Discrete  Probability Distributions
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Discrete Probability Distributions

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Discrete  Probability Distributions Discrete Probability Distributions Presentation Transcript

  • Business Statistics, 6th ed. by Ken Black Chapter 5 Discrete Probability Distributions Copyright 2010 John Wiley & Sons, Inc. 1Copyright 2010 John Wiley & Sons, Inc.
  • 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.Copyright 2010 John Wiley & Sons, Inc. 2
  • Discrete vs. Continuous Distributions Discrete distributions – constructed from discrete (individually distinct) random variables Continuous distributions – based on continuous random variables Random Variable - a variable which contains the outcomes of a chance experimentCopyright 2010 John Wiley & Sons, Inc. 3
  • Discrete vs. Continuous Distributions Categories of Random Variables Discrete Random Variable - the set of all possible values is at most a finite or a countable infinite number of possible values Continuous Random Variable - takes on values at every point over a given intervalCopyright 2010 John Wiley & Sons, Inc. 4
  • Describing a Discrete Distribution A discrete distribution can be described by constructing a graph of the distribution Measures of central tendency and variability can be applied to discrete distributions Discrete values of outcomes are used to represent themselvesCopyright 2010 John Wiley & Sons, Inc. 5
  • Describing a Discrete Distribution Mean of discrete distribution – is the long run average If the process is repeated long enough, the average of the outcomes will approach the long run average (mean) Requires the process to eventually have a number which is the product of many processes Mean of a discrete distribution µ = ∑ (X * P(X)) where (X) is the long run average; X = outcome, P = Probability of XCopyright 2010 John Wiley & Sons, Inc. 6
  • Describing a Discrete Distribution Variance and Standard Deviation of a discrete distribution are solved by using the outcomes (X) and probabilities of outcomes (P(X)) in a manner similar to computing a mean Standard Deviation is computed by taking the square root of the varianceCopyright 2010 John Wiley & Sons, Inc. 7
  • Some Special Distributions Discrete binomial Poisson Hypergeometric Continuous normal uniform exponential t chi-square FCopyright 2010 John Wiley & Sons, Inc. 8
  • Discrete Distribution -- Example Observe the discrete distribution in the following table. An executive is considering out-of-town business travel for a given Friday. At least one crisis could occur on the day that the executive is gone. The distribution contains the number of crises that could occur during the day the executive is gone and the probability that each number will occur. For example, there is a .37 probability that no crisis will occur, a .31 probability of one crisis, and so on.Copyright 2010 John Wiley & Sons, Inc. 9
  • Discrete Distribution -- Example Distribution of Daily Crises P 0.5 Number of r Probability o 0.4 Crises b a 0.3 0 0.37 b 1 0.31 i 0.2 2 0.18 l i 0.1 3 0.09 t 4 0.04 y 0 5 0.01 0 1 2 3 4 5 Number of CrisesCopyright 2010 John Wiley & Sons, Inc. 10
  • Variance and Standard Deviation of a Discrete Distribution 2 2 2 X P( X ) 1.2 12 110 . . 2 X P(X) 2 X (X ) (X ) P( X ) -1 .1 -2 4 .4 0 .2 -1 1 .2 1 .4 0 0 .0 2 .2 1 1 .2 3 .1 2 4 .4 1.2Copyright 2010 John Wiley & Sons, Inc. 11
  • Requirements for a Discrete Probability Function -- Examples X P(X) X P(X) X P(X) -1 .1 -1 -.1 -1 .1 0 .2 0 .3 0 .3 1 .4 1 .4 1 .4 2 .2 2 .3 2 .3 3 .1 3 .1 3 .1 1.0 1.0 1.2 : YES NO NOCopyright 2010 John Wiley & Sons, Inc. 12
  • Mean of a Discrete Distribution E X X P( X ) X P(X) X P( X) -1 .1 -.1 0 .2 .0 1 .4 .4 2 .2 .4 3 .1 .3 1.0 = 1.0Copyright 2010 John Wiley & Sons, Inc. 13
  • Mean of the Crises Data Example E X X P( X ) 115 . X P(X) X P(X) P r 0.5 0 .37 .00 o 0.4 b 1 .31 .31 a 0.3 b 2 .18 .36 0.2 i l 0.1 3 .09 .27 i t 0 4 .04 .16 0 1 2 3 4 5 y Number of Crises 5 .01 .05 1.15Copyright 2010 John Wiley & Sons, Inc. 14
  • Variance and Standard Deviation of Crises Data Example 2 2 2 X P( X ) 141 . 141 119 . . X P(X) (X- ) (X- (X- 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 1.41Copyright 2010 John Wiley & Sons, Inc. 15
  • Binomial Distribution Probability n! X n X function P( X ) X! n X ! p q for 0 X n Mean value n p Variance and 2 n p q Standard 2 Deviation n p qCopyright 2010 John Wiley & Sons, Inc. 16
  • Binomial Distribution: Demonstration Problem 5.3 According to the U.S. Census Bureau, approximately 6% of all workers in Jackson, Mississippi, are unemployed. In conducting a random telephone survey in Jackson, what is the probability of getting two or fewer unemployed workers in a sample of 20?Copyright 2010 John Wiley & Sons, Inc. 17
  • Binomial Distribution: Demonstration Problem 5.3 In the following example, 6% are unemployed => p The sample size is 20 => n 94% are employed => q x is the number of successes desired What is the probability of getting 2 or fewer unemployed workers in the sample of 20? The hard part of this problem is identifying p, n, and x – emphasis this when studying the problems.Copyright 2010 John Wiley & Sons, Inc. 18
  • Binomial Distribution: Demonstration Problem 5.3 n 20 p . 06 q . 94 P( X 2 ) P( X 0) P( X 1) P( X 2) . 2901 . 3703 . 2246 . 8850 20! 0 20 0 P( X 0) 0!(20 0)! .06 .94 (1)(1)(. 2901 ) .2901 20! 1 20 1 P( X 1) 1!(20 1)! .06 .94 (20)(.06)(.3086) .3703 20! 2 20 2 P( X 2) 2!(20 2)! .06 .94 (190)(.0036)(.3283) .2246Copyright 2010 John Wiley & Sons, Inc. 19
  • Binomial Distribution Table: Demonstration Problem 5.3 n = 20 PROBABILITY n 20 X 0.05 0.06 0.07 p . 06 0 0.3585 0.2901 0.2342 q . 94 1 0.3774 0.3703 0.3526 P( X 2 ) P( X 0) P( X 1) P( X 2) 2 0.1887 0.2246 0.2521 . 2901 . 3703 . 2246 . 8850 3 0.0596 0.0860 0.1139 4 0.0133 0.0233 0.0364 P( X 2) 1 P( X 2) 1 . 8850 .1150 5 0.0022 0.0048 0.0088 6 0.0003 0.0008 0.0017 n p (20)(. 06) 1. 20 2 7 0.0000 0.0001 0.0002 n p q ( 20 )(. 06 )(. 94 ) 1.128 8 0.0000 0.0000 0.0000 2 … … … … 1.128 1. 062 20 0.0000 0.0000 0.0000Copyright 2010 John Wiley & Sons, Inc. 20
  • 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)Copyright 2010 John Wiley & Sons, Inc. 21
  • Minitab’s Binomial Function X P(X =x) 0 0.000000 1 0.000000 2 0.000000 3 0.000001 4 0.000006 5 0.000037 6 0.000199 7 0.000858 Binomial with n = 23 and p = 0.64 8 0.003051 9 0.009040 10 0.022500 11 0.047273 12 0.084041 13 0.126420 14 0.160533 15 0.171236 16 0.152209 17 0.111421 18 0.066027 19 0.030890 20 0.010983 21 0.002789 22 0.000451 23 0.000035Copyright 2010 John Wiley & Sons, Inc. 22
  • Mean and Std Dev of Binomial Distribution Binomial distribution has an expected value or a long run average denoted by µ (mu) If n items are sampled over and over for a long time and if p is the probability of success in one trial, the average long run of successes per sample is expected to be np => Mean µ = np => Std Dev = √(npq)Copyright 2010 John Wiley & Sons, Inc. 23
  • Poisson Distribution The Poisson distribution focuses only on the number of discrete occurrences over some interval or continuum Poisson does not have a given number of trials (n) as a binomial experiment does Occurrences are independent of other occurrences Occurrences occur over an intervalCopyright 2010 John Wiley & Sons, Inc. 24
  • Poisson Distribution If Poisson distribution is studied over a long period of time, a long run average can be determined The average is denoted by lambda (λ) Each Poisson problem contains a lambda value from which the probabilities are determined A Poisson distribution can be described by λ aloneCopyright 2010 John Wiley & Sons, Inc. 25
  • Poisson Distribution Probability function X P( X ) e for X 0,1,2,3,... X! where : long run average e 2.718282... (the base of natural logarithms)  Mean value  Variance  Standard deviationCopyright 2010 John Wiley & Sons, Inc. 26
  • Poisson Distribution: Demonstration Problem 5.7 Bank customers arrive randomly on weekday afternoons at an average of 3.2 customers every 4 minutes. What is the probability of having more than 7 customers in a 4-minute interval on a weekday afternoon?Copyright 2010 John Wiley & Sons, Inc. 27
  • Poisson Distribution: Demonstration Problem 5.7 Solution λ = 3.2 customers>minutes X > 7 customers/4 minutes The solution requires obtaining the values of x = 8, 9, 10, 11, 12, 13, 14, . . . . Each x value is determined until the values are so far away from λ = 3.2 that the probabilities approach zero. The exact probabilities are summed to find x 7. If the bank has been averaging 3.2 customers every 4 minutes on weekday afternoons, it is unlikely that more than 7 people would randomly arrive in any one 4-minute period. This answer indicates that more than 7 people would randomly arrive in a 4-minute period only 1.69% of the time. Bank officers could use these results to help them make staffing decisions.Copyright 2010 John Wiley & Sons, Inc. 28
  • Poisson Distribution: Demonstration Problem 5.7 3.2 customers/ minutes 4 3. 2 customers/ 4 minutes X = 10 customers/ minutes 8 X = 6 customers/ 8 minutes Adjusted Adjusted = 6.4 customers/ minutes 8 = 6. 4 customers/ 8 minutes X X P(X) = e P(X) = e X! X! P( X = 6) = 6.4 e 10 6.4 6 6. 4 P( X = 10) = 6.4 e 0.0528 0.1586 10! 6!Copyright 2010 John Wiley & Sons, Inc. 29
  • Poisson Distribution: Using the Poisson Tables 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 1.6 P( X 5) P ( X 6) P ( X 7) P ( X 8) P ( X 9) .0047 .0011 .0002 .0000 .0060Copyright 2010 John Wiley & Sons, Inc. 30
  • Poisson Distribution: Using the Poisson Tables 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 1.6 P( X 2) 1 P ( X 2) 1 P( X 0) P ( X 1) 1 .2019 .3230 .4751Copyright 2010 John Wiley & Sons, Inc. 31
  • Excel’s Poisson Function = 1.6 X P(X) 0 =POISSON(D5,E$1,FALSE) 1 =POISSON(D6,E$1,FALSE) 2 =POISSON(D7,E$1,FALSE) 3 =POISSON(D8,E$1,FALSE) 4 =POISSON(D9,E$1,FALSE) 5 =POISSON(D10,E$1,FALSE) 6 =POISSON(D11,E$1,FALSE) 7 =POISSON(D12,E$1,FALSE) 8 =POISSON(D13,E$1,FALSE) 9 =POISSON(D14,E$1,FALSE)Copyright 2010 John Wiley & Sons, Inc. 32
  • Minitab’s Poisson Function X P(X =x) 0 0.149569 1 0.284180 2 0.269971 Poisson with mean = 1.9 3 0.170982 4 0.081216 5 0.030862 6 0.009773 7 0.002653 8 0.000630 9 0.000133 10 0.000025Copyright 2010 John Wiley & Sons, Inc. 33
  • Mean and Std Dev of a Poisson Distribution Mean of a Poisson Distribution is λ Understanding the mean of a Poisson distribution gives a feel for the actual occurrences that are likely to happen Variance of a Poisson distribution is also λ Std Dev = Square root of λCopyright 2010 John Wiley & Sons, Inc. 34
  • Poisson Approximation of the Binomial Distribution Binomial problems with large sample sizes and small values of p, which then generate rare events, are potential candidates for use of the Poisson Distribution Rule of thumb, if n > 20 and np < 7, the approximation is close enough to use the Poisson distribution for binomial problemsCopyright 2010 John Wiley & Sons, Inc. 35
  • Poisson Approximation of the Binomial Distribution Procedure for Approximating binomial with Poisson Begin with the computation of the binomial mean distribution µ = np Because µ is the expected value of the binomial, it becomes λ for Poisson distribution Use µ as the λ, and using the x from the binomial problem allows for the approximation of the probabilities from the Poisson table or Poisson formulaCopyright 2010 John Wiley & Sons, Inc. 36
  • 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. If n 20 and n p 7, the approximation is acceptable . Poisson approximation Use n p.Copyright 2010 John Wiley & Sons, Inc. 37
  • Hypergeometric Distribution Sampling without replacement from a finite population The number of objects in the population is denoted N. Each trial has exactly two possible outcomes, success and failure. Trials are not independent X is the number of successes in the n trials The binomial is an acceptable approximation, if n < 5% N. Otherwise it is not.Copyright 2010 John Wiley & Sons, Inc. 38
  • Hypergeometric Distribution Probability function ACx N A Cn x N is population size P( x ) n is sample size N Cn A is number of successes in population x is number of successes in sample Mean A n Value N Variance and standard 2 A( N A)n( N n) deviation N 2 (N 1) 2Copyright 2010 John Wiley & Sons, Inc. 39
  • Hypergeometric Distribution: Probability Computations N = 24 ACx N A Cn x X=8 P( x 3) N Cn n=5 8 C3 24 8 C5 3 P(x) x C5 24 0 0.1028 1 0.3426 56 120 2 0.3689 42,504 3 0.1581 4 0.0264 .1581 5 0.0013Copyright 2010 John Wiley & Sons, Inc. 40
  • Excel’s Hypergeometric Function N = 24 A= 8 n= 5 X P(X) 0 =HYPGEOMDIST(A6,B$3,B$2,B$1) 1 =HYPGEOMDIST(A7,B$3,B$2,B$1) 2 =HYPGEOMDIST(A8,B$3,B$2,B$1) 3 =HYPGEOMDIST(A9,B$3,B$2,B$1) 4 =HYPGEOMDIST(A10,B$3,B$2,B$1) 5 =HYPGEOMDIST(A11,B$3,B$2,B$1) =SUM(B6:B11)Copyright 2010 John Wiley & Sons, Inc. 41
  • Minitab’s Hypergeometric Function X P(X =x) 0 0.102767 1 0.342556 Hypergeometric with N = 24, A = 8, n = 5 2 0.368906 3 0.158103 4 0.026350 5 0.001318Copyright 2010 John Wiley & Sons, Inc. 42