The normal distribution

Slide 1

The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

Shakeel NoumanM.Phil Statistics

The Normal Distribution

Slide 2

Using Statistics The Normal Probability Distribution The Standard Normal Distribution The Transformation of Normal Random

Variables The Inverse Transformation The Normal Distribution as an

Approximation to Other Probability Distributions

Summary and Review of Terms

The Normal Distribution4


Slide 3

As n increases, the binomial distribution approaches a ...n = 6 n = 14n = 10

Normal Probability Density Function:

6543210

0.3

0.2

0.1

0.0

x

P(x)

Binomial Dis tribution: n=6, p=.5

109876543210

0.3

0.2

0.1

0.0

x

P(x)

Binomial Distribution: n=10, p=.5

14131211109876543210

0.3

0.2

0.1

0.0

x

P(x)

Binomial Dis trib ution: n=14, p=.5

50-5

0.4

0.3

0.2

0.1

0.0

x

f( x)

Normal Distribution: = 0, = 1

f x ex

x

e

( )

. ... . ...

12 2

2

2 2

2 7182818 314159265

for

where and

4-1 Introduction


Slide 4

The normal probability density function:

50-5

0.4

0.3

0.2

0.1

0.0

x

f( x)


f x ex

x

e

( )

. ... . ...

12 2

2

2 2

2 7182818 314159265

for

where and

The Normal Probability Distribution


Slide 5

• The normal is a family ofBell-shaped and symmetric distributions. because the

distribution is symmetric, one-half (.50 or 50%) lies on either side of the mean.

Each is characterized by a different pair of mean, , and variance, . That is: [X~N()].

Each is asymptotic to the horizontal axis.The area under any normal probability density

function within k of is the same for any normal distribution, regardless of the mean and variance.

Properties of the Normal Probability Distribution


Slide 6

• If several independent random variables are normally distributed then their sum will also be normally distributed.

• The mean of the sum will be the sum of all the individual means.

• The variance of the sum will be the sum of all the individual variances (by virtue of the independence).

Properties of the Normal Probability Distribution (continued)


Slide 7

• If X1, X2, …, Xn are independent normal random variable, then their sum S will also be normally distributed with

• E(S) = E(X1) + E(X2) + … + E(Xn)• V(S) = V(X1) + V(X2) + … + V(Xn)• Note: It is the variances that can be added above and

not the standard deviations.

Properties of the Normal Probability Distribution

(continued)


Slide 8

Example 4.1: Let X1, X2, and X3 be independent random variables that are normally distributed with means and variances as shown.

Properties of the Normal Probability Distribution – Example

4-1

Mean VarianceX1 10 1

X2 20 2

X3 30 3

Let S = X1 + X2 + X3. Then E(S) = 10 + 20 + 30 = 60 and V(S) = 1 + 2 + 3 = 6. The standard deviation of S is

= 2.45.6


Slide 9

• If X1, X2, …, Xn are independent normal random variable, then the random variable Q defined as Q = a1X1 + a2X2 + … + anXn + b will also be normally distributed with

• E(Q) = a1E(X1) + a2E(X2) + … + anE(Xn) + b• V(Q) = a1

2 V(X1) + a22 V(X2) + … + an

2 V(Xn)• Note: It is the variances that can be added above and

not the standard deviations.

Properties of the Normal Probability Distribution (continued)


Slide 10

Example 4.3: Let X1 , X2 , X3 and X4 be independent random variables that are normally distributed with means and variances as shown. Find the mean and variance of Q = X1 - 2X2 + 3X2 - 4X4 + 5

Properties of the Normal Probability Distribution – Example 4-3

Mean VarianceX1 12 4

X2 -5 2

X3 8 5

X4 10 1

E(Q) = 12 – 2(-5) + 3(8) – 4(10) + 5 = 11V(Q) = 4 + (-2)2(2) + 32(5) + (-4)2(1) = 73SD(Q) =

544.873


Slide 11Computing the Mean, Variance and Standard Deviation for the Sum of Independent Random Variables Using the Template


Slide 12

All of these are normal probability density functions, though each has a different mean and variance.

Z~N(0,1)

50-5

0.4

0.3

0.2

0.1

0.0

z

f(z)

Normal Distribution: =0, =1

W~N(40,1) X~N(30,25)

454035

0.4

0.3

0.2

0.1

0.0

w

f(w)


6050403020100

0.2

0.1

0.0

x

f(x)


Y~N(50,9)

65554535

0.2

0.1

0.0

y

f(y)


50

Consider:

P(39 W 41)P(25 X 35)P(47 Y 53)P(-1 Z 1)

The probability in each case is an area under a

normal probability density function.

Normal Probability Distributions


Slide 13Computing Normal Probabilities

Using the Template


Slide 14

The standard normal random variable, Z, is the normal random variable with mean = 0 and standard deviation = 1:

Z~N(0,12).

543210- 1- 2- 3- 4- 5

0 . 4

0 . 3

0 . 2

0 . 1

0 . 0

Z

f(z )

Standard Normal Distribution

=0

=1{

4-3 The Standard Normal Distribution


Slide 15

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .090.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.15170.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.18790.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.22240.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.25490.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.28520.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.31330.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.33891.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.36211.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.38301.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.40151.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.41771.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.43191.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.44411.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.45451.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.46331.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.47061.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.47672.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.48172.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.48572.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.48902.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.49162.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.49362.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.49522.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.49642.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.49742.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.49812.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.49863.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990

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0.4

0.3

0.2

0.1

0.0

Z

f( z)


1.56{

Standard Normal Probabilities

Look in row labeled 1.5 and column labeled .06 to

find P(0 z 1.56) = .4406

Finding Probabilities of the Standard Normal Distribution: P(0

< Z < 1.56)


Slide 16

To find P(Z<-2.47):Find table area for 2.47

P(0 < Z < 2.47) = .4932

P(Z < -2.47) = .5 - P(0 < Z < 2.47) = .5 - .4932 = 0.0068

543210-1-2-3-4-5

0.4

0.3

0.2

0.1

0.0

Z

f(z)


Table area for 2.47P(0 < Z < 2.47) = 0.4932

Area to the left of -2.47P(Z < -2.47) = .5 - 0.4932

= 0.0068

Finding Probabilities of the Standard Normal Distribution: P(Z

< -2.47)z ... .06 .07 .08

. . . .

. . . .

. . . .2.3 ... 0.4909 0.4911 0.49132.4 ... 0.4931 0.4932 0.49342.5 ... 0.4948 0.4949 0.4951

.

.

.


Slide 17

z .00 ... . . . . . .

0.9 0.3159 ...1.0 0.3413 ...1.1 0.3643 ...

. . . . . .

1.9 0.4713 ...2.0 0.4772 ...2.1 0.4821 ...

. . . . . .

To find P(1 Z 2):1. Find table area for 2.00F(2) P(Z 2.00) .5 + .4772 .97722. Find table area for 1.00F(1) P(Z 1.00) .5 + .3413 .8413

3. P(1 Z 2.00) P(Z 2.00) P(Z 1.00) .9772 .8413 .1359

543210-1-2-3-4-5

0.4

0.3

0.2

0.1

0.0

Z

f(z)


Area between 1 and 2P(1 Z 2) .9772 .8413 0.1359

Finding Probabilities of the Standard Normal Distribution:

P(1< Z < 2)


Slide 18

To find z such that

P(0 Z z) = .40:

1. Find a probability as close as possible to .40 in the table of standard normal probabilities.

2. Then determine the value of z from the corresponding row

and column.

P(0 Z 1.28) .40

Also, since P(Z 0) = .50

P(Z 1.28) .90543210-1-2-3-4-5

0.4

0.3

0.2

0.1

0.0

Z

f(z)


Area = .40 (.3997)

Z = 1.28

Area to the left of 0 = .50P(z 0) = .50

Finding Values of the Standard Normal Random Variable: P(0 < Z

< z) = 0.40z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.15170.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.18790.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.22240.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.25490.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.28520.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.31330.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.33891.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.36211.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.38301.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.40151.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


Slide 19

z .04 .05 .06 .07 .08 .09. . . . . . .

. . . . . . .

. . . . . . .2.4 ... 0.4927 0.4929 0.4931 0.4932 0.4934 0.49362.5 ... 0.4945 0.4946 0.4948 0.4949 0.4951 0.49522.6 ... 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964

. . . . . . .

. . . . . . .

. . . . . . .

To have .99 in the center of the distribution, there should be (1/2)(1-.99) = (1/2)(.01) = .005 in each

tail of the distribution, and (1/2)(.99) = .495 in each half of the .99 interval. That is:

P(0 Z z.005) = .495

Look to the table of standard normal probabilities to find that:

z.005 z.005

P(-.2575 Z ) = .99

543210-1-2-3-4-5

0.4

0.3

0.2

0.1

0.0

Z

f(z)

-z.005 z.005

Area in right tail = .005Area in left tail = .005

Area in center right = .495

Area in center left = .495

2.575-2.575

Total area in center = .99

99% Interval around the Mean


Slide 20

The area within k of the mean is the same for all normal random variables. So an area under any normal distribution is equivalent to an area under the standard normal. In this

example: P(40 X P(-1 Z since m = 50 and s = 10.

1009080706050403020100

0.07

0.06

0.05

0.04

0.03

0.02

0.01

0.00

X

f(x)


=10{

543210-1-2-3-4-5

0.4

0.3

0.2

0.1

0.0

Z

f(z)


1.0{

Transformation

(2) Division by x)

The transformation of X to Z:

ZX x

x

The inverse transformation of Z to X:

X x Z x +

4-4 The Transformation of Normal Random Variables

(1) Subtraction: (X - x)


Slide 21

Example 4-9 X~N(160,302)

( )

P X

PX

P Z

P Z

( )

.. . .

100 180100 180

100 160

30

180 160

302 6667

0 4772 0 2475 0 7247

+

Example 4-10X~N(127,222)

( )

P X

PX

P Z

P Z

( )

.. . .

+

150150

150 127

221 045

0 5 0 3520 0 8520

Using the Normal Transformation


Slide 22

Example 4-11 X~N(383,122)

( )

P X

PX

P Z

P Z

( )

. .. . .

394 399394 399

394 383

12

399 383

120 9166 1 333

0 4088 0 3203 0 0885

440390340

0.05

0.04

0.03

0.02

0.01

0.00

X

f( X)


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0.2

0.1

0.0

Z

f(z)


Equivalent areas

Using the Normal Transformation - Example 4-11

Template solution


Slide 23

The transformation of X to Z:

ZX x

x

The inverse transformation of Z to X:X

xZ

x +

The transformation of X to Z, where a and b are numbers::

P X a P Z a

P X b P Z b

P a X b P a Z b

( )

( )

( )

The Transformation of Normal Random Variables


Slide 24

543210-1-2-3-4-5

0 .4

0 .3

0 .2

0 .1

0 .0

Z

f(z)

S t a n d a rd N o rm a l D is trib u tio n• The probability that a normal random variable will be within 1 standard deviation from its mean (on either side) is 0.6826, or approximately 0.68.

• The probability that a normal random variable will be within 2 standard deviations from its mean is 0.9544, or approximately 0.95.

• The probability that a normal random variable will be within 3 standard deviation from its mean is 0.9974.

Normal Probabilities (Empirical Rule)


Slide 25

z .07 .08 .09 . . . . . . . . . . . . . . .

1.1 . . . 0.3790 0.3810 0.38301.2 . . . 0.3980 0.3997 0.40151.3 . . . 0.4147 0.4162 0.4177

. . . . . . . . . . . . . . .

The area within k of the mean is the same for all normal random variables. To find a probability associated with any interval of values for any normal random variable, all that

is needed is to express the interval in terms of numbers of standard deviations from the mean. That is the purpose of the standard normal transformation. If X~N(50,102),

That is, P(X >70) can be found easily because 70 is 2 standard deviations above the mean of X: 70 = + 2. P(X > 70) is equivalent to P(Z > 2), an area under the standard

normal distribution.

P X Px

P Z P Z( ) ( )

70

70 70 5010

2

Example 4-12 X~N(124,122)P(X > x) = 0.10 and P(Z > 1.28)

0.10x = + z = 124 + (1.28)(12) =

139.36

18013080

0.04

0.03

0.02

0.01

0.00

X

f( x)


4-5 The Inverse Transformation

0.01

139.36


Slide 26

Example 4-12 X~N(124,122)P(X > x) = 0.10 and P(Z > 1.28)

0.10x = + z = 124 + (1.28)(12) =

139.36

Template Solution for Example 4-12


Slide 27

z .02 .03 .04 . . . . . . . . . . . . . . .

2.2 . . . 0.4868 0.4871 0.48752.3 . . . 0.4898 0.4901 0.49042.4 . . . 0.4922 0.4925 0.4927

. . . . . . . . . . . . . . .

z .05 .06 .07 . . . . . . . . . . . . . . .

1.8 . . . 0.4678 0.4686 0.46931.9 . . . 0.4744 0.4750 0.47562.0 . . . 0.4798 0.4803 0.4808

. . . . . . . . . .

Example 4-13 X~N(5.7,0.52)P(X > x)=0.01 and P(Z > 2.33)

0.01x = + z = 5.7 + (2.33)(0.5) = 6.865

Example 4-14 X~N(2450,4002)P(a<X<b)=0.95 and P(-1.96<Z<1.96)0.95

x = z = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234)

P(1666 < X < 3234) = 0.95

8.27.26.25.24.23.2

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.08.27.26.25.24.23.2

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

X

f(x)

Normal Distribution: = 5.7 = 0.5

543210-1-2-3-4-5

z Z.01 = 2.33

Area = 0.49

Area = 0.01

4000300020001000

0.0015

0.0010

0.0005

0.0000

X

f(x)

Normal Distribution: = 2450 = 400

4000300020001000

0.0015

0.0010

0.0005

0.0000

543210-1-2-3-4-5

Z

.4750.4750

.0250.0250

-1.96 1.96

The Inverse Transformation (Continued)

X.01 = +z = 5.7 + (2.33)(0.5) = 6.865


Slide 28

4000300020001000

0.0012

0.0010

0.0008

0.0006

0.0004

0.0002

0.0000

X

f( x)


.

.

.

.

.

.

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0.4

0.3

0.2

0.1

0.0

Z

f(z)

S ta nd ard Norm al Dis trib utio n

1. Draw pictures of the normal distribution in question and of the standard normal distribution.

Finding Values of a Normal Random Variable, Given a

Probability


Slide 29


2. Shade the area corresponding to the desired probability.


Probability

4000300020001000

0.0012

0.0010

0.0008

0.0006

0.0004

0.0002

0.0000

X

f( x)


.

.

.

.

.

.

.4750.4750

.9500

543210-1-2-3-4-5

0.4

0.3

0.2

0.1

0.0

Z

f(z)


.4750.4750

.9500


Slide 30

z .05 .06 .07 . . . . . . . . . . . . . . .

1.8 . . . 0.4678 0.4686 0.46931.9 . . . 0.4744 0.4750 0.47562.0 . . . 0.4798 0.4803 0.4808

. . . . . . . . . .

3. From the table of the standard normal distribution, find the z value or values.

1. Draw pictures of the normal

distribution in question and of the standard normal

distribution.



Probability

4000300020001000

0.0012

0.0010

0.0008

0.0006

0.0004

0.0002

0.0000

X

f( x)

Norm al Distribution: = 2450, = 400

.

.

.

.

.

.

.4750.4750

.9500

543210-1-2-3-4-5

0.4

0.3

0.2

0.1

0.0

Z

f(z)


.4750.4750

.9500

-1.96 1.96


Slide 31

4. Use the transformation from z to x to get value(s) of the original random variable.

x = z = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234)


Probability

z .05 .06 .07 . . . . . . . . . . . . . . .1.8 . . . 0.4678 0.4686 0.46931.9 . . . 0.4744 0.4750 0.47562.0 . . . 0.4798 0.4803 0.4808 . . . . . . . . . .

3. From the table of the standard normal distribution, find the z value or values.



4000300020001000

0.0012

0.0010

0.0008

0.0006

0.0004

0.0002

0.0000

X

f( x)


.

.

.

.

.

.

.4750.4750

.9500

543210-1-2-3-4-5

0.4

0.3

0.2

0.1

0.0

Z

f(z)


.4750.4750

.9500

-1.96 1.96


Slide 32

1050

0.3

0.2

0.1

0.0

X

f(x)

Normal Distribution: = 3.5, = 1.323

76543210

0.3

0.2

0.1

0.0

X

P(x)

Binomial Distribution: n = 7, p = 0.50

The normal distribution with = 3.5 and = 1.323 is a close approximation to the binomial with n = 7 and p = 0.50.

P(x<4.5) = 0.7749

MTB > cdf 4.5;SUBC> normal 3.5 1.323.

Cumulative Distribution Function

Normal with mean = 3.50000 and standard deviation = 1.32300

x P( X <= x) 4.5000 0.7751

MTB > cdf 4;SUBC> binomial 7,.5.


Binomial with n = 7 and p = 0.500000

x P( X <= x) 4.00 0.7734

P( x 4) = 0.7734


Probability


Slide 33

1050

0.3

0.2

0.1

0.0

X

f(x)

Normal Distribution: = 5.5, = 1.6583

11109876543210

0.2

0.1

0.0

X

P(x)

Binomial Distribution: n = 11, p = 0.50

The normal distribution with = 5.5 and = 1.6583 is a closer approximation to the binomial with n = 11 and p = 0.50.

P(x < 4.5) = 0.2732P(x 4) = 0.2744

MTB > cdf 4.5;SUBC> normal 5.5 1.6583.


Normal with mean = 5.50000 and standard deviation = 1.65830

x P( X <= x) 4.5000 0.2732

MTB > cdf 4;SUBC> binomial 11,.5.


Binomial with n = 11 and p = 0.500000

x P( X <= x) 4.00 0.2744

4-6 The Normal Approximation of Binomial Distribution


Slide 34

P a X b P a npnp p

Z b npnp p

( ) ( ) ( )

1 1

for large (n 50) and not too close to 0 or 1.00n p

P a X b P a npnp p

Z b npnp p

( ) .( )

.( )

+

0 51

0 51

for moderately large (20 n < 50).n

or:

If p is either small (close to 0) or large (close to 1), use the Poisson approximation.

Approximating a Binomial Probability Using the Normal

Distribution


Slide 35

Using the Template for Normal Approximation of the Binomial

Distribution


Slide 36

M.Phil (Statistics)

GC University, . (Degree awarded by GC University)

M.Sc (Statistics) GC University, . (Degree awarded by GC University)

Statitical Officer (BS-17) (Economics & Marketing Division)

Livestock Production Research Institute Bahadurnagar (Okara), Livestock & Dairy Development

Department, Govt. of Punjab

Name Shakeel NoumanReligion ChristianDomicile Punjab (Lahore)Contact # 0332-4462527. 0321-9898767E.Mail [email protected] [email protected]


mailto:[email protected]

mailto:[email protected]

Date post:	16-Jan-2017
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The normal distribution

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