Pertemuan 10 Sebaran Binomial dan Poisson

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Pertemuan 10 Sebaran Binomial dan Poisson

• Matakuliah I0134 Metoda Statistika
• Tahun 2005
• Versi Revisi

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Learning Outcomes

• Pada akhir pertemuan ini, diharapkan mahasiswa
• akan mampu
• Mahasiswa dapat menghitungpeluang, rataan dan
  varians peubah acak Binomial dan Poisson.

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Outline Materi

• Sebaran Peluang Binomial
• Nilai harapan dan varians sebaran Binomial
• Sebaran peluang Poisson
• Nilai harapan dan varians sebaran Poisson

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Binomial and Poisson Probability Distributions
Binomial Probability Distribution l Consider a
situation where there are only two possible
outcomes (a Bernoulli trial) Example u
flipping a coin head or tail u rolling a
dice 6 or not 6 (i.e. 1, 2, 3, 4, 5) Label
the possible outcomes by the variable k We
want to find the probability P(k) for event k to
occur Since k can take on only 2 values we
define those values as k 0 or k 1 u let
P(k 0) q (remember 0 q 1) u something
must happen so P(k 0) P(k 1) 1
(mutually exclusive events) P(k 1) p 1
- q u We can write the probability
distribution P(k) as P(k) pkq1-k
(Bernoulli distribution) u coin toss define
probability for a head as P(1) P(1) 0.5
and P(0tail) 0.5 too! u dice rolling define
probability for a six to be rolled from a six
sided dice as P(1) P(1) 1/6 and P(0not a
six) 5/6.
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• l What is the mean (?) of P(k)?
• l What is the Variance (?2) of P(k)?
• l Suppose we have N trials (e.g. we flip a coin
  N times)
• what is the probability to get m successes (
  heads)?
• l Consider tossing a coin twice. The possible
  outcomes are
• no heads P(m 0) q2
• one head P(m 1) qp pq (toss 1 is a
  tail, toss 2 is a head or toss 1 is head, toss 2
  is a tail)
• 2pq
• two heads P(m 2) p2

discrete distribution
we don't care which of the tosses is a head so
there are two outcomes that give one head
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l If we look at the three choices for the coin
flip example, each term is of the
form CmpmqN-m m 0, 1, 2, N 2 for our
example, q 1 - p always! coefficient Cm
takes into account the number of ways an outcome
can occur without regard to order. for m 0
or 2 there is only one way for the outcome (both
tosses give heads or tails) C0 C2 1 for m
1 (one head, two tosses) there are two ways
that this can occur C1 2. l Binomial
coefficients number of ways of taking N things m
at time 0! 1! 1, 2! 12 2, 3!
123 6, m! 123m Order of occurrence
is not important u e.g. 2 tosses, one head
case (m 1) n we don't care if toss 1
produced the head or if toss 2 produced the
head Unordered groups such as our example are
called combinations Ordered arrangements are
called permutations For N distinguishable
objects, if we want to group them m at a time,
the number of permutations u example If we
tossed a coin twice (N 2), there are two ways
for getting one head (m 1) u example
Suppose we have 3 balls, one white, one red, and
one blue. n Number of possible pairs we
could have, keeping track of order is 6 (rw, wr,
rb, br, wb, bw) n If order is not important
(rw wr), then the binomial formula gives
number of two color combinations
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• l Binomial distribution the probability of m
  success out of N trials
• p is probability of a success and q 1 - p is
  probability of a failure
• l To show that the binomial distribution is
  properly normalized, use Binomial Theorem

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• Mean of binomial distribution
• A cute way of evaluating the above sum is to take
  the derivative

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• Example Suppose you observed m special events
  (success) in a sample of N events
• u The measured probability (efficiency) for
  a special event to occur is
• What is the error on the probability ("error
  on the efficiency")
• The sample size (N) should be as large as
  possible to reduce certainty in the probability
  measurement
• Lets relate the above result to Lab 2 where we
  throw darts to measure the value of p.
• If we inscribe a circle inside a square with
  sides then the ratio of the area of the circle
• to the rectangle is

we will derive this later in the course

• So, if we throw darts at random at our rectangle
  then the probability (ºe) of a dart landing
  inside the
• circle is just the ratio of the two areas, p/4.
  The we can determine p using
• p 4e.
• The error in p is related to the error in e by

We can estimate how well we can measure p by
this method by assuming that ep/4 (3.14159)/4
This formula says that to improve our estimate
of p by a factor of 10 we have to throw 100 (N)
times as many darts! Clearly, this is an
inefficient way to determine p.
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• Example Suppose a baseball player's batting
  average is 0.300 (3 for 10 on average).
• u Consider the case where the player either
  gets a hit or makes an out (forget about walks
  here!).
• prob. for a hit p 0.30
• prob. for "no hit q 1 - p 0.7
• u On average how many hits does the player get
  in 100 at bats?
• ?? Np 1000.30 30 hits
• u What's the standard deviation for the number
  of hits in 100 at bats?
• ? (Npq)1/2 (1000.300.7)1/2 4.6 hits
• we expect 30 5 hits per 100 at bats
• u Consider a game where the player bats 4
  times
• probability of 0/4 (0.7)4 24
• probability of 1/4 4!/(3!1!)(0.3)1(0.7)3
  41
• probability of 2/4 4!/(2!2!)(0.3)2(0.7)2
  26
• probability of 3/4 4!/(1!3!)(0.3)3(0.7)1
  8
• probability of 4/4 4!/(0!4!)(0.3)4(0.7)0
  1
• probability of getting at least one hit 1
  - P(0) 1-0.2476

Pete Roses lifetime batting average 0.303
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• Poisson Probability Distribution
• l The Poisson distribution is a widely used
  discrete probability distribution.
• l Consider the following conditions
• p is very small and approaches 0
• u example a 100 sided dice instead of a 6
  sided dice, p 1/100 instead of 1/6
• u example a 1000 sided dice, p 1/1000
• N is very large and approaches 8
• u example throwing 100 or 1000 dice instead
  of 2 dice
• The product Np is finite
• l Example radioactive decay
• Suppose we have 25 mg of an element
• very large number of atoms N 1020
• Suppose the lifetime of this element t 1012
  years 5x1019 seconds
• probability of a given nucleus to decay in
  one second is very small p 1/t 2x10-20/sec
• Np 2/sec finite!
• The number of decays in a time interval is a
  Poisson process.
• l Poisson distribution can be derived by taking
  the appropriate limits of the binomial
  distribution

u radioactive decay u number of Prussian soldiers
kicked to death by horses per year per
army corps! u quality control, failure rate
predictions
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• u m is always an integer 0
• u ??does not have to be an integer
• It is easy to show that
• ?? Np mean of a Poisson distribution
• ?2 Np ? variance of a Poisson
  distribution
• l Radioactivity example with an average of 2
  decays/sec
• i) Whats the probability of zero decays in
  one second?
• ii) Whats the probability of more than one
  decay in one second?
• iii) Estimate the most probable number of
  decays/sec?
• u To solve this problem its convenient to
  maximize lnP(m, ?) instead of P(m, ?).

The mean and variance of a Poisson distribution
are the same number!
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• u In order to handle the factorial when take
  the derivative we use Stirling's Approximation
• The most probable value for m is just the
  average of the distribution
• u This is only approximate since Stirlings
  Approximation is only valid for large m.
• u Strictly speaking m can only take on integer
  values while ? is not restricted to be an
  integer.
• If you observed m events in a counting
  experiment, the error on m is

ln10!15.10 10ln10-1013.03
14 ln50!148.48 50ln50-50145.601.9
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Comparison of Binomial and Poisson distributions
with mean m 1
Not much difference between them!
N
N
For N large and m fixed Binomial Þ Poisson
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Uniform distribution and Random Numbers
What is a uniform probability distribution
p(x)? p(x)constant (c) for a x b p(x)zero
everywhere else Therefore p(x1)dx1 p(x2)dx2 if
dx1dx2 Þ equal intervals give equal
probabilities For a uniform distribution with
a0, b1 we have p(x)1
What is a random number generator ? A number
picked at random from a uniform distribution with
limits 0,1 All major computer languages
(FORTRAN, C) come with a random number
generator. FORTRAN RAN(iseed) The following
FORTRAN program generates 5 random numbers
iseed12345 do I1,5 yran(iseed)
type , y enddo end
0.1985246 0.8978736 0.2382888
0.3679854 0.3817045
If we generate a lot of random numbers all
equal intervals should contain the same amount of
numbers. For example generate 106 random
numbers expect 105 numbers 0.0, 0.1 105
numbers 0.45, 0.55
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• Selamat Belajar Semoga Sukses.