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STAT 400 UIUC 2.6 – Poisson Distribution Stepanov

Dalpiaz Nguyen

Poisson Distribution: X = the number of occurrences of a particular event in an interval of time or space.

P( X = x ) = , x = 0, 1, 2, 3, … .

E( X ) = l, Var( X ) = l. Table III ( pp. 580 – 582 ) gives P( X £ x ) R: dpois(x, lambda) gives P( X = x ) ppois(q, lambda) gives P( X £ x ) Example 1: Traffic accidents at a particular intersection follow Poisson distribution with an average rate of 1.4 per week. a) What is the probability that the next week is accident-free? b) What is the probability that there will be exactly 3 accidents next week?

!λ λ

x

x e -×

X- Ioi sson (d)pmf ::

O O

X -- O E(X) = 1.4⇒ D= I . 4

X = # of accidents happenend inaweekx- Poisson ( D= I. 4)

IC x -- o) =d°o = ""°e, = 0.24660

IC X -- 3) = d3ge = lh4)%e eo

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c) What is the probability that there will be at most 2 accidents next week? d) What is the probability that there will be at least 2 accidents during the next two weeks? e) What is the probability that there will be exactly 5 accidents during the next four weeks? f) What is the probability that there will be exactly 2 accidents tomorrow?

X -- fo , I , 2 , 3, . . . }

X E 2

I ( X f 2) = I (X -- O) t ICX -- l) tI ( X --2)= d°e t Ne, t HL = 0.83350

OR I ( Xf 2) = 0.8330 by Cdf table . Iweek → X-- I . 4

2 weeks → D= 2.8

Y = # of accidents happened in 2 weeks

Y N Poisson ( D= 2 (1.47--2.8)- aye- X

I ( Y 32) = I- = . - .

9=2 Y != , - [If y -- o) t ICY -- l)) - 0.76890

= I - I ( Y 22 ) = I - I (Y f D= I - O . 231=0.7690

Z = # of accidents happened in 4 weeks 1 week → D= 1.4

Z N Poisson ( D= 5.6) 4 weeks → D= 4 ( 1.47= 5.6

IC Z = 5) = d5g = l56? = 0.16970 ICE f 5)

I (Zf4) 4 5

I week → A = 1.4

X = # of accidents in a day 1 day → D= 1¥ = 0.2

XN Poisson ( D= 0.2)

I ( x -- D= Mg = l0⇒} = 0.01640

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g) What is the probability that the next accident will not occur for three days? h) What is the probability that there will be exactly three accident-free weeks during the next eight weeks? i) What is the probability that there will be exactly five accident-free days during the next week?

4=0

Y = # of accidents in 3 days l week → D= 1.4

Yu Ioisson ( D= 0.6)3 days → D= ( 'II) × 3

=

do e- A

=

0.6°

e- O- G = 0.6

If 4=0) - to = 0.548800 !

# of successes =3 n --8

"success

"= an accident - free week

week) =1.40 e- 1.4

p = I ( no accident in a - = 0.2466O !

X = # of accident -free weeks in 8 weeks

XN Binom ( n -- 8, p = 0.2466)

I ( X =3) = ( § ) (o. 2466) ' ( I - 0.2466) 5=0.203840n -- F

usuccess

"= an accident

-

free day 0.20 e-0.2

p = I ( no accident in a day )=- = 0.81873- o !D= = 0.2

X = # of accident -free days in a week

Xu Binom ( n -- F, p = 0.81873)

I ( X = 5) = ( t ) ( 0.818735 ( l - 0.818735 = 0.253850

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Note: When n is large ( n ³ 20 ) and p is small ( p £ 0.05 ) and n × p £ 5, Binomial probabilities can be approximated by Poisson probabilities. For this, set l = n × p. Example 2: Suppose the defective rate at a particular factory is 1%. Suppose 50 parts were selected from the daily output of parts. Let X denote the number of defective parts in the sample. a) Find the probability that the sample contains exactly 2 defective parts. b) Use Poisson approximation to find the probability that the sample contains exactly 2 defective parts. c) Find the probability that the sample contains at most 1 defective part. d) Use Poisson approximation to find the probability that the sample contains at most 1defective part.

P Poisson

X -- #of defective parts in the sampleX ~ Binom ( n = 50

, p = 0.01)

I ( X = 2) = ( 520) ( 0.01)'

( I - 0.01748=0.0756180

n = 50720 p= 0.01 hp-- 0.51 5

X E Poisson ( D= rip = 0.5)IC x -- 2) = Q5{? = 0.0758160

I (X Sl) = I ( X -- O) t I ( X -- I)

= ( 500) (0.015 ( 0.993"

t ( 5,0) (o.oD ' ( aaa)"

" 0.910570

I ( X fl) = I ( X -- O) t IC X =D

= a5°% + a5'e! = 0.90980