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This lecture is from Advanced Project Management. Key important points are: Discrete Distributions, Continuous Random Variables, Variance of Discrete Distribution, Type of Statistical Experiments, Binomial Distribution, Poisson Distribution, Standard Deviation

Typology: Slides

2012/2013

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Download Discrete Distributions - Advanced Project Management - Lecture Slides and more Slides Project Management in PDF only on Docsity! Discrete Distributions 1 Learning Objectives • Distinguish between discrete random variables and continuous random variables. • Know how to determine the mean and variance of a discrete distribution. • Identify the type of statistical experiments that can be described by the binomial distribution, and know how to work such problems. 2 5 The numerical values are determined by some characteristic of the elementary outcome, and typically it will vary from outcome to outcome. The word random serves to emphasize the fact that before the experiment is performed, we do not know the specific outcome and, consequently, its associated value of X. The following examples illustrate the concept of a random variable. Outcome Value of X HHH 3 HHT 2 HTH 2 HTT 1 THH 2 THT 1 TTH 1 TTT 0 6 7 For each elementary outcome, there is only one value for X. However, several elementary outcomes may yield the same value. Scanning our list, we now identify the events (the collections of elementary outcomes) that correspond to distinct values of X. Discrete vs Continuous Distributions • Discrete Random Variable -- the set of all possible values is at most a finite or a countably infinite number of possible values – Number of new subscribers to a magazine – Number of bad checks received by a restaurant – Number of absent employees on a given day • Continuous Random Variable -- takes on values at every point over a given interval – Current Ratio of a motorcycle distributorship – Elapsed time between arrivals of bank customers 10 Discrete Probability Distributions • A discrete random variable is a variable that can assume only a countable number of values Many possible outcomes: – number of complaints per day – number of TV’s in a household – number of rings before the phone is answered Only two possible outcomes: – gender: male or female – defective: yes or no – spreads peanut butter first vs. spreads jelly first 11 Discrete Distribution -- Example 12 0 1 2 3 4 5 0.37 0.31 0.18 0.09 0.04 0.01 Number of Crises Probability Distribution of Daily Crises 0 0.1 0.2 0.3 0.4 0.5 0 1 2 3 4 5 P r o b a b i l i t y Number of Crises Variance and Standard Deviation of a Discrete Distribution 15 2.1)(22 XPX 2 12 110. . X -1 0 1 2 3 P(X) .1 .2 .4 .2 .1 -2 -1 0 1 2 X 4 1 0 1 4 .4 .2 .0 .2 .4 1.2 )( 2 X 2 ( ) ( )X P X Mean Example 16 15.1)( XPXXE X P(X) X P(X) 0 .37 .00 1 .31 .31 2 .18 .36 3 .09 .27 4 .04 .16 5 .01 .05 1.15 0 0.1 0.2 0.3 0.4 0.5 0 1 2 3 4 5 P r o b a b i l i t y Number of Crises 17 Variance and Standard Deviation Example 2 2 141 X P X( ) . 19.141.1 2 X P(X) (X- ) (X- ) 2 (X- ) 2 P(X) 0 .37 -1.15 1.32 .49 1 .31 -0.15 0.02 .01 2 .18 0.85 0.72 .13 3 .09 1.85 3.42 .31 4 .04 2.85 8.12 .32 5 .01 3.85 14.82 .15 1.41 Binomial Distribution • Probability function • Mean value • Variance and standard deviation 20 n. 1,2,,0Xfor !! ! )( qp XnX XnX n XP n p 2 2 n p q n p q Binomial Distribution: Development • Experiment: randomly select, with replacement, two families from the residents of Tiny Town • Success is ‘Children in Household:’ p = 0.75 • Failure is ‘No Children in Household:’ q = 1- p = 0.25 • X is the number of families in the sample with ‘Children in Household’ 21 Family Children in Household Number of Automobiles A B C D Yes Yes No Yes 3 2 1 2 Listing of Sample Space (A,B), (A,C), (A,D), (A,A), (B,A), (B,B), (B,C), (B,D), (C,A), (C,B), (C,C), (C,D), (D,A), (D,B), (D,C), (D,D) Binomial Distribution: Development Continued • Families A, B, and D have children in the household; family C does not • Success is ‘Children in Household:’ p = 0.75 • Failure is ‘No Children in Household:’ q = 1- p = 0.25 • X is the number of families in the sample with ‘Children in Household’ 22 (A,B), (A,C), (A,D), (A,A) (B,A), (B,B), (B,C), (B,D), (C,A), (C,B), (C,C), (C,D), (D,A), (D,B), (D,C), (D,D) Listing of Sample Space 2 1 2 2 2 2 1 2 1 1 0 1 2 2 1 2 X 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 P(outcome) Binomial Distribution: Development Continued 25 X Possible Sequences 0 1 1 2 (F,F) (S,F) (F,S) (S,S) P(sequence) (. ) (. ) ( . ) 25 25 2 25 (. ) (. ) 25 75 (. ) (. ) 75 25 (. ) (. ) ( . ) 75 75 2 75 X 0 1 2 P(X) (. ) (. ) 25 75 2 =0.375 (. ) (. ) ( . ) 75 75 2 75 =0.5625 (. ) (. ) ( . ) 25 25 2 25 =0.0625 P X n X n X x n x p q( ) ! ! ! P X( ) ! ! .. . 0 2 0! 2 0 0 0625 0 2 0 75 25 P X( ) ! ! ! .. . 1 2 1 2 1 0 375 1 2 1 75 25 P X( ) ! ! ! .. . 2 2 2 2 2 05625 2 2 2 75 25 i l ( , ) ( , ) ( , ) ( , ) ( ) (. ) (. ) . 2 (. )(. ) (. )(. ) (. ) (. ) . 2 ( ) (. )(. ) . (. ) (. ) . 2 . (. ) (. ) . 2 . x n x . 0 2 0 . . 2 2 2 Binomial Distribution Example: The Gallup survey discussed in the Decision Dilemma found that 65% of all financial consumers were very satisfied with their primary financial institution. If this figure still holds true today, suppose 40 financial consumers are sampled randomly. What is the probability that exactly 23 of the 40 are very satisfied with their primary financial institution? 26 Solution: 27 The value of p is .65 (very satisfied), the value of q = 1 – p =1 - 0.65 = 0.35 (not very satisfied), n=40, and X=23. the binomial formula yields the final answer. 40C23(.65)23(.35)17=(88732378800)(.000049775) (.000000018)=(.0784) P X n X n X x n x p q( ) ! ! ! If 65% of the financial consumers are very satisfied, about 7.84% of the time the researcher would get exactly 23 out of 40 financial consumers who are very satisfied with their financial institution. Solution: 30 n p q P X P X P X P X 20 06 94 2 0 1 2 2901 3703 2246 8850 . . ( ) ( ) ( ) ( ) . . . . P X( ) )! ( )( )(. ) .. . 0 20! 0!(20 0 1 1 2901 2901 0 20 0 06 94 P X( ) !( )! ( )(. )(. ) .. . 1 20! 1 20 1 20 06 3086 3703 1 20 1 06 94 P X( ) !( )! ( )(. )(. ) .. . 2 20! 2 20 2 190 0036 3283 2246 2 20 2 06 94 31 If 6% of workers in Jackson, Mississippi, are unemployed, the telephone surveyor would get zero, one, or two unemployed workers 88.5% of the time in a random sample of 20 workers. The requirement of getting two or fewer is satisfied by getting zero, one, or two unemployed workers. Thus this problem is the union of three probabilities. Whenever the binomial formula is used to solve for cumulative success (not an exact number), the probability of each x value must be solved and the probabilities summed. If an actual survey produced such a result, it would serve to validate the census figures. Binomial Table 32 n = 20 PROBABILITY X 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0 0.122 0.012 0.001 0.000 0.000 0.000 0.000 0.000 0.000 1 0.270 0.058 0.007 0.000 0.000 0.000 0.000 0.000 0.000 2 0.285 0.137 0.028 0.003 0.000 0.000 0.000 0.000 0.000 3 0.190 0.205 0.072 0.012 0.001 0.000 0.000 0.000 0.000 4 0.090 0.218 0.130 0.035 0.005 0.000 0.000 0.000 0.000 5 0.032 0.175 0.179 0.075 0.015 0.001 0.000 0.000 0.000 6 0.009 0.109 0.192 0.124 0.037 0.005 0.000 0.000 0.000 7 0.002 0.055 0.164 0.166 0.074 0.015 0.001 0.000 0.000 8 0.000 0.022 0.114 0.180 0.120 0.035 0.004 0.000 0.000 9 0.000 0.007 0.065 0.160 0.160 0.071 0.012 0.000 0.000 10 0.000 0.002 0.031 0.117 0.176 0.117 0.031 0.002 0.000 11 0.000 0.000 0.012 0.071 0.160 0.160 0.065 0.007 0.000 12 0.000 0.000 0.004 0.035 0.120 0.180 0.114 0.022 0.000 13 0.000 0.000 0.001 0.015 0.074 0.166 0.164 0.055 0.002 14 0.000 0.000 0.000 0.005 0.037 0.124 0.192 0.109 0.009 15 0.000 0.000 0.000 0.001 0.015 0.075 0.179 0.175 0.032 16 0.000 0.000 0.000 0.000 0.005 0.035 0.130 0.218 0.090 17 0.000 0.000 0.000 0.000 0.001 0.012 0.072 0.205 0.190 18 0.000 0.000 0.000 0.000 0.000 0.003 0.028 0.137 0.285 19 0.000 0.000 0.000 0.000 0.000 0.000 0.007 0.058 0.270 20 0.000 0.000 0.000 0.000 0.000 0.000 0.001 0.012 0.122 Binomial Distribution using Table: U.S. Census Bureau Problem 35 n p q P X P X P X P X 20 06 94 2 0 1 2 2901 3703 2246 8850 . . ( ) ( ) ( ) ( ) . . . . P X P X( ) ( ) . . 2 1 2 1 8850 1150 n p ( )(. ) .20 06 1 20 2 2 20 06 94 1 128 1 128 1 062 n p q ( )(. )(. ) . . . n = 20 PROBABILITY X 0.05 0.06 0.07 0 0.3585 0.2901 0.2342 1 0.3774 0.3703 0.3526 2 0.1887 0.2246 0.2521 3 0.0596 0.0860 0.1139 4 0.0133 0.0233 0.0364 5 0.0022 0.0048 0.0088 6 0.0003 0.0008 0.0017 7 0.0000 0.0001 0.0002 8 0.0000 0.0000 0.0000 … … … 20 0.0000 0.0000 0.0000 … Excel’s Binomial Function n = 20 p = 0.06 X P(X) 0 =BINOMDIST(A5,B$1,B$2,FALSE) 1 =BINOMDIST(A6,B$1,B$2,FALSE) 2 =BINOMDIST(A7,B$1,B$2,FALSE) 3 =BINOMDIST(A8,B$1,B$2,FALSE) 4 =BINOMDIST(A9,B$1,B$2,FALSE) 5 =BINOMDIST(A10,B$1,B$2,FALSE) 6 =BINOMDIST(A11,B$1,B$2,FALSE) 7 =BINOMDIST(A12,B$1,B$2,FALSE) 8 =BINOMDIST(A13,B$1,B$2,FALSE) 9 =BINOMDIST(A14,B$1,B$2,FALSE) 36 Graphs of Selected Binomial Distributions 37 n = 4 PROBABILITY X 0.1 0.5 0.9 0 0.656 0.063 0.000 1 0.292 0.250 0.004 2 0.049 0.375 0.049 3 0.004 0.250 0.292 4 0.000 0.063 0.656 P = 0.1 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000 0 1 2 3 4 X P (X ) P = 0.5 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000 0 1 2 3 4 X P (X ) P = 0.9 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000 0 1 2 3 4 X P (X ) Solution: 40 a) n = 20 p = .78 x = 14 20C14 (.78)14(.22) 6 = 38,760(.030855)(.00011338) = .1356 b) n = 20 p = .75 x = 20 20C20 (.75) 20(.25)0 = (1)(.0031712)(1) = .0032 c) n = 20 p = .70 x < 12 Use table A.2: P(x=0) + P(x=1) + . . . + P(x=11) = 0.000 + 0.000 + 0.000 + 0.000 + 0.000 + 0.000 + 0.000 + 0.001 + 0.004 + 0.012 +0 .031 + 0.065 = .113 Example: 41 The Wall Street Journal reported some interesting statistics on the job market. One statistic is that 40% of all workers say they would change jobs for “slightly higher pay”. In addition, 88% of companies say that there is a shortage of qualified job candidates. Suppose 16 workers are randomly selected and asked if they would change jobs for “slightly higher pay”. What is the probability that nine or more say yes? What is the probability that three, four, five, or six say yes? If 13 companies are contacted, what is the probability that exactly 10 say there is a shortage of qualified job candidates? What is the probability that all of the companies say there is a shortage of qualified job candidates? What is the expected number of companies that would say there is a shortage of qualified job candidates? 42 n = 16 p = .40 P(x > 9): from Table A.2: x Prob 9 .084 10 .039 11 .014 12 .004 13 .001 .142 Question 5.23 45 Diane Burns is the mayor of a large city. Lately, she has become concerned about the possibility that large numbers of people who are drawing unemployment checks are secretly employed. Her assistants estimate that 40 percent of unemployment beneficiaries fall into this category, but Ms. Bruns is not convinced. She asks one of her aides to conduct a quite investigation of 10 randomly selected unemployment beneficiaries. (a) If the mayer’s assistants are correct, what is the probability that more than eight of individuals investigated have jobs? (b) If the mayor’s assistants are correct, what is the probability that only three of the investigated have jobs? Poisson Distribution • Describes discrete occurrences over a continuum or interval • A discrete distribution • Describes rare events • Each occurrence is independent any other occurrences. • The number of occurrences in each interval can vary from zero to infinity. • The expected number of occurrences must hold constant throughout the experiment. 46 Poisson Distribution: Applications • Arrivals at queuing systems – airports -- people, airplanes, automobiles, baggage – banks -- people, automobiles, loan applications – computer file servers -- read and write operations • Defects in manufactured goods – number of defects per 1,000 feet of extruded copper wire – number of blemishes per square foot of painted surface – number of errors per typed page 47 Solution: 50 3 2 6 4 10 10 0 0528 6 4 . ! ! . . customers / 4 minutes X = 10 customers / 8 minutes Adjusted = . customers / 8 minutes P(X) = ( = ) = X 10 6.4 e e X P X 3 2 6 4 6 6 0 1586 6 4 . ! ! . . customers / 4 minutes X = 6 customers / 8 minutes Adjusted = . customers / 8 minutes P(X) = ( = ) = X 6 6.4 e e X P X Poisson Distribution: Probability Table 51 X 0.5 1.5 1.6 3.0 3.2 6.4 6.5 7.0 8.0 0 0.6065 0.2231 0.2019 0.0498 0.0408 0.0017 0.0015 0.0009 0.0003 1 0.3033 0.3347 0.3230 0.1494 0.1304 0.0106 0.0098 0.0064 0.0027 2 0.0758 0.2510 0.2584 0.2240 0.2087 0.0340 0.0318 0.0223 0.0107 3 0.0126 0.1255 0.1378 0.2240 0.2226 0.0726 0.0688 0.0521 0.0286 4 0.0016 0.0471 0.0551 0.1680 0.1781 0.1162 0.1118 0.0912 0.0573 5 0.0002 0.0141 0.0176 0.1008 0.1140 0.1487 0.1454 0.1277 0.0916 6 0.0000 0.0035 0.0047 0.0504 0.0608 0.1586 0.1575 0.1490 0.1221 7 0.0000 0.0008 0.0011 0.0216 0.0278 0.1450 0.1462 0.1490 0.1396 8 0.0000 0.0001 0.0002 0.0081 0.0111 0.1160 0.1188 0.1304 0.1396 9 0.0000 0.0000 0.0000 0.0027 0.0040 0.0825 0.0858 0.1014 0.1241 10 0.0000 0.0000 0.0000 0.0008 0.0013 0.0528 0.0558 0.0710 0.0993 11 0.0000 0.0000 0.0000 0.0002 0.0004 0.0307 0.0330 0.0452 0.0722 12 0.0000 0.0000 0.0000 0.0001 0.0001 0.0164 0.0179 0.0263 0.0481 13 0.0000 0.0000 0.0000 0.0000 0.0000 0.0081 0.0089 0.0142 0.0296 14 0.0000 0.0000 0.0000 0.0000 0.0000 0.0037 0.0041 0.0071 0.0169 15 0.0000 0.0000 0.0000 0.0000 0.0000 0.0016 0.0018 0.0033 0.0090 16 0.0000 0.0000 0.0000 0.0000 0.0000 0.0006 0.0007 0.0014 0.0045 17 0.0000 0.0000 0.0000 0.0000 0.0000 0.0002 0.0003 0.0006 0.0021 18 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0001 0.0002 0.0009 Poisson Distribution: Using the Poisson Tables 52 If a real estate office sells 1.6 houses on an average weekday and sales of houses on weekdays are Poisson distributed, what is the probability of selling exactly four houses in one day? What is the probability of selling no houses in one day? What is the probability of selling more than five houses in one day? What is the probability of selling 2 or more houses in one day? Poisson Distribution: Using the Poisson Tables 55 1 6 2 1 2 1 0 1 1 2019 3230 4751 . ( ) ( ) ( ) ( ) . . . P X P X P X P X X 0.5 1.5 1.6 3.0 0 0.6065 0.2231 0.2019 0.0498 1 0.3033 0.3347 0.3230 0.1494 2 0.0758 0.2510 0.2584 0.2240 3 0.0126 0.1255 0.1378 0.2240 4 0.0016 0.0471 0.0551 0.1680 5 0.0002 0.0141 0.0176 0.1008 6 0.0000 0.0035 0.0047 0.0504 7 0.0000 0.0008 0.0011 0.0216 8 0.0000 0.0001 0.0002 0.0081 9 0.0000 0.0000 0.0000 0.0027 10 0.0000 0.0000 0.0000 0.0008 11 0.0000 0.0000 0.0000 0.0002 12 0.0000 0.0000 0.0000 0.0001 Poisson Distribution: Graphs 56 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0 1 2 3 4 5 6 7 8 1 6. 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0 2 4 6 8 10 12 14 16 6 5. Problem: 57 According to the United National Environmental Program and World Health Organization, in Bombay, India, air pollution standards for particulate matter are exceeded an average of 5.6 days in every three- week period. Assume that the distribution of number of days exceeding the standards per three-week period is Poisson distributed. What is the probability that the standard is not exceeded on any day during a three-week period? What is the probability that the standard is exceeded exactly 6 days of a three-week period? What is the probability that the standard is exceeded exactly 15 or more days during a three-week period? If this outcome actually occurred, what might you conclude? Solution: n = 300, p = .01, = n(p) = 300(.01) = 3 a) Prob(x = 5): Using = 3 and Table A.3 = .1008 b) Prob (x < 4) = Prob.(x = 0) + Prob.(x = 1) + Prob.(x = 2) + Prob.(x = 3) = .0498 + .1494 + .2240 + .2240 = .6472 c) The expected number = µ = = 3 60 Problem: 61 The average number of annual trips per family to amusement parks in the United State is Poisson distributed, with a mean of 0.6 trips per year. What is the probability of randomly selecting an American family and finding the following: a) The family did not make a trip to an amusement park last year? b) The family took exactly one trip to an amusement park last year? c) The family took two or more trips to amusement parks last year? d) The family took three or fewer trips to amusement parks over a three-year period? e) The family took exactly four trips to amusement parks during a six-year period? Solution: 62 = 0.6 trips 1 year a) Prob (x=0 = 0.6): from Table A.3 = .5488 b) Prob (x=1 = 0.6): from Table A.3 = .3293 65 Prob(x=4 6 years): The interval has been increased (6 times) New Lambda = = 3.6 trips6 years Prob(x=4 = 3.6): from Table A.3 = .1912 Problem: 66 Ship collisions in the Houston Ship Channel are rare. Suppose the numbers of collisions are Poisson distributed, with the mean of 1.2 collisions every four months. What is probability of having no collisions occur over a four- month period? What is probability of having exactly two collisions in a two month period? What is probability of having one or fewer collisions in a six month period? If this outcome occurred, what might you conclude about ship channel conditions during this period? What might you conclude about ship channel safety awareness during this period? What might you conclude about weather conditions during this period? What might you conclude about lambda? Solution: 67 = 1.2 collisions4 months a) Prob(x=0 = 1.2): from Table A.3 = .3012 b) Prob(x=2 2 months): The interval has been decreased (by ½) New Lambda = = 0.6 collisions2 months Prob(x=2 = 0.6): from Table A.3 = .0988 70 = 1.2 penscarton a) Prob(x=0 = 1.2): from Table A.3 = .3012 b) Prob(x > 8 = 1.2): from Table A.3 = .0000 71 Prob(x > 3 = 1.2): from Table A.3 x Prob. 4 .0260 5 .0062 6 .0012 7 .0002 8 .0000 x > 3 .0336 Question: 72 A high percentage of people who fracture or dislocate a bone see a doctor for that condition. Suppose the percentage is 99%. Consider a sample in which 300 people are randomly selected who have fractured or dislocated a bone. What is the probability that exactly five of them did not see a doctor? What is the probability that fewer than four of them did not see a doctor? What is the expected number of people who would not see a doctor? Poisson Approximation of the Binomial Distribution • Binomial probabilities are difficult to calculate when n is large. • Under certain conditions binomial probabilities may be approximated by Poisson probabilities. • Poisson approximation 75 If and the approximation is acceptable.n n p 20 7, Use n p. Poisson Approximation of the Binomial Distribution 76 X Error 0 0.2231 0.2181 -0.0051 1 0.3347 0.3372 0.0025 2 0.2510 0.2555 0.0045 3 0.1255 0.1264 0.0009 4 0.0471 0.0459 -0.0011 5 0.0141 0.0131 -0.0010 6 0.0035 0.0030 -0.0005 7 0.0008 0.0006 -0.0002 8 0.0001 0.0001 0.0000 9 0.0000 0.0000 0.0000 Poisson 1 5. Binomial n p 50 03. X Error 0 0.0498 0.0498 0.0000 1 0.1494 0.1493 0.0000 2 0.2240 0.2241 0.0000 3 0.2240 0.2241 0.0000 4 0.1680 0.1681 0.0000 5 0.1008 0.1008 0.0000 6 0.0504 0.0504 0.0000 7 0.0216 0.0216 0.0000 8 0.0081 0.0081 0.0000 9 0.0027 0.0027 0.0000 10 0.0008 0.0008 0.0000 11 0.0002 0.0002 0.0000 12 0.0001 0.0001 0.0000 13 0.0000 0.0000 0.0000 Poisson 3 0. Binomial n p 10 000 0003 , . Question 5-31 77 Concert pianist Donna Prima has become quite upset at the number of coughs occurring in the audience just before she begins to play. On her latest tour, Donna estimates that on average eight coughs occur just before the start of her performance. Ms. Prima has sworn to her conductor that if she hears more than five coughs at tonight’s performance, she will refuse to play. What is the Probability that she will play tonight?