1 Pertemuan 10 Sebaran Binomial dan Poisson Matakuliah: I0134 – Metoda Statistika Tahun: 2005 Versi: Revisi

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

3 Outline Materi Sebaran Peluang Binomial Nilai harapan dan varians sebaran Binomial Sebaran peluang Poisson Nilai harapan dan varians sebaran Poisson

4 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) = p k q 1-k (Bernoulli distribution) u coin toss: define probability for a head as P(1) P(1) = 0.5 and P(0=tail) = 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(0=not a six) = 5/6.

5 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) = q 2 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) = p 2 Note: P(m=0)+P(m=1)+P(m=2)=q 2 + qp + pq +p 2 = (p+q) 2 = 1 (as it should!) l We want the probability distribution function P(m, N, p) where: m = number of success (e.g. number of heads in a coin toss) N = number of trials (e.g. number of coin tosses) p = probability for a success (e.g. 0.5 for a head) we don't care which of the tosses is a head so there are two outcomes that give one head discrete distribution

6 l If we look at the three choices for the coin flip example, each term is of the form: C m p m q N-m m = 0, 1, 2, N = 2 for our example, q = 1 - p always! coefficient C m 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): C 0 = C 2 = 1 for m = 1 (one head, two tosses) there are two ways that this can occur: C 1 = 2. l Binomial coefficients: number of ways of taking N things m at time 0! = 1! = 1, 2! = 1·2 = 2, 3! = 1·2·3 = 6, m! = 1·2·3···m 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

7 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:  binomial distribution is properly normalized

8 Mean of binomial distribution: A cute way of evaluating the above sum is to take the derivative: Variance of binomial distribution (obtained using similar trick):

9 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 Let’s relate the above result to Lab 2 where we throw darts to measure the value of . If we inscribe a circle inside a square with side=s then the ratio of the area of the circle to the rectangle is: So, if we throw darts at random at our rectangle then the probability (  ) of a dart landing inside the circle is just the ratio of the two areas,  /4. The we can determine  using: . The error in  is related to the error in  by: We can estimate how well we can measure  by this method by assuming that  = ( …)/4: This formula “says” that to improve our estimate of  by a factor of 10 we have to throw 100 (N) times as many darts! Clearly, this is an inefficient way to determine . we will derive this later in the course

10 Example: Suppose a baseball player's batting average is (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 = 100·0.30 = 30 hits u What's the standard deviation for the number of hits in 100 at bats?  = (Npq) 1/2 = (100·0.30·0.7) 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) = =76% Pete Rose’s lifetime batting average: 0.303

11 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 ∞ 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 ≈ Suppose the lifetime of this element  = years ≈ 5x10 19 seconds probability of a given nucleus to decay in one second is very small: p = 1/  = 2x /sec Np = 2/secfinite! 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  radioactive decay  number of Prussian soldiers kicked to death by horses per year per army corps!  quality control, failure rate predictions

12 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) What’s the probability of zero decays in one second? ii) What’s 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!

13 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!= ln10-10=13.03  14% ln50!= ln50-50=  1.9%

14 Comparison of Binomial and Poisson distributions with mean  = 1 Not much difference between them! For N large and  fixed: Binomial  Poisson N N

15 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(x 1 )dx 1 = p(x 2 )dx 2 if dx 1 =dx 2  equal intervals give equal probabilities For a uniform distribution with a=0, b=1 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: iseed=12345 do I=1,5 y=ran(iseed) type *, y enddo end If we generate “a lot” of random numbers all equal intervals should contain the same amount of numbers. For example: generate: 10 6 random numbers expect: 10 5 numbers [0.0, 0.1] 10 5 numbers [0.45, 0.55]

16 Selamat Belajar Semoga Sukses.