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Lecture 6

Lecture 6. Negative Binomial, Geometric, Hypergeometric and Poisson Distributions. Negative Binomial (nb) Distribution

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Lecture 6

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  1. Lecture 6 Negative Binomial, Geometric, Hypergeometric and Poisson Distributions

  2. Negative Binomial (nb) Distribution • Consider Bernoulli experiments performed one after another. nb random variable occurs when the interest lies in the rth success (r1) occurring at the zth trial, z=r, r+1, r+2, r+3, … • The special case of the nb distribution with r=1 is called the Geometric distribution. (The 1st success occurs at the zth trial.) • P(occurrence) = p

  3. Example 3.37 (page 132) • A doctor needs to recruit 5 couples for a research. • P(a random couple agrees to participate) = 0.20 • P(15 couples are interviewed to recruit 5) = ? • z=15, x=10, r=5 • nb(15;5,0.20)= (z-1)C(r-1)(0.20)r(0.80)z-r =14C4(0.20)5(0.80)10=0.0343941 • P(at most 15 couples will be interviewed) = ?

  4. Examples Rocket launching: P(success)=0.95 (a) Compute P(3rd failure occurs at the 15th launch) Event of interest is launch failure.  P(failure) = 1  P(success) = 0.05 3rd failure occurs at the 15th launch; 2 failures will occur in 14 trials. The 15th trial will fail. nb(15;3,0.05) = 14C2(0.05)3(0.95)12= 0.0061466 (b) P(The 1st failure occurs at the 8th trial), if P(failure) = 0.05  Z=8, r=1, p=0.05  nb(8,1,0.05)=7C0(0.05)1(0.95)7 = 0.034917 Since r=1, this is geometric distribution with the following parameters: g(z;p), where z is the number of trials, and p is the P(event of interest) g(z;p) = (1  p)z-1p g(8;0.05) = (1-0.05)7(0.05) = (0.95)7(0.05) = 0.034917

  5. Example Rocket launching: P(success)=0.95 (a) Compute P(1st failure occurs within the first seven launches) P(failure) = 1  P(success) = 0.05 P(Z7) = g(1;p)+ g(2;p)+ g(3;p)+ g(4;p)+ g(5;p)+ g(6;p)+ g(7;p) = 0.05 + 0.0475 + 0.045125 + 0.042869 + 0.040725 + 0.038689 + 0.036755 = 0.301663 OR P(Z7) = 1  P(the first seven launches are successful) = 1  (0.95)7 = 0.301663 (b)Compute P(1st failure occurs after the 10th launch) P(Z>10) = P(the first 10 launches are successful) = (0.95)10 = 0.598737

  6. Example An aircraft has 3 computers, only one of which is needed to operate the aircraft. The other 2 computers are on standby redundancy that are activated one at a time in case the on-line computer fails. From past experience it has been determined that the probability of failure (for the on-line computer) during any one hour is roughly 0.0004. Let Z1, Z2 and Z3 denote the number of hours of operation before the failure of the 1st, 2nd, and the 3rd computer, respectively. Then the time to failure (TTF) of the system is: Z = Z1 + Z2 + Z3 hours. (a) Assuming that failures can only occur at the end of each hour and hours can be thought of independent Bernoulli trials, compute the probability of system failure during a 7 hour flight. (b) Compute the mean TTF (MTTF). (c) Compute the variance of TTF. (a) P(the third failure occurs  7th trial) = nb(3;3,0.0004)+ nb(4;3,0.0004)+ nb(5;3,0.0004)+ nb(6;3,0.0004)+ nb(7;3,0.0004) = 2.2373110-9 (b) E(Z) = r / p = 3 / 0.0004 = 7500 hrs (c) V(Z) = r(1-p) / p2 = 18,742,500 hours2 = (4329.261 hours)2

  7. The Hypergeometric Distribution If X is the number of S’s in a completely random sample of size n drawn (with no replacement) from a population with size N, consisting of M S’s and (N-M) F’s, then the probability distribution of X is called the hypergeometric distribution. for x an integer such that max(0, n-N+M)  x  min(n, M) i.e., if sample size n is smaller than M, then the largest possible X = n if M<n, then the largest possible X = M if number of failures (N-M)>n, then the smallest possible X = 0 if (N-M)<n, then the smallest possible value of X = n-(N-M) • The mean and variance of hypergeometric distribution:

  8. Example Lots of size N = 200 pump shafts are inspected before shipment. The outgoing average lot quality (AOQ) is p = 0.02 over many lots. The inspection plan calls for random samples of n = 25 units (w/o replacement) and accepting an outgoing lot if the number of defectives in the sample is 1 or less. (a) Compute acceptance pr of a lot, denoted by Pa. (b) Compute Pa if FNC increases to 5%. (c) Repeat part (a) if sampling is done with replacement. ANS: (c) Pa = 0.911355. For parts (a) and (b) it is sufficient to give answers in terms of nCx. The approximate binomial pr for part (b) is Pa = 0.642376. (a) P(defective)=0.02, M = 0.02·200 = 4, N=200, n=25 P(number of defectives  1) = h(0;25,4,200) + h(1;25,4,200) = (4)C(0)·(200-4)C(25-0) / (200)C(25) + (4)C(1)·(200-4)C(25-1) / (200)C(25) = 0.583643104 + 0.339327386 = 0.92297049 (b) P(defective)=0.05, M = 0.02·200 = 10, N=200, n=25 P(number of defectives  1) = h(0;25,10,200) + h(1;25,10,200) = (10)C(0)·(200-10)C(25-0) / (200)C(25) + (10)C(1)·(200-10)C(25-1) / (200)C(25) = 0.637696247

  9. The Poisson Distribution Consider extra many Bernoulli trials such that occurrence probability of success at each trial is small (p<0.15) and average number of successes per time unit,  = n·p =  is constant. Let the rv X denote the number of Poisson events that occur during one time interval of unit length. Then: • The mean and variance of Poisson distribution:

  10. The Poisson process: Important applications of the Poisson distribution arise when the events of interest occur over time. E.g. customer arrivals (10 per hr. etc.), traffic (cars per hr.), etc. Example: The number of accidents per day in a city is Poisson distributed with  = 5 accidents / day. (a) Compute the probability that no accidents occur in the city the next day. (b) Compute the probability that at most 4 accidents happen during the next day. (c) Compute the probability that an odd number of accidents happen during the next day. (a) p(0;5) = e(-5) ·(5)0 / 0! = 0.006737947 (b) p(0;5) + p(1;5) + p(2;5) + p(3;5) + p(4;5) = 0.440493285 (c) Hint: Sinh(x) = ½ (ex-e-x) = (x/1!)+(x3/3!)+(x5/5!)+… p(odd number of accidents) = p(1;5) + p(3;5) + p(5;5) + p(7;5) + p(9;5) + … = e(-5) ·(5)1 / 1! + e(-5) ·(5)3 / 3! + e(-5) ·(5)5 / 5! + e(-5) ·(5)7 / 7! + … = e(-5) [(5)1 / 1! +(5)3 / 3! + (5)5 / 5! + (5)7 / 7! + …] = e(-5) · Sinh(5) = 0.4999773 OR Using excel :

  11. The Poisson distribution can be used to compute probabilities over intervals of length other than 1 unit of time. Let Y=number of Poisson events occurring during an interval of length t, where the average number of events occurring during 1 unit of time is . Then the average number of Poisson events occurring per interval t is E(Y)= t. As a result, the pmf for the rv Y is given by:

  12. Example For the car accident example; (a) Compute the probability that there will be exactly 11 accidents in the city during the next 3 days. (b) Compute probability that there will be at least 11 accidents in the next 3 days. (c) Compute the probability that there will be at most 18 accidents in the next 4 days. (a) =5 accidents / day p(11 accidents in 3 days) = p(11;5·3) = p(11;15) = (15)11· e(-15) / 11! = 0.066287387 (b) p(at least 11 accidents in 3 days) = 1  [p(0;15)+p(1;15)+p(2;15)+p(3;15)+p(4;15)…+p(10;15)] = 1  0.118464412 = 0.881535588 (c) p(at most 18 accidents in 4 days)  t = 20 p(Y18) = p(0;20) + p(1;20) + p(2;20) + p(3;20) +….+ p(18;20) p(Y18) = 0.381421949

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