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Probabilitas dan Distribusi
Densiti
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PROBABILITAS
SUMBER PROBABILITAS: Model
Nilai ditaksir dari model (sederhana)
Contoh: Coin simetri, diundi menghasilkan probabilitas
Data Hasil experiment
Contoh:
John Kerrich mengundi coin 10000 kali, hasil ~
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PROBABILITAS
SUMBER PROBABILITAS:
Subjective
Perkiraan seseorang berdasar
pengetahuan/pengalaman Contoh:
Fatality rates of nuclear reactors accidents
1 minggu: 1 dalam 300,000,000
20 minggu: 1 dalam 16,000,000
Dasar data tsb. chain mechanism + subjectiveestimate
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PROBABILITAS
MODEL PROBABILITAS SEDERHANA:
Gambaran mekanisme suatuperistiwa/kejadian secara sederhana.
Outcomes
Sample Spaces
Events
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Outcomes
Keluaran/hasil suatu peristiwa atau
experiment Contoh:
outcomes undian coin: Hatau T
outcomes undian dadu: 1,2,3,4,5,6 outcomes WTC ditabrak pesawat: terbakar,
roboh sebagian, runtuh total, .
PROBABILITAS
MODEL PROBABILITAS SEDERHANA:
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PROBABILITAS
MODEL PROBABILITAS SEDERHANA:
Sample spaces (Ruang sample), S
himpunan (set) seluruh keluaran yang
mungkin (all possible outcomes) dariexperimen.
Contoh: 2 coin diundi bersama
1kali: S={HH,TH,HT,TT} 2 kali: S={HHHH, HHHT, HHTH, HTHH, THHH,
HHTT, HTTH, TTHH, HTHT, THHT, THTH,HTTT, THTT, TTHT, TTTH, TTTT}
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PROBABILITAS
MODEL PROBABILITAS SEDERHANA:
Events
kumpulan keluaran (outcomes)
terjadi bila keluaran (outcome) yangmendukung terjadi
Contoh:
Event A=satu Hkeluar dari undian sekali 2coin A={TH, HT}
Event A terjadi bila outcome THatau HTterjadi
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PROBABILITAS
MODEL PROBABILITAS SEDERHANA:
Sample space: event lengkap
Event subset dari sample space.
A
Sample space, S
Event, A
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PROBABILITAS
MODEL PROBABILITAS SEDERHANA:
Sebuah event dapat hanya terdiri dari
satu keluaran (outcome) Contoh:
A={TT} B={HH}
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PROBABILITAS
MODEL PROBABILITAS SEDERHANA:
Komplemen dari event A, dengan notasiA, terjadi bila A tidak terjadi
S
AA
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PROBABILITAS
MODEL PROBABILITAS SEDERHANA:
Gabungan Event
Union,
AB
A B
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PROBABILITAS
MODEL PROBABILITAS SEDERHANA:
Gabungan event
Intersection
A B
AB
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PROBABILITAS
MODEL PROBABILITAS SEDERHANA:
Gabungan event
Mutually exclusive
A B
Mutually exclusive
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HUKUM PROBABILITAS
AXIOMA:
Probability sample space, S
P[S] =1 pasti terjadi
Probability event, A
0 < P[A] < 1 mungkin terjadi
Event kosong
P[] = 0
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Partisi
Event C1, C2, C3, .., Ck membentuk
partisi (pembagian) sample space bilaevent-event tersebut mutually exclusivedan menghabiskan seluruh sample space.
C1C
2C
3...C
k= S
P[C1C2C3...Ck] = P[C1] + P[C2] +
P[C3] + + P[Ck]
HUKUM PROBABILITAS
TEOREMA:
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Partisi
P[A] = P[AC1] + P[AC2] + + P[ACk]
= P[ACi]
HUKUM PROBABILITAS
TEOREMA:
C6
C3C2
C1
C5C4
Ci
AC1
C2 C3
C4C5
C6
A Ci (diarsir gelap)
i=1
k
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HUKUM PROBABILITAS
TEOREMA: Probabilitas Bersyarat (Conditional
Probability)
Event A terjadi bila event B terjadi Probability-nya
P[A|B] =
P[AB]
P[B] syarat P[B
]
0
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Example of Conditional Probability:Proportion of women with the given eyesight
grades
Grade of Left Eye
GradeofRightEye
Highest ThirdSecond Lowest
Highest
Third
Second
Lowest
Total
Total
.237
.058
.017
.024
.027
.010
.009
.112
.066
.328
.301
.265
1
.106
.336
.203
.016
.005
.255
.031
.036
.048
.011
.291
.202
Anggapan: sample cukup banyak dan mewakilipopulasi, proporsi mewakiliprobabilitas
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What is the probability of women with righteye 2nd grade given her left eye highest
grade ?
P[right eye 2nd grade|left eye highest grade] =
P[right eye 2nd grade left eye highest grade]
P[left eye highest grade]
Example of Conditional Probability:Proportion of women with the given eyesight
grades
0.0310.225
= 0.1216
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Consider 3 special cards below
Example of Conditional Probability:3 Cards Trick
Randomized& draw oneblue
red
blue
blue
red
red
blue
?
Probability of blue bottom = ?
Probability of red bottom = ?
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Probability of blue top faces:
P[blue top] = 3/4
Probability of blue bottom:P[blue bottom] = 1/2
Probability of a card having blue bottom face
given blue top face:P[blue bottom|blue top] = (1/2)/(3/4) = 2/3
Example of Conditional Probability:3 Cards Trick
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Example of Conditional Probability:3 Cards Trick
Top
Bottom
Blue Red
Blue
Red 1/6 2/6
1/62/6 1/2
1/2
1/2 1/2 1total
total
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Example of Conditional Probability:3 Cards Trick
blue top
red top
3/6
3/6
blue bottom
blue bottom
red bottom
red bottom
2/3
1/3
2/3
1/3
3/6 x 2/3 = 2/6
3/6 x 1/3 = 1/6
3/6 x 2/3 = 2/6
3/6 x 1/3 = 1/6
P[A] P[B|A] P[AB]
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Independence and Multiplication
Rule Independent Events
Two events are independent if one may
occur irrespective of the other
Event A and B are independent if and
only if P[AB] = P[A] P[B]
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Example:
A1 = sample contains Pb P[A1] = 0.32
A2 = sample contains Hg P[A2] = 0.16 sample contains both P[A1A2] = 0.10
What is the probability of a samplecontains Pb will also contains Hg?
Independence and Multiplication Rule
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Answer
P[A2|A1] = P[A1A2] / P[A1]
= 0.10/0.32 = 0.31
P[A2] = 0.16
A1 and A2 are not independent
Independent if
P[A2|A1] = P[A2]
Independence and Multiplication Rule
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Bayes Theorem
Let A1, A2, A3, , Anbe a collection ofevents which partition S.
Let B be event such that P[B] 0 For any events Aj, j= 1,2,3,,n then
ni
ii
jjj
APABP
APABPBAP
1
][]|[
][]|[]|[
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Example Blood type distribution
in a country is
type A = 41%
type B = 9%
type AB = 4%
type O = 46%
Bayes Theorem
Estimated that during handlingemergency there are
probability of wrong blood typeidentification
type O identified A = 4%
type A identified A =
88% type B identified A = 4%
type AB identified A =10%
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Someone having a traffic accident his/herblood identified as type A. What is the
probability that this is a correct type? Let
A : someone has type A blood
B : someone has type B blood
AB : someone has type AB bloodO : someone has type O blood
TA : someone identified as type A blood
Bayes Theorem
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Bayes Theorem
TA
OABB
A OTAABTA
BTAATA
TA = (ATA)(BTA)(ABTA )(OTA )
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We want to find P[A|TA] =?
Given
P[A] = 0.41 P[TA|A] = 0.88P[B] = 0.09 P[TA|B] = 0.04
P[AB] = 0.04 P[TA|AB] = 0.10
P[O] = 0.46 P[TA|O] = 0.04
Bayes Theorem
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Lets try to use conditional probability
Where are P[ATA] and P[TA] ??P[ATA] = P[TA|A] P[A]
= (0.88)(0.41) = 0.36P[TA] = P[ATA] + P[BTA] + P[ABTA]
+ P[OTA]
Bayes Theorem
][
][]|[
TAP
TAAPTAAP
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P[TA] = P[ATA] + P[BTA] + P[ABTA]
+ P[OTA]
= P[TA|A] P[A] + P[TA|B] P[B] +P[TA|AB] P[AB] + P[TA|O] P[O]
= (0.88)(0.41) + (0.04)(0.09) + (0.10)(0.04)
+ (0.04)(0.46)
= 0.39 Finally P[A|TA] = 0.36/0.39 = 0.92
Bayes Theorem
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DISTRIBUSI PROBABILITAS DISKRIT(Discrete Probability Distribution)
DISKRIT
VARIABLE & VALUE
RANDOM VARIABLE & RANDOMVALUE
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DISTRIBUSI PROBABILITAS DISKRIT(Discrete Probability Distribution)
DISKRIT
terpisah jelas satu dengan lainnya dapat dihitung/cacah (countable)
masing-masing bernilai sendiri dan
terpisah tidak dapat dipecah atau dibagi
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DISTRIBUSI PROBABILITAS DISKRIT(Discrete Probability Distribution)
VARIABLE & NILAI (VALUE)
Variable: suatu kuantitas (event) yang
mempunyai satu nilai tertentu daribanyak kemungkinan nilai
Nilai (Value): ukuran yang dikaitkandengan variable
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DISTRIBUSI PROBABILITAS DISKRIT(Discrete Probability Distribution)
RANDOM VARIABLE & RANDOMVALUE
Random variable: nilainya random
Random value (nilai random):
besar pastinya tidak diketahui kisaran nilainya diketahui dari sample
spacenya
terkait probabilitas
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DISTRIBUSI PROBABILITAS DISKRIT(Discrete Probability Distribution)
Cara penulisan
Nilai: x
Variabel: X Probabilitas variabel Xbernilai x:
P[X=x]
Probabilitas bila xbermacam-macam:f(x)= P[X=x] x= macam-macam
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DISTRIBUSI PROBABILITAS DISKRIT(Discrete Probability Distribution)
Distribusi Probabilitas (ProbabilityDistribution)
gambaran/fungsi sebaran probabilitas adalah f(x)
dinyatakan dalam bentuk
distribusi densitas distribusi kumulatif
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DISTRIBUSI PROBABILITAS DISKRIT(Discrete Probability Distribution)
Distribusi Densitas Probabilitas
menyatakan nilai probabilitas pada setiap
nilai xf(x) =P[X=x]
x
f(x)
Total probabilitas =
f(x)= 1 utk seluruh xBila xkontinyuf(x)dx=1
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Distribusi Kumulatif Probabilitas
menyatakan nilai kumulatif probabilitas
sampai dengan nilai x
F(x) =P[X < x]
limit F(x) = 1
x
F(x)
DISTRIBUSI PROBABILITAS DISKRIT(Discrete Probability Distribution)
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PARAMETER DISTRIBUSI:
NILAI HARAPAN (Expected Value)
Definisi Nilai Harapan (Expected Value),
H(x) AndaikanX : variabel random diskrit
dengan distribusi densiti f(x)
AndaikanH(X) : variabel random (yanglain)
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Nilai harapandari H(X), yi. E[H(X)]:
xsemua
H(x)f(x)E[H(X)]
PARAMETER DISTRIBUSI:NILAI HARAPAN (Expected Value)
semua xSyaratH(X)f(x) bernilai tertentu
Perjumlahan untuk semua nilai x dengan P[X=x]0
Statistik:nilai harapan = mean = matau
mx
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PARAMETER DISTRIBUSI:NILAI HARAPAN (Expected Value)
TEOREMA AndaikanXdanYadalah variabel
random
Andaikancadalah sebarang bilanganreal E[X + Y] = E[X] + E[Y]
E[cX ] = cE[X]
E[c] = c
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Definisi Variance
Andaikan X adalah random variable
denganMeanm Variance X
Var X =s2 = E[(X-m)2] hitungmdulu
ataus2 = E[X2] - (E[X])2 lebih praktis
PARAMETER DISTRIBUSI:
VARIANCE dan STANDARD DEVIASI
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Defini Standard Deviasi, s Penyimpangan standar dari mean
Akar dari variances=Var X =s2
Derajad Kebebasan (Degrees of
Freedom), nn= n - 1 n = banyaknya X
PARAMETER DISTRIBUSI:
VARIANCE dan STANDARD DEVIASI
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AndaikanXdanYadalah variabelrandom dancadalah bilangan real
Var c = 0Var cX = c2Var X
Var(X + Y) = Var X + Var Yuntuk X dan Y tak saling terikat
(independent)
PARAMETER DISTRIBUSI:
VARIANCE dan STANDARD DEVIASI
TEOREMA
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CIRI
Outcomes: sukses dan gagal Undian identik dan independent
Probabilitas sukses,p, tetappada setiap undian
Undian Bernoulli
(Bernoulli Trial)
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Undian Bernoulli
(Bernoulli Trial)
Trial
n kali trial hinggasukses pertama
kali
jml suksesdalam n kali
trial
DistribusiGeometrik
DistribusiBinomial
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Distribusi Geometrik
Variabel random Xberdistribusigeometrik dengan parameterpapabila
densitas probabilitasnya
f(x) = (1- p)x-1puntuk 0
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Berlaku Bernoulli trial
Tiap trial identik dan independent Probability sukses ptetap
Xmenyatakan jumlah trial sampai
sukses yang pertama kali
Distribusi Geometrik
Ciri
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Trial Geometrik:
Coba terus sampai berhasil!!
s
p = 3/4 q = 1 - p = 1/4
f
f(1)=3/4
s
s
s
f
f
f(3)=1/4 x 1/4 x 3/4 = 3/64
f(4)=1/4 x 1/4 x 1/4 x 3/4 = 3/256
f(2)=1/4 x 3/4 = 3/16
1x
2x
3x
4x
Probabilitas kumulatif s/d 4x =255/256 = 0.9961
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Trial Geometrik:Coba terus sampai berhasil!!
0.0000
0.2000
0.4000
0.6000
0.8000
1.0000
0 2 4 6
trial
probability
density
cuml
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Distribusi Binomial
Definisi Variabel random Xberdistribusi
geometrik dengan parameterndanp
apabila densitas probabilitasnyaf(x) = (1- p)n-xpx
untuk 0
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Kombinasi
Distribusi Binomial
Definisi
nx = C(n,x) = n!x! (n-x)!
C(7,3) = 7!3! (7-3)!
= 7.6.53.2.1
= 35
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Andaikan Xvariabel random denganparameter ndan p, maka
Expected Value, E(X)=m= np
-Varian, s2(X)
= npq q= 1 -p
Distribusi Binomial
Teorema
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Berlaku Bernoulli trial
Jumlah ntetap pada setiap trial
Tiap trial identik dan independent
Probability sukses ptetap
Xmenyatakan jumlah sukses dalamsetiap trial n
Distribusi Binomial
Ciri
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Anggap probabilitas hujan setiap hari, p
Anggap ptetap = 0.15
Berapa probabilitas 3 hari hujan dalamseminggu?
n = 7 p = 0.15 q = 0.85
x = 3
P[3] = C(7,3)(0.85)4(0.15)3=0.062
Distribusi Binomial
Contoh
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Sampling
Sampling
withreplacement
withoutreplacement
BinomialDistribution
HypergeometricDistribution
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draws or trials are not independent
probability of success, p, is not constant
variable X, the number of success in ntrials follows hypergeometric distribution
Sampling without Replacement
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A random variable Xhas a hypergeometricdistribution with parameters N, n, and rif itsdensity is given by
Hypergeometric Distribution
rx
N
n
N - rn- xf(x) =
max[0, n- (N - r)] < x< min(n, r)N, r, n positive integers
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Example
There are 20 small balls inside a bowl.
12 of them are black and 8 are white All balls are well mixed inside the bowl
5 balls are sampled from the bowl withoutreplacement
Xrepresents the number of black balls in thesample
Hypergeometric Distribution
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What is the probability of X?
N = 20, r = 12, n = 5
P[X=0] = C(12,0)C(8,5)/C(20,5) = 0.0036P[X=1] = C(12,1)C(8,4)/C(20,5) = 0.0542
P[X=2] = C(12,2)C(8,3)/C(20,5) = 0.2384
P[X=3] = C(12,3)C(8,2)/C(20,5) = 0.3973P[X=4] = C(12,4)C(8,1)/C(20,5) = 0.2554
P[X=5] = C(12,5)C(8,0)/C(20,5) = 0.0511
Hypergeometric Distribution
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Limiting case
if n
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Example of limiting case
100 transistors are sampled from 5000
the number of defective transistors are5%
What is the probability of X, the number
of defective transistors in the sample?
Hypergeometric Distribution
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Hypergeometric solution N = 5000 r = 250 n = 100
P[X=0] = C(250,0)C(4750,100)/C(5000,100) = 0.0056P[X=1] =C(250,1)C(4750,99)/C(5000,100) = 0.0302
P[X=2] = C(250,2)C(4750,98)/C(5000,100) = 0.0800
P[X=3] =C(250,3)C(4750,97)/C(5000,100) = 0.1392
P[X=4] = C(250,4)C(4750,96)/C(5000,100) = 0.1792
P[X=5] =C(250,5)C(4750,95)/C(5000,100) = 0.1818
Hypergeometric Distribution
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Binomial solution
n = 100, p = 0.05
P[X=0] = C(100,0)(0.05)0(0.95)100= 0.0059P[X=1] =C(100,1)(0.05)1(0.95)99= 0.0312
P[X=2] = C(100,2)(0.05)2(0.95)98= 0.0812
P[X=3] =C(100,3)(0.05)3(0.95)97= 0.1396
P[X=4] = C(100,4)(0.05)4(0.95)96= 0.1781
P[X=5] =C(100,5)(0.05)5(0.95)95= 0.1800
Hypergeometric Distribution
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Hypergeometric
P[X=0] = 0.0056
P[X=1] = 0.0302
P[X=2] = 0.0800
P[X=3] = 0.1392
P[X=4] = 0.1792P[X=5] = 0.1818
Hypergeometric Distribution
Limiting Case Comparison
Binomial
P[X=0] = 0.0059
P[X=1] = 0.0312P[X=2] = 0.0812
P[X=3] = 0.1396
P[X=4] = 0.1781P[X=5] = 0.1800
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Poisson Distribution
Definition French mathematician: Simeon Denis
Poisson (1781-1840)
A random variable X has a Poissondistribution with parameter k if itsdensity is
e-kkx
f(x) = ------------x! x = 0,1,2,k > 0
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Let Xbe a Poisson random variablewith parameter k
E[X] = k
Var[X] = k
Poisson Distribution
Theorem
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Connected to Poisson processes
Poisson process:
observing discrete and infrequent eventsin a continuous interval of time, length,or space
Xis the number of occurrences of theevent in the interval of sunits
k= ls, l= rates of occurrence per
units
Poisson Distribution
Features
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Average white blood-cell count of a healthyperson = 6000/mm3 of blood.
To detect white blood-cell deficiency, a 0.001mm3 of blood is taken and the number ofwhite cells Xis counted.
How many white cells are expected in a
healthy person? If at most 2 cells are found, is there evidence
of white blood-cell deficiency?
Poisson Distribution
Example
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Solutionunit = mm3 s= 0.001l= 6000 ls= (6000)(0.001) = 6
Expected value, E[X]= ls= 6 (for a healthyperson)
How rare is it to see at most two?P[X
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Worksheet Functions
BINOMIAL DISTRIBUTION
BINOMDIST(number_s,trials,probability_s,cumulative)
Returns the individual term binomial distribution
probability.
Number_s : the number of successes in trials.
Trials : the number of independent trials.
Probability_s : the probability of success oneach trial.
Cumulative : a logical value. TRUE for thecumulative distribution function, FALSE for theprobability density/mass function.
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Example
The flip of a coin can only result in heads ortails. The probability of the first flip beingheads is 0.5, and the probability of exactly 6 of10 flips being heads is:
BINOMDIST(6,10,0.5,FALSE) equals 0.205078
Worksheet Functions
BINOMIAL DISTRIBUTION
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Remarks
Number_s and trials are truncated to integers.
If number_s, trials, or probability_s isnonnumeric, BINOMDIST returns the#VALUE! error value.
If number_s < 0 or number_s > trials,BINOMDIST returns the #NUM! error value.
If probability_s < 0 or probability_s > 1,BINOMDIST returns the #NUM! error value.
Worksheet Functions
BINOMIAL DISTRIBUTION
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COMBIN(number,number_chosen) Returns the number of combinations for a given number
of items.
Number is the number of items. Number_chosen is the number of items in each
combination.
Example
Suppose you want to form a two-person teamfrom eight candidates, and you want to knowhow many possible teams can be formed.COMBIN(8, 2) equals 28 teams.
Worksheet Functions
COMBINATION
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Remarks
Numeric arguments are truncated to integers.
If either argument is nonnumeric, COMBINreturns the #NAME? error value.
If number < 0, number_chosen < 0, or number
number_population, HYPGEOMDIST returnsthe #NUM! error value.
If population_s < 0 or population_s >number_population, HYPGEOMDIST returnsthe #NUM! error value.
Worksheet Functions
HYPERGEOMETRIC DISTRIBUTION
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If number_population < 0, HYPGEOMDISTreturns the #NUM! error value.
HYPGEOMDIST is used in sampling withoutreplacement from a finite population.
Worksheet Functions
HYPERGEOMETRIC DISTRIBUTION
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POISSON(x,mean,cumulative) Returns the Poisson distribution.
X is the number of events.
Mean is the expected numeric value. Cumulative is a logical value. TRUE for
cumulative Poisson probability and FALSE forprobability density/mass function
Examples
POISSON(2,5,FALSE) equals 0.084224
POISSON(2,5,TRUE) equals 0.124652
Worksheet Functions
POISSON DISTRIBUTION
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Remarks
If x is not an integer, it is truncated.
If x or mean is nonnumeric, POISSON returnsthe #VALUE! error value.
If x 0, POISSON returns the #NUM! errorvalue.
If mean 0, POISSON returns the #NUM!error value.
Worksheet Functions
POISSON DISTRIBUTION
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Continue to Probability2
http://c/Kuliah/Statistik/Probability2.ppthttp://c/Kuliah/Statistik/Probability2.ppt