Normal Random Variable | Its Important Properties

Normal Random variable and Normal distribution

      The random variable with uncountable set of values is known to be continuous random variable, and the probability density function with the help of integration as area under the curve gives the continuous distribution, Now we will focus one of the most used and frequent continuous random variable viz normal random variable which has another name as Gaussian random variable or Gaussian distribution.

Normal random variable

      Normal random variable is the continuous random variable with probability density function

f(x) =\frac{1}{\sqrt{2\pi }\sigma} e^{-(x-\mu )^{^{2}/2\sigma ^{^{2}}}} \ \ -\infty < x < \infty

having mean μ and variance σ2 as the statistical parameters and geometrically the probability density function has the bell shaped curve which is symmetric about the mean μ.

Normal Random variable
Normal Random variable

We know that probability density function has the total probability as one so

\frac{1}{\sqrt{2\pi }\sigma}\int_{-\infty}^{\infty} e^{-(x-\mu )^{^{2}/2\sigma ^{^{2}}}} dx =1

by putting y= (x-μ)/σ

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

\int_{-\infty}^{\infty} e^{-y^{^{2}/2}} dy = {\sqrt{2\pi}}

let \ \ I= \int_{-\infty}^{\infty} e^{-y^{^{2}/2}} dy \ \ then

I^{2}= \int_{-\infty}^{\infty} e^{-y^{^{2}/2}} dy \int_{-\infty}^{\infty} e^{-x^{^{2}/2}} dx

= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^-({y^2+x^2})/2 \ \ dy dx

this double integration can be solved by converting it  into polar form

x= r\cos \theta , y= r\sin \theta , dydx= rdrd\theta

I^{^{2}}= \int_{0}^{\infty}\int_{0}^{2\pi } e^-{r^{2}/2} rd\theta dr

= 2\pi \int_{0}^{\infty} r e^-{r^{2}/2} \ \ dr

=- 2\pi e^{-r^{2}/2}\lvert_{\infty }^{0} =2\pi

which is the required value so it is verified for the integral I.

  • If X is normally distributed with parameter μ  and σ2 then Y=aX+b is also normally distributed with the parameters  aμ+b and a2μ2

Expectation and Variance of Normal Random variable

The Expected value of the normal random variable and the variance we will get with the help of

Z = \frac{X-\mu}{\sigma }

where X is normally distributed with the parameters mean μ and standard deviation σ.

E [Z] = \int_{-\infty}^{\infty} xfZ(x) dx

=\frac{1}{\sqrt{2\pi }} \int_{-\infty}^{\infty} x e^{-x^{2}/2} dx

=-\frac{1}{\sqrt{2\pi }} e^{-x^{2}/2} \lvert_{-\infty }^{\infty} =0

since mean of Z is zero so we have the variance as

Var (Z) = E[Z^{2}]

=\frac{1}{\sqrt{2\pi }} \int_{-\infty}^{\infty} x^{2} e^{-x^{2}/2} dx

by using integration by parts

Var(Z) = \frac{1}{\sqrt{2\pi }}(-xe^{-x^{2}/2} \lvert_{-\infty }^{\infty} + \int_{-\infty}^{\infty} e^{-x^{2}/2} dx)

= \frac{1}{\sqrt{2\pi }} \int_{-\infty}^{\infty} e^{-x^{2}/2} dx =1

for the variable Z the graphical interpretation is as follows

Normal Random variable
Normal Random variable

and the area under the curve for this variable Z which is known as standard normal variable, it is calculated for the reference (given in the table), as the curve is symmetric so for negative value the area will be same as that of positive values

P \left { Z\leq -x \right } = P \left { Z > x \right } \ \ -\infty < x < \infty

z0.000.010.020.030.040.050.060.070.080.09
0.00.500000.503990.507980.511970.515950.519940.523920.527900.531880.53586
0.10.539830.543800.547760.551720.555670.559620.563560.567490.571420.57535
0.20.579260.583170.587060.590950.594830.598710.602570.606420.610260.61409
0.30.617910.621720.625520.629300.633070.636830.640580.644310.648030.65173
0.40.655420.659100.662760.666400.670030.673640.677240.680820.684390.68793
0.50.691460.694970.698470.701940.705400.708840.712260.715660.719040.72240
0.60.725750.729070.732370.735650.738910.742150.745370.748570.751750.75490
0.70.758040.761150.764240.767300.770350.773370.776370.779350.782300.78524
0.80.788140.791030.793890.796730.799550.802340.805110.807850.810570.81327
0.90.815940.818590.821210.823810.826390.828940.831470.833980.836460.83891
1.00.841340.843750.846140.848490.850830.853140.855430.857690.859930.86214
1.10.864330.866500.868640.870760.872860.874930.876980.879000.881000.88298
1.20.884930.886860.888770.890650.892510.894350.896170.897960.899730.90147
1.30.903200.904900.906580.908240.909880.911490.913080.914660.916210.91774
1.40.919240.920730.922200.923640.925070.926470.927850.929220.930560.93189
1.50.933190.934480.935740.936990.938220.939430.940620.941790.942950.94408
1.60.945200.946300.947380.948450.949500.950530.951540.952540.953520.95449
1.70.955430.956370.957280.958180.959070.959940.960800.961640.962460.96327
1.80.964070.964850.965620.966380.967120.967840.968560.969260.969950.97062
1.90.971280.971930.972570.973200.973810.974410.975000.975580.976150.97670
2.00.977250.977780.978310.978820.979320.979820.980300.980770.981240.98169
2.10.982140.982570.983000.983410.983820.984220.984610.985000.985370.98574
2.20.986100.986450.986790.987130.987450.987780.988090.988400.988700.98899
2.30.989280.989560.989830.990100.990360.990610.990860.991110.991340.99158
2.40.991800.992020.992240.992450.992660.992860.993050.993240.993430.99361
2.50.993790.993960.994130.994300.994460.994610.994770.994920.995060.99520
2.60.995340.995470.995600.995730.995850.995980.996090.996210.996320.99643
2.70.996530.996640.996740.996830.996930.997020.997110.997200.997280.99736
2.80.997440.997520.997600.997670.997740.997810.997880.997950.998010.99807
2.90.998130.998190.998250.998310.998360.998410.998460.998510.998560.99861
3.00.998650.998690.998740.998780.998820.998860.998890.998930.998960.99900
3.10.999030.999060.999100.999130.999160.999180.999210.999240.999260.99929
3.20.999310.999340.999360.999380.999400.999420.999440.999460.999480.99950
3.30.999520.999530.999550.999570.999580.999600.999610.999620.999640.99965
3.40.999660.999680.999690.999700.999710.999720.999730.999740.999750.99976
3.50.999770.999780.999780.999790.999800.999810.999810.999820.999830.99983

since we have used the substitution

Z =\frac{X-\mu }{\sigma } \ \ X =\mu + \sigma Z

E[X] =\mu + \sigma E[Z] = \mu

Var(X) = \sigma ^{2} Var(Z) = \sigma ^{2}

Here keep in mind that Z is standard normal variate where as continuous random variable X is normally distributed normal random variable with mean μ and standard deviation σ .

So to find the distribution function for the random variable we will use the conversion to the standard normal variate as

FX(a)= P\left { X\leq a \right } = P (\frac{X-\mu }{\sigma } \leq \frac{a-\mu }{\sigma }) =\Phi \left (\frac{a-\mu }{\sigma } \right )

for any value of a.

Example: In the standard normal curve find the area between the points 0 and 1.2.

If we follow the table the value of 1.2 under the column 0 is 0.88493 and value of 0 is 0.5000 ,

Normal Random variable
Normal Random variable

P\left ( 0\leq Z \leq 1.2 \right ) =\Phi \left ( 1.2 \right ) - \Phi \left ( 0 \right ) =0.88493 -0.50000 =0.38493

Example: find the area for the standard normal curve within -0.46 to 2.21.

Normal Random variable
Normal Random variable

From the shaded region we can bifurcate this region from -0.46 to 0 and 0 to 2.21 because the normal curve is symmetric about y axis so the area from -0.46 to 0 is same as the are from 0 to 0.46 thus from the table

P\left ( -0.46 \leq Z \leq 0 \right ) = P\left ( 0 \leq Z \leq 0.46 \right ) =0.1772

and

P\left ( 0 \leq Z \leq 2.21 \right ) =0.4864

so we can write it as

Total Area =(area between z = -0.46 and z=0 ) + (area between z =0 and z=2.21)

= 0.1722 + 0.4864

= 0.6586

Example: If X is normal random variable with mean 3 and variance 9 then find the following probabilities

P\left { 2 < X < 5 \right }

P \left \{ X > 0 \right \}

P \left { \left | X - 3 \right | > 6 \right }

Solution:  since we have

F X(a) =P\left { X \leq a \right } = P \left ( \frac{X-\mu }{\sigma } \leq \frac{a-\mu }{\sigma } \right ) = \Phi \left ( \frac{a-\mu }{\sigma } \right )

P \left { 2< X < 5 \right } =P \left { \frac{2-3}{3} < \frac{X-3}{3}< \frac{5-3}{3} \right }

=P \left { -\frac{1}{3} < Z < \frac{2}{3} \right }

Normal Random variable
Normal Random variable

so bifurcating into the intervals -1/3 to 0 and 0 to 2/3 we will get the solution from the tabular values

P \left { -\frac{1}{3} < Z < \frac{2}{3} \right } = P \left { -\frac{1}{3} < Z < 0 \right } + P \left ( 0 < Z < \frac{2}{3} \right )

or

 = \Phi \left ( \frac{2}{3} \right ) - \Phi \left ( - \frac{1}{3} \right )

= \Phi \left ( \frac{2}{3} \right ) - \left [ 1- \Phi \left ( \frac{1}{3} \right ) \right ]

=0.74537 -1 + 0.62930 =0.37467

and

P \left { X > 0 \right } =P\left { \frac{X-3}{3} > \frac{0-3}{3} \right } = P\left { Z> -1 \right }

= 1- \Phi \left ( -1 \right )= \Phi \left ( 1 \right ) \approx 0.8413

Normal Random variable
Normal Random variable

P \left { \left | X -3 \right | > 6 \right } =P\left { X > 9 \right } + P\left { X < -3 \right }

=P\left { \frac{X-3}{3} > \frac{9-3}{3} \right } + P\left { \frac{X-3}{3} < \frac{-3-3}{3} \right }

=P \left { Z > 2 \right } + P \left { Z < -2 \right }

= 1- \Phi \left ( 2 \right ) + \Phi \left ( -2 \right )

=2[1- \Phi \left ( 2 \right )] \approx .0456

Normal Random variable
Normal Random variable

Example: An observer in paternity case states that the length (in days) of human growth

is normally distributed with parameters mean  270 and variance 100. In this case the suspect  who is father of the child provided the proof that he was out of the country during a period that started 290 days before the birth of the child and ended 240 days earlier the birth. Find the probability that the mother could have had the very long or very short pregnancy indicated by the witness?

Let X denote the normally distributed random variable for gestation and consider the suspect is the father of the child. In that case the birth of the child happened within the specified time has the probability

P \left { X > 290 \ \ or \ \ X  290 \right } + P\left { X < 240 \right }

= P\left { \frac{X-270}{10}> 2 \right } + P\left { \frac{X-270}{10}< -3 \right }

= 1- \Phi \left ( 2 \right ) +1 - \Phi \left ( 3 \right ) \approx .0241

Relation between Normal random variable and Binomial random variable

      In case of Binomial distribution the mean is np and the variance is npq so if we convert such binomial random variable with such  mean and standard deviation having n very large and p or q are very small going nearer to zero then standard normal variable Z with the help of these mean and variance is

Z = \frac{X - np}{\sqrt{npq}}

here in terms of Bernouli trials X considers the number of successes in n trials. As n is increases and goes nearer to infinity this normal variate goes in the same way to become standard normal variate.

The relation of binomial and standard normal variate we can find with the help of following theorem.

DeMoivre Laplace limit theorem

If Sn denotes the number of successes that occur when n  independent trials, each resulting in a success with probability p , are performed, then, for any a < b ,

P\left { a\leq \frac{S_{n}-np}{\sqrt{np\left ( 1-p \right )}} \leq b \right } \rightarrow \Phi \left ( b \right ) -\Phi \left ( a \right )

as \ \ n \rightarrow \infty

 In other words

\lim_{n \to \infty} P \left ( a\leq \frac{X -np}{\sqrt{np\left ( 1 -p \right )}} \leq b \right ) =\frac{1}{\sqrt{2\pi}} \int_{a}^{b} e^{-u^{2}/2} du

Example: With the help of normal approximation to the binomial random variable find the probability of occurrence of 20 times tail when a fair coin tossed 40 times.

Solution: Suppose the random variable X represents the occurrence of tail, since the binomial random variable is discrete random variable and normal random variable is continuous random variable so to convert the discrete into the continuous, we write it as

P\left ( X = 20 \right ) =P\left { 19.5\leq X \leq 20.5 \right }

=P\left { \frac{19.5-20}{\sqrt{10}} < \frac{X-20}{\sqrt{10}} < \frac{20.5-20}{\sqrt{10}} \right }

\approx P\left { -.16< \frac{X-20}{\sqrt{10}} < .16 \right }

\approx \Phi \left ( .16 \right ) -\Phi \left ( -.16 \right ) \approx .1272

and if we solve the given example with the help of binomial distribution we will get it as

P\left { X=20 \right } = \binom{40}{20}\left ( \frac{1}{2} \right )^{40}\approx .1254

Example: To decide the efficiency of a definite nourishment in decreasing the extent of cholesterol in the blood circulation, 100 people are placed on the nourishment. The cholesterol count were observed for the define time after providing the nourishment. If from this sample 65 percent have low cholesterol count then nourishment will be approved. What is the probability that the nutritionist approves the new nourishment if, actually, it has no consequence on the cholesterol level?

solution:  Let the random variable express the cholesterol level if down by the nourishment so the probability for such random variable will be ½ for each person, if X denotes the low level number of people then the probability that result approved even there is no effect of nourishment to reduce the level of cholesterol is

\sum_{i=65}^{100}\binom{100}{i}\left ( \frac{1}{2} \right )^{100} =P\left { X\geq 64.5 \right }

=P\left { \frac{X-(100)\left ( \frac{1}{2} \right )}{\sqrt{100 \left ( \frac{1}{2} \right )\left ( \frac{1}{2} \right )}} \geq 2.9 \right }

\approx 1-\Phi \left ( 2.9 \right ) \approx .0019

Conclusion:

   In this article the concept of continuous random variable namely normal random variable and its distribution with probability density function were discussed and the statistical parameter mean, variance for the normal random variable is given. The conversion of normally distributed random variable to the new standard normal variate and area under the curve for such standard normal variate is given in tabulated form one of the relation with discrete random variable is also mentioned with example ,if you want further reading then go through:

Schaum’s Outlines of Probability and Statistics

https://en.wikipedia.org/wiki/Probability.

For more topics on mathematics please check this page.

About DR. MOHAMMED MAZHAR UL HAQUE

I am DR. Mohammed Mazhar Ul Haque , Assistant professor in Mathematics. Having 12 years of experience in teaching. Having vast knowledge in Pure Mathematics , precisely on Algebra. Having the immense ability of problem designing and solving. Capable of Motivating candidates to enhance their performance.
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