chapter 07rger

8/19/2019 Chapter 07rger

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Discrete & ContinuousProbability Distributions

Sunu Wibirama

Basic Probability and Statistics

Department of Electrical Engineering and Information Technology

Faculty of Engineering, Universitas Gadjah Mada


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OUTLINE

! Basic Probability Distributions

!

Binomial Distributions! Poisson Distributions

! Normal Distributions (!)


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SOME IMPORTANT THINGS

! Please read Walpole 8th Chapter 5 (for Binomial andPoisson) and Chapter 6 (forNormal Distribution)

! You can use Table A.1, A.2,and A.3 from Walpole book(p. 751)

! Some exercises from thesechapters (5 and 6) will beused in Midterm Exam


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Binomial Distribution


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CASE 1

! Suppose that 80% of the jobs submitted to adata-processing center are of a statisticalnature.

! Selecting a random sample of 10 submittedjobs would be analogous to tossing anunbalanced coin 10 times, with the probability

of observing a head (drawing a statistical job)on a single trial equal to 0.80.


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CASE 2

! Test for impurities commonly found in drinkingwater from private wells showed that 30% of allwells in a particular country have impurity A.

! If 20 wells are selected at random then it would beanalogous to tossing an unbalanced coin 20 times,with the probability of observing a head (selecting

a well with impurity A) on a single trial equal to0.30.


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BINOMIAL EXPERIMENT

!

Inspection of chip(defective or non-defective)

!

Inspection of public opinian(approve or not)

! Inspection of digging wells

(success or failed)

!

Etc….


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BERNOULLI PROCESS

! An experiment often consists of repeated trials,each with two possible outcome that may belabeled success and failed

!

The process is called Bernoulli Process:

! Consists of n repeated trials

! The outcome can be classified as success offailed (i.e., only 2 possible outcomes)

!

The probability of success (p) remains constantfrom trial to trial

! The repeated trials are independent


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BINOMIAL DISTRIBUTION

! The number X of successes in n Bernoulli trials iscalled binomial random variable

!

The probability of success is denoted by p

! The probability distribution of binomial randomvariable is called binomial distribution

b(x; n, p)


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! Each success has probability, p

! Each failure has probability, q

q = 1 - p

! x = number of successes

! n - x = numbe of failures

!

Since independent, combination probabilities aremultiplied

b ( x ;n , p ) =n !

x !(n ! x )!" # $

% & ' p x q n ! x


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GRAPH OF BINOMIAL DISTRIBUTION

b(x;n,p)

65 87 109 121121 43 130


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BINOMIAL CUMULATIVE DISTRIBUTION

!=

==

x

x

pn xb pn x B

0

1),;(),;(

• As with all other distributions,the sum of all probabilities must equal unity:


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Table A.1: Binomial Cumulative DistributionProportion, p

Trials Success 0.1 0.2 0.25 0.3 0.4 0.5 0.6 0.7 0.8 0.9

1

0

0.9000

0.8000

0.7500

0.7000

0.6000

0.5000

0.4000

0.3000

0.2000

0.1000

1 1 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

2 0 0.8100 0.6400 0.5625 0.4900 0.3600 0.2500 0.1600 0.0900 0.0400 0.0100

2 1 0.9900 0.9600 0.9375 0.9100 0.8400 0.7500 0.6400 0.5100 0.3600 0.1900

2 2 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

3 0 0.7290 0.5120 0.4219 0.3430 0.2160 0.1250 0.0640 0.0270 0.0080 0.0010

3 1 0.9720 0.8960 0.8438 0.7840 0.6480 0.5000 0.3520 0.2160 0.1040 0.0280

3 2 0.9990 0.9920 0.9844 0.9730 0.9360 0.8750 0.7840 0.6570 0.4880 0.2710

3 3 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

4

0

0.6561

0.4096

0.3164

0.2401

0.1296

0.0625

0.0256

0.0081

0.0016

0.0001

4 1 0.9477 0.8192 0.7383 0.6517 0.4752 0.3125 0.1792 0.0837 0.0272 0.0037

4 2 0.9963 0.9728 0.9492 0.9163 0.8208 0.6875 0.5248 0.3483 0.1808 0.0523

4

3

0.9999

0.9984

0.9961

0.9919

0.9744

0.9375

0.8704

0.7599

0.5904

0.3439

4 4 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000


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! Mean

! Variance

µ = np

! 2= npq


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EXAMPLE

! Test for impurities commonly found in drinkingwater from private wells showed that 30% of allwells in a particular country have impurity A.

! If a random sample of 5 wells is selected from thelarge number of wells in the country, what is theprobability that:

!

Exactly 3 will have impurity A?! At least 3?

! Fewer than 3?


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SOLUTION (Preliminary)

!

Confirm that this experiment is a binomial experiment.

! This experiment consists of n = 5 trials, onecorresponding to each random selected well.

! Each trial results in an S (the well contains impurity A)or an F (the well does not contain impurity A).

! Since the total number of wells in the country is large,

the probability of drawing a single well and finding thatit contains impurity A is equal to 0.30 and thisprobability will remain the same for each of the 5selected wells.


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SOLUTION (a)

! Therefore, the sampling process represents abinomial experiment with n = 5 and p = 0.30.

a) The probability of drawing exactly x = 3 wellscontaining impurity A, with n = 5, p = 0.30 and x = 3

b(3;5, 0.30) =5!

3!2!(0.30)3(1

!0.30)5

!

3 = 0.1323


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SOLUTION (b)

b) The probability of observing at least 3 wellscontaining impurity A is:

P (x ! 3) = p(3)+p(4)+p(5).

We have calculated p(3) = 0.1323 p(4) = 0.02835, p(5) = 0.00243.

In result,

P (x ! 3) = 0.1323+0.02835+0.00243 = 0.16380.


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SOLUTION (c)

c) Probability of observing less than 3 wellscontaining impurity A is P (x


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Poisson Distribution


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POISSON DISTRIBUTION

! The experiment consists of counting the number x of times a particular event occurs during a givenunit of time

!

The probability that an event occurs in a given unitof time is the same for all units

! The number of events that occur in one unit oftime is independent of the number that occur inother units.

! The mean number of events in each unit will bedenoted by the Greek letter !


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EXAMPLE

! Suppose that we are investigating the safety of adangerous intersection.

!

Past police records indicate a mean of 5 accidentsper month at this intersection.

! Suppose the number of accidents is distributedaccording to a Poisson distribution. Calculate the

probability in any month of exactly 0, 1, 2, 3 or 4accidents.


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SOLUTION

!

Since the number of accidents is distributed accordingto a Poisson distribution and the mean number ofaccidents per month is 5, we have the probability of

happening accidents in any month

! By this formula we can calculate:

p(x) =5 xe!5

x!

p(0) = 0.00674, p(1) = 0.03370, p(2) = 0.08425,

p(3) = 0.14042, p(4) = 0.17552.


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POISSON CUMULATIVE DISTRIBUTION

! Consider the previous example. Suppose we want toknow cumulative distribution of probability less than 3accidents per month

!

Than we have p(x < 3) = p(0) + p(1) + p(2)p(x < 3) = 0.00674+0.03370+0.08425= 0.12469

!

Simplify your computation using Table A.2

P(r;! ) = p( x=0

r

! x;! )


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Normal Distribution


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28

NORMAL DISTRIBUTION

!

Most important continuous probability distribution

! Graph called the “normal curve” (bell-shaped). Totalarea under the curve = 1

!

Derived by DeMoivre and Gauss. Called the “Gaussian”distribution.

!

Describes many phenomena in nature, industry andresearch

!

Random variable, Xf(x)

2 3 54 6


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0.0000

0.0500

0.1000

0.1500

0.2000

0.2500

0.3000

0.3500

0.4000

0.4500

-4.0 -2.0 0.0 2.0 4.0

f ( x )

Z-value

2

NORMAL DISTRIBUTION

! Definition: Density function of the normal

random variable, X, with mean and variancesuch that:

f ( x) = n( x; µ ,! ) =e

!0.5 ( x! µ )

!

"

#$%

&'

2

! 2" , ! (


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30

DIFFERENT NORMAL DISTRIBUTION

Different ________ f(x)

0 1 32 4

Different means

f(x)

0 1 32 4 5 6

Different meansand _______

f(x)

0 1 32 4


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!"##$%$&' &)%*+, !"-'%"./'")&-


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32

AREAS UNDER THE CURVE

! Area under the curve between x= x1 and x= x2

equals P(x1 < x < x2)

dxedx xn x X x P

x

x

x x

x

! ! "#$

%&' (

(

==


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STANDARD NORMAL DISTRIBUTION

• General density function:

• For standardized normaldistribution:

f ( x) = n( x; µ ,! ) =e

! x

2

"

#$%

&'

2

2"

f ( x) = n( x; µ ,! ) =e

!0.5 ( x! µ )

!

"

#$%

&'

2

! 2"


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STANDARD NORMAL DISTRIBUTION

! "0ʼ1 2314 523367 !"#$%&'$()*+,

! "0 821 2 962: 4; < 2:7 102:72=7 76>?2@4: 4; A

! '=2:1;4=9?:B :4=923 =2:749 >2=?2C36 ! 04 102:72=7 :4=923

=2:749 >2=?2C36 " x z

µ

!

"

=


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35

THE ORIGIN OF Z-DISTRIBUTION

! To enable use of z-table (Table A.3), transform to

unit values! Mean = 0

! Variance =1!

µ "=

X Z

{ }

2

2

1

2 2

1

2

1

( )

0.51 2

0.5

1( )2

1

2

( ; 0,1)

x x

x

z

z

z

z

z

P x X x e dx

e dz

n z dz

µ

!

! "

! "

#$ %#

& '( )

#

< < =

=

=

*

*

*


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Find area under normal curveExample: (P < -2.13) = 0.0166


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37

EXAMPLE

! Given the standard normal distribution, find the area

under the curve that lies to the LEFT of z = 1.84

! Solution: (use Table A.3)

f(x)

0

z 0.0000 0.0100 0.0200 0.0300 0.0400 0.0500

1.7

0.9554

0.9564

0.9573

0.9582

0.9591

0.9599

1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678


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SOLVING CASE WITH Z-DISTRIBUTION

! Suppose we have a problem finding probability of acase in X domain " p(X)

!

We can use Z-Distribution to solve the problem

! We should transform X " Z

! “Finding X from Z”


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#?:7?:B ! ;=49 "


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EXAMPLE


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-),/'")&


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-),/'")&


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-),/'")&


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TUGAS INDIVIDU(sekaligus latihan untuk Ujian MIDTERM)

! D6=E2F2: 72=? 6C44F G23H436 IJ08 $7?@4:KL

! $M6=5?16 NOP IH2B6 ANAQ F21R1 0=R5FK

! $M6=5?16 NONJ IH2B6 ASNQ F21R1 0=2T5 255?76:01K

!

$M6=5?16 SOAA IH2B6 AJSQ F21R1 32UV6=K

! $M6=5?16 SOAW IH2B6 AJXQ F21R1 10R76:01K

! D6=E2F2: 76:B2: 0R3?12: V2:B E6321Q C?32 7?H6=3RF2:Q029C28F2: ?3R10=21? H272 E2U2C2: +:72O Y2:B2: 3RH252:0R9F2: &+*+ 72: &"*

! .2021 U2F0R H6:BR9HR32:Q 82=? -6:?:


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THANK YOU

andGOOD LUCK

chapter 07rger

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