Normal Distribution With Examples

Probability / Thursday, October 11th, 2018
(Last Updated On: November 16, 2018)

In the previous two articles of this unit we learned Binomial distribution and Poisson Distribution, in this article we will learn Normal distribution.

Normal Distribution

Definition

Let µ and σ be two real constants such that -∞ <µ< ∞ and σ > 0. Then the probability distribution for which
$f(x)=N\left( \mu ,\sigma ,x \right)=\frac{1}{\sigma \sqrt{2\pi }}{{e}^{-\frac{1}{2}{{\left( \frac{x-\mu }{\sigma } \right)}^{2}}}}$
is the density function is called the Normal distribution and the corresponding continuous random variable X is called the normal variate.

Mean

$E\left( X \right)=\int\limits_{-\hbar }^{\infty }{xf(x)dx=\int\limits_{-\infty }^{\infty }{x}}\frac{1}{\sigma \sqrt{2\pi }}{{e}^{-\frac{1}{2}{{\left( \frac{x-\mu }{\sigma } \right)}^{2}}}}dx$
$Let,\frac{x-\mu }{\sigma }=z\Rightarrow x=\mu +\sigma z$
$\Rightarrow dx=\sigma dz$
$\therefore E\left( X \right)=\frac{1}{\sigma \sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{\left( \mu +\sigma z \right)}{{e}^{-\frac{{{z}^{2}}}{2}}}\sigma dz$
$\Rightarrow E\left( X \right)=\frac{2\mu }{\sqrt{2\pi }}\int\limits_{0}^{\infty }{{{e}^{-\frac{{{z}^{2}}}{2}}}dz}\left[ \because \int\limits_{-\infty }^{\infty }{\sigma z}{{e}^{-\frac{{{z}^{2}}}{2}}}dz=0 \right]$
$Put,t=\frac{{{z}^{2}}}{2}\Rightarrow dt=zdz$
$E\left( X \right)=\frac{2\mu }{\sqrt{2\pi }}\int\limits_{0}^{\infty }{{{e}^{-t}}\frac{dt}{\sqrt{2t}}}$
$\Rightarrow E\left( X \right)=\frac{\mu }{\sqrt{\pi }}\int\limits_{0}^{\infty }{{{e}^{-t}}{{t}^{\frac{1}{2}-1}}dt}$
$\therefore E\left( X \right)=\frac{\mu }{\sqrt{\pi }}\Gamma \left( \frac{1}{2} \right)=\mu \left[ As,\Gamma \left( \frac{1}{2} \right)=\sqrt{\pi } \right]$

Variance

$V\left( X \right)=E{{\left[ X-E\left( X \right) \right]}^{2}}=E{{\left[ X-\mu \right]}^{2}}$
$\therefore V\left( X \right)={{\int\limits_{-\infty }^{\infty }{\left( x-\mu \right)}}^{2}}\frac{1}{\sigma \sqrt{2\pi }}{{e}^{-\frac{1}{2}{{\left( \frac{x-\mu }{\sigma } \right)}^{2}}}}dx$
$Let,\frac{x-\mu }{\sigma }=z\Rightarrow x=\mu +\sigma z$
$\Rightarrow dx=\sigma dz$
$\Rightarrow V\left( X \right)=\frac{1}{\sigma \sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{{{\sigma }^{2}}{{z}^{2}}}{{e}^{-\frac{{{z}^{2}}}{2}}}\sigma dz$
$Put,t=\frac{{{z}^{2}}}{2}\Rightarrow dt=zdz$
$\therefore V\left( X \right)=\frac{\sqrt{2}{{\sigma }^{2}}}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{{{t}^{\frac{1}{2}}}}{{e}^{-t}}dt={{\sigma }^{2}}$

Mode

Mode is the value for which the density function f(x) is maximum, i.e. mode is a solution of
$f'(x)~=0~~\text{and}~f”(x)<0$
For normal distribution
$f\left( x \right)=\frac{1}{\sigma \sqrt{2\pi }}{{e}^{-\frac{1}{2}{{\left( \frac{x-\mu }{\sigma } \right)}^{2}}}}$
$\Rightarrow \log f(x)=\log \left( \frac{1}{\sigma \sqrt{2\pi }}{{e}^{-\frac{1}{2}{{\left( \frac{x-\mu }{\sigma } \right)}^{2}}}} \right)$
$\Rightarrow \log f(x)=\log \frac{1}{\sigma \sqrt{2\pi }}-\frac{1}{2}{{\left( \frac{x-\mu }{\sigma } \right)}^{2}}$
$\Rightarrow \frac{f'(x)}{f(x)}=-\frac{x-\mu }{\sigma }\Rightarrow f'(x)=-\frac{x-\mu }{\sigma }f(x)$
$And~~also$
$f”(x)=-\frac{1}{{{\sigma }^{2}}}f(x)-\frac{1}{{{\sigma }^{2}}}\left( x-\mu \right)f'(x)$
$\Rightarrow f”(x)=-\frac{1}{{{\sigma }^{2}}}\left[ f(x)-\left( x-\mu \right)f'(x) \right]$
$\therefore f”(x)=-\frac{f(x)}{{{\sigma }^{2}}}\left[ 1-\frac{{{\left( x-\mu \right)}^{2}}}{{{\sigma }^{2}}} \right]$
$\text{Now}~f'(x)=0~~gives~~x=\mu$
$and~~also~~f”(x)~~at~~x=\mu ~~is~~-\frac{1}{{{\sigma }^{2}}.\sigma \sqrt{2\pi }}<0$
$Hence~~x=\mu ~~is~~the~~\text{mode}~of~~the~~distribution.$

Median

Since the probability distribution function of the normal distribution is symmetric about x = µ the median will also be at x = µ.
Hence for normal distribution mean = median = mode.

Standard Normal Distribution

The Normal distribution for which the mean is zero and the variance 1 is called the standard normal distribution and the corresponding random variable z is called the standard normal variate.
The density function for the standard normal distribution is
$\Phi \left( z \right)=\frac{1}{\sqrt{2\pi }}{{e}^{-\frac{{{z}^{2}}}{2}}}$
The curve represented by this function is called standard normal curve and it is symmetrical about the line z = 0.

Note:
1.The cumulative distribution function of the standard normal distribution will be denoted by Φ.
That is
$\Phi \left( s \right)=\frac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{s}{{{e}^{-\frac{{{x}^{2}}}{2}}}}dx$
2.
$\text{If Z}\sim N\left( 0,1 \right)$
$P\left( a\le z\le b \right)=\Phi \left( b \right)-\Phi \left( a \right)$
3.
$\text{If X}\sim N\left( \mu ,{{\sigma }^{2}} \right)then$
$z=\frac{X-\mu }{\sigma }\sim N\left( 0,1 \right)$
$\text{Hence}~~P\left( a\le X\le b \right)=P\left( \frac{a-\mu }{\sigma }\le z\le \frac{b-\mu }{\sigma } \right)$
$=\Phi \left( \frac{b-\mu }{\sigma } \right)-\Phi \left( \frac{a-\mu }{\sigma } \right)$
4.
$\Phi \left( -x \right)=1-\Phi \left( x \right)$

 Example 01

Suppose X has normal distribution X ~ N(2, 0.16). Use the normal distribution table and evaluate (i) P(X ≥ 2.3) (ii) P(1.8 ≤ X ≤ 2.1)

Solution:
$(i)~~\mu =2,~~{{\sigma }^{2}}=0.16,~~\sigma =0.4$
$Put~~Z=\frac{X-\mu }{\sigma },~~then~~Z\sim N(0,1)$
$P\left( X\ge 2.3 \right)=P\left( Z\ge \frac{2.3-2}{0.4} \right)$
$=P\left( Z\ge 0.75 \right)=1-P\left( Z<0.75 \right)$
$=1-\Phi \left( 0.75 \right)=0.2266$
$(ii)P\left( 1.8\le X\le 2.1 \right)=P\left( \frac{1.8-2}{0.4}\le Z\le \frac{2.1-2}{0.4} \right)$
$=P\left( -0.5\le Z\le 0.25 \right)$
$=\Phi \left( 0.25 \right)-\Phi \left( -0.5 \right)=0.2902$