Lecture 7-3-2014 - Normal Distribution II(1)

8/12/2019 Lecture 7-3-2014 - Normal Distribution II(1)

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Special Continuous ProbabilityDistributions

Normal Distribution II

Instructor: Dr. Gaurav Bhatnagar

22001: PROBABILITY AND STATISTICS


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Normal Distribution - Definition

Probability estimation for Normal distribution.

a b

f(x) P(a


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a

f(x)

P(X


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a

f(x)

P(X> a)

P( ) = ( ) 1 ( ) 1 P( )a

X a f x dx F a X a


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Standard Normal Distribution - Definition

A random variable is said to have a standard normal

distribution if the mean and standard deviation are 0 and 1

respectively.

The probability density function is given by

Notation:

Z

2

21( ) , (2)

2

z

f z e z

~ (0,1)X N


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Moments:

Mean: r= 1

Variance:

21

21

' ( ) (3)2

zr r

r z f z dz z e dz

21

2

1

1' 0 (4)

2

z

ze dz

2 22

2 1( ) ( ) ' ' 1 (5)V X E X E X


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Moment Generating Function:

2

2

1

2

1

2

( )

1

2

(6)

zt

Z

zzt

t

M t E e

e e dz

e


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Limitation of Normal Distribution

Each normal distribution with its own values ofand

would need its own calculation of the area under various

points on the curve.

100 200

(= 100, = 50)

(= 0, = 1)

0 2

Possible Solution: Standardizing the normal distribution


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Standardizing or Z-score

Suppose:X ~ N(,)

Form a new random variable (Z) by subtracting the mean

() fromXand dividing by the standard deviation (), i.e.,

It is clear that the mean and standard deviation of new

random variable (Z) is 0 and 1, i.e.,Z ~ N(0,1)

This process is called standardizing the random variable

X.

Z-scores make it easier to compare data values

measured on different scales.

XZ


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Meaning of Z-score

Suppose marks of Probability and Statistics among

students are normally distributed with a mean of 25 and a

standard deviation of 10. If a student scores a 45, whatwould be the z-score?

- 45 252

10

Xz

Soln:TheZ-score is

The z-score would be 2 which means her score is two

standard deviations above the mean.


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Meaning of Z-score

Suppose mid-sem 1 marks of students in Probability and

Statistics has a mean of 9 and a standard deviation of 4

whereas their Signal, System and Networks marks has amean of 12 and a standard deviation of 8. For which course

18 marks have higher standing?

18 92.25

4

Pz

18 120.75

8Sz

Soln:TheZ-score is

The P&S mark would have the highest standing since it is 2

standard deviation above the mean while the SSN mark is

only 0.75 standard deviation above the mean.


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P( ) = ( ) ( ) ( )b

a

a Z b f z dz F b F a

a b

f(z)

P(a


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Cumulative Distribution Function:

Analytical solution is very hard to obtain. Therefore, the

above equation needs to solve numerically. The numericalvalues lie from 0 to 3 are reported in a Table usually called

Z-tables.

Typically negative values are not reportedo Symmetrical, therefore area below negative value = Area above

its positive value

Always helps to draw a sketch!

Probabilit ies and z scores: z tables

21

21( ) ( ) 2

zu

F z P Z z e du


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Draw a picture of standard normal distribution

depicting the area of interest.

Re-express the area in terms of shapes like the one

on top of the Standard Normal Table

Look up the areas using the table.

Do the necessary addition and subtraction.

Strategies for finding probabilities for N(0,1)

P(Z>b)=P(Z


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P(0


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P(Z>0.78)?



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P(-1.2


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Meaning of Z-score Revisited

Suppose mid-sem 1 marks of students in Probability and

Statistics has a mean of 9 and a standard deviation of 4

whereas their Signal, System and Networks marks has amean of 12 and a standard deviation of 8. For which course

18 marks have higher standing?

Soln:TheZ-score is

The P&S mark would have the highest standing.

18 92.25

4

( ) 0.9878

P

P

z

P Z z

18 120.75

8

( ) 0.7734

S

S

z

P Z z


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Example 1

IfX~N(109, 13), Then find out (i) P (X141)

(i)

(ii)

120( 120) ( 120 )

109 120 1090.85 0.8023

13 13

XP X P X P

XP P Z

141( 141) ( 141 )

109 141 1092.46 1 ( 2.46)

13 13

1 0.9931 0.0069

XP X P X P

XP P Z P Z


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Example 2

The P&S exam is given to 147 students. The scores

have a normal distribution with a mean of 78 and a

standard deviation of 5.

(i) How many students have scores between 82 and 90?

(ii) What percent of the students have scores above 60?(iii) How many students have scores above 70?

Ans.:(i) 0.2037

(ii) 0.6406

(iii) 0.9452


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Normal and Binomial Distribution

Theorem: IfXis a random variable with distribution B(n,

p), then for sufficiently large n, it can be approximated by

Normal Distribution, i.e.,

2~ (0,1), with and (1 )X

Z N np np p

~ ( , ) ~ , (1 )X B n p X N np np p

Observations:

(i)

(ii) The normal distribution is a good approximation for the

binomial distribution whennp 5andn(1 p) 5.


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Example 1: Is it possible to obtain normal distribution

approximation from the binomial distribution where n= 20

andp= 0.25?

220 0.25 5

(1 ) 20 0.25 0.75 3.75

3.75 1.94

npnp p

Solution: According to theorem, define the following

parameters:

Sincenp= 5 5andn(1 p)= 15 5, based on the previous

theorem, we can conclude thatB(20,0.25) ~ N(5,1.94).


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Example 2: A sample of 100 items is taken at random

from a batch known to contain 40% defectives. What is the

probability that the sample contains: (i) at least 44

defectives (ii) exactly 44 defectives?

2

100 0.40 40(1 ) 100 0.4 0.6 24

24 4.89

np

np p

Solution: According to theorem, define the following

parameters:

Sincenp= 40 5andn(1 p)= 60 5, based on the previous

theorem, we can conclude thatB(100,0.4) ~ N(40,4.89).

Ans.:(i) 0.2376 (ii) 0.584

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Lecture 7-3-2014 - Normal Distribution II(1)

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