphically, we first plot the following lines formed by changing the constraints to equations. \[x_1+2x_2=2000\] \[x_1+x_2=1500\] \[x_2=600\] Then in order to locate the feasible region of the problem we put \(x_1=x_2=0\) and check if the statements are either true or false as follows: \[0\leq 0(TRUE)\] \[0\leq 0(TRUE)\] If the statements are true then we draw arrows from the respective lines (pointing) towards the origin, otherwise we draw arrows pointing in the opposite direction. The feasible region (in our case, the blue shaded polygon OABCD) is the region where all the arrows, merge. It is to be noted that our LPP problem restricts the values of x1 and x2 to be in the first quadrant.

As the Fundamental Theorem[Geometrical Approach] of LPP states that if an LPP admits an optimal solution, then the objective function assumes optimum value of an extreme point of the convex set generated by the set of all feasible solutions, we first proceed with finding the values of the objective function at the extreme points O,A,B,C,D as:


Now we draw the lines(given below) when the objective function takes on the above values. \[z_0=x_1+x_2=0\] \[z_1=x_1+x_2=600\] \[z_2=x_1+x_2=1400\]


By viewing the lines(dotted green lines) it is quite easy to see that as how the solution set moves around as we increase the value of the objective function. Also it is quite evident that our objective function can take a maximum value of 1500 in the feasible region (as a value more than 1500 would simply imply that the solution will be outside the feasible region). Hence, the optimum value of the objective function is 1500. Also, as there were two extreme points C and D where the objective function can take a value of 1500, all the points in the line segment CD may be viewed as the solution leading us to conclude that there are infinite solutions to the given LPP.

 x = seq(-100,2000,1)
y1 = (2000-x)/2  
y2 = (1500-x)
y3 = rep(600,length(x))
z = function(x,y) x+y
       ylim=c(-100,800),col="blue",main="Figure 1")
for(i in 1:length(opt.values)){
  lines(x,opt.values[i]-x,lty="longdash",col="dark green",lwd=2)