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Unformatted text preview: Chapter 6 Discrete Probability Distributions 6.1 Random Variables Random Variable A random variable is a function that assigns a numerical value to each random outcome of an experiment. outcome Example 6.1.1: We flip a coin two times and are interested in the Example number of heads. number The outcomes are HH, HT, TH and TT. Let x be the “total number of heads observed”. Then x = 0, 1, or 2. Let Therefore, the above outcomes can be assigned these numbers as follows. Therefore, Outcomes TT Value of x (i.e., total number of heads) 0 1 1 2 2 TH HT HH The variable “total number of heads observed” in this experiment is called a The random variable. random 6.1 Random Variables (cont.) Discrete Random Variable A random variable that can assume a finite number of possible values is called a discrete random variable. That is, the possible values of a discrete random variable can be listed or counted. Example 6.1.2: X is a discrete random variable if X represents the number of he people out of 200 who will make an airline reservation and then fail to show up. people Example 6.1.3: You roll two dice, a red die and a blue die. Suppose X is the total of the two dice. X is a discrete random variable because X has finite number of possible values.
Outcome
3 Red Die 1 1 1 . . 6 Blue Die 1 2 3 . . 6 Value of X 2 3 4 . . 12 1 2 3 . . 36 6.1 Random Variables (cont.) Continuous Random Variable A random variable that can take any value over some continuous range of values is called a continuous random variable. Example 6.1.4: Let Y be the amount of rainfall during the month of September. Y is a continuous random variable because the amount of rainfall can be any nonnegative value. 4 6.1 Random Variables (cont.) Probability Distribution A probability distribution is the list of all possible outcomes of a random variable and their associated probabilities. Example 6.1.5: Toss three fair coins and let X equal the number of tails observed. The outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. X = 0, 1, 2, and 3.
Value of X 0 1 Frequency (i.e., number of outcomes) 1 3 3 1 8 P(X) 0.125 0.375 0.375 0.125 1.000 5 2 3 P(X) is between 0 and 1 (inclusive) ∑P(X) = 1 6.2 Probability Distributions for Discrete Random Variables Three popular methods of describing probabilities associated with a discrete random variable. List each value of X and its corresponding probability. Use a histogram to convey the probabilities corresponding to the various values of X. Use a function that assigns a probability to each value of X. 6 6.2 Probability Distributions for Discrete Random Variables (cont.) List each value of X and its corresponding probability. Used in Example 6.1.5. Use a histogram to convey the probabilities corresponding to the various values of X. 3/8 – Probability 2/8 – 1/8 – – 0 1 2 x = number of tails 3 7 6.2 Probability Distributions for Discrete Random Variables (cont.) Probability Mass Function (PMF) A PMF is a function that assigns a probability to each value of X. P(X = x) = some expression (usually containing x) that produces a some that probability of observing x = P(x). probability P(x) is between 0 and 1 (inclusive) for each x is ∑P(x) = 1 Example 6.2.1: The function P(X=x) = x/30 for x = 0, 10, and 20 (and Example P(X=x) zero elsewhere) is a probability mass function because: zero P(X=x) = x/30 assigns a probability to each value of x. ∑P(X=x) = 1
8 Value of X 0 10 20 Others P(X = x) = x/30 0/30 = 0 10/30 = 1/3 20/30 = 2/3 0 1.00 6.2 Probability Distributions for Discrete Random Variables (cont.) Mean of Discrete Random Variables The mean of a discrete random variable represents the average value of the The random variable if you were to observe this variable over an indefinite period of time. time. The mean of a discrete random variable is written as µ and µ = ∑xP(x). The ∑xP(x). Example 6.2.2: Toss three fair coins and let X equal the number of tails observed. Find the mean number of tails observed.
Value of X 0 1
9 P(X) 0.125 0.375 0.375 0.125 1.000 XP(X) 0(0.125) = 0 1(0.375) = 0.375 2(0.375) = 0.75 3(0.125) = 0.375 ∑XP(X) = 1.500 2 3 6.2 Probability Distributions for Discrete Random Variables (cont.) Variance of Discrete Random Variables The variance of a discrete random variable, X, is a parameter describing the The variation of the corresponding population. variation The symbol used is σ 2. The σ 2 = ∑(x  µ)2 • P(x) = ∑x2P(x)  µ2 Example 6.2.3: Toss three fair coins and let X equal the number of tails observed. Find the variance for the number of tails observed.
Value of X 0
10 P(X) 0.125 0.375 1.000 XP(X) 0(0.125) = 0 2(0.375) = 0.75 1.500 Xµ (X  µ)2 (X  µ)2P(X) 0  1.5 = 1.5 (1.5)2 = 2.25 2.25(0.125) = 0.28125 2  1.5 = 0.5 (0.5)2 = 0.25 (1.5)2 = 2.25 0.25(0.375) = 0.09375 2.25(0.125) = 0.28125 ∑(X  µ)2P(X) = 0.75 1 2 3 0.375 1(0.375) = 0.375 1  1.5 = 0.5 (0.5)2 = 0.25 0.25(0.375) = 0.09375 0.125 3(0.125) = 0.375 3  1.5 = 1.5 6.2 Probability Distributions for Discrete Random Variables (cont.) Variance of Discrete Random Variables Finding variance using the formula σ 2 = ∑x2P(x)  µ2 Example 6.2.3: Toss three fair coins and let X equal the number of tails observed. Find the variance for the number of tails observed.
Value of X 0 1 2 3 P(X) 0.125 0.375 1.000 XP(X) 0(0.125) = 0 2(0.375) = 0.75 1.500 X2 02 = 0 22 = 4 X2P(x) 0(0.125) = 0 1(0.375) = 0.375 4(0.375) = 1.500 9(0.125) = 1.125 ∑x2P(x) = 3.000 0.375 1(0.375) = 0.375 12 = 1 0.125 3(0.125) = 0.375 32 = 9 11 µ2 = (1.500)2 = 2.25 (1.500) σ 2 = ∑x2P(x)  µ2 = 3.00 – 2.25 = 0.75 6.3 Binomial Random Variable Binomial Random Variable A discrete random variable that can assume one of two possible outcomes in each trial of an experiment comprising of n independent trials. An experiment in which each trial results in one of two mutually exclusive outcomes is called a binomial experiment. 12 6.3 Binomial Random Variable (cont.) Characteristics of a binomial experiment The experiment consists of n repetitions, called trials. Each trial has two mutually exclusive possible outcomes, referred to as Each success and failure. success The n trials are independent. The probability for a success for each trial is denoted p; and remains the The same for each trial. same The random variable x is the number of successes out of n trials. Example 6.3.1: Flip a coin five times. Let X be the number of heads. Is it a binomial situation?
13 n = 5. Success = head, failure = tail. The results on one flip do not affect the results on another flip. p = the probability of flipping a head on a particular flip = ½. X = the number of heads out of five flips. A Container Container with 20 poker chips 6 red chips 14 blue chips Select 4 chips with replacement
14 Let X = number of red chips Is X Binomial?
1. 2. 1. 1. 2. 15 There are n = 4 trials There are 2 possible outcomes on each trial • drawing a red chip ← success • drawing a blue chip ← failure What is p (chance of a success in each trial)? p is 6/20 = .3 Are the trials independent? Yes, since the chips are selected with replacement Yes, X is binomial with n = 4 and p = .3 With or Without Replacement When selecting items, your choice is with replacement or without 16 replacement To be a binomial random variable, items should be selected with replacement But with this procedure, you could select one of the items more than once, generally not something you want to happen For this reason, sampling without replacement is the usual procedure It turns out that it doesn’t really matter whether you are sampling with replacement or without replacement provided your sample size (n) is less than 5% of the population size (N)   that is, n/N < .05 With or Without Replacement In the previous example, N = number of chips in the container (20) In the chips example, 4/20 = .2 > .05 and this rule doesn’t apply For this situation, the chips must be selected with replacement for X to be binomial For many situations, you are selecting a sample from a huge population and the sample size is less than 5% of the population size In this case, whether you select the items with or without replacement doesn’t really matter
17 The Combination Function How to find: a! a Cb = b!(a − b)! a! is read “a factorial” For example, 5C2 counts the number of ways of getting 2 H’s (and 3 T’s) in 5 flips of a coin
5 C2 = This is
18 5! 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 5·4·3! 5·4 = = = = 10 2!3! 2!3! 2!3! 2 I would suggest you figure out how to find combinations on your calculator!! 6.3 Binomial Random Variable (cont.) Binomial Distribution The probability distribution of a binomial random variable is called a binomial distribution. Probability Mass Function (PMF) of a binomial random variable is: n C x p x (1 − p ) n − x for x = 0, 1, 2,...., n P( X = x) = elsewhere 0 = (number of ways of getting X = x)(probability of each one) 19 Example 6.3.2: A lawyer estimates that 40% of the cases in which she represented the defendant were won. If the lawyer is presently representing 10 defendants, what is the probability that 5 of the cases will be won? n = 10; x = 5; p = 0.4 P ( X = 5)= n C x p x (1 − p ) n − x =10 C5 (0.4) 5 (1 − 0.4)10−5 = 0.201 6.3 Binomial Random Variable (cont.) Using Binomial Table A.1 to Determine Probabilities The binomial PMFs have been tabulated in Table A.1 for various values The of n and p. of If n = 4 and p = 0.3 and you wish to find the P(2) locate n = 4 and x = 2. Go across to p = 0.3 and you will find the corresponding probability Go (after inserting the decimal in front of the number). This probability is 0.265. 0.265. 20 6.3 Binomial Random Variable (cont.) Cumulative Binomial Probabilities Finding P(X ≤ k), that is, finding the total probability of all successes up to and including k. P(X ≤ k) = P(X = 0) + P(X = 1) + P(X = 2) + ….. + P(X = k). P(X < k) = P(X = 0) + P(X = 1) + P(X = 2) + ….. + P(X = k  1). P(X > k) = 1  P(X ≤ k). P(X ≥ k) = 1  P(X < k). Buzz Words (No more than and atleast) No more than “k” implies P(X ≤ k) Atleast “k” implies P(X ≥ k) 21 Using Binomial Table A.2 If n = 4 and p = 0.3 and you wish to find the P(x ≤ 2) locate n = 4 and x = 2. If 2) Go across to p = 0.3 and you will find the corresponding probability (after Go inserting the decimal in front of the number). This probability is 0.916. inserting 6.3 Binomial Random Variable (cont.) Example 6.3.3: Suppose that 60% of all the employees of ABC Example Company favor unionization. A poll of 20 employees is taken to determine the number who favor unionization. determine Find the probability that at most 10 employees favor unionization. p = 0.6, n = 20, P(x ≤ 10) = 0.245 [Using Table A.2] 10) Find the probability that more than 12 employees favor unionization. P(x > 12) = 1 – P(x ≤ 12) = 1 – 0.584 (From Table A.2) = 0.416 P(x 12) 22 6.3 Binomial Random Variable (cont.) Mean and Variance of a Binomial Random Variable µ = np np σ 2 = np(1  p) np(1 Example 6.3.4: Suppose that 60% of all the employees of ABC Example Company favor unionization. A poll of 20 employees is taken to determine the number who favor unionization. Find the mean and standard deviation of those who favor unionization. standard p = 0.6, n = 20, µ = np = 20(0.6) = 12. np σ 2 = np(1  p) = 20(0.6)(1 – 0.6) = 4.8, σ = sqrt(4.8) = 2.19. np(1 20(0.6)(1 23 Using Excel Excel has a builtin function to determine binomial probabilities You can obtain either “=“ probabilities (Table A1) or “≤” probabilities (Table A2) The name of this function is BINOMDIST (stands for binomial distribution) This will be illustrated using the X = number of red chips example 24 Excel’s BINOMDIST 25 Excel’s BINOMDIST 26 Excel’s BINOMDIST Put where 0 is (cell A1) This is n This is p 27 Put “false” if you don’t want cumulative probabilities (Table A1) or “true” if you do (Table A2) Excel’s BINOMDIST
Put your cursor on the black square and drag down through cell B5 28 Excel’s BINOMDIST 29 These probabilities agree with the Table A1 results obtained earlier Excel’s BINOMDIST 30 These probabilities are obtained by putting “true” in the bottom box and agree with Table A2 6.4 Poisson Distribution The Poisson distribution is useful for counting the number of times a The particular event occurs over a specified period of time or over a specified area. specified For example, arrivals of customers at a service facility follow Poisson For Distribution. Conditions for the Poisson Distribution: 31 The number of occurrences in one measurement unit are independent The of the number of occurrences in any other nonoverlapping measurement unit. measurement The expected number of occurrences in any given measurement unit The are proportional to the size of the measurement unit. are Events can not occur at exactly the same point in the measurement Events unit. unit. 6.4 Poisson Distribution (cont.) PMF of Poisson Distribution: µ xe−µ P( X = x) = x! for x = 0, 1, 2, 3, ...... Mean and Variance of a Poisson Random Variable Mean of X = ∑xP(x) = µ = expected number of occurrences. Variance of X = σ 2 = ∑x2P(x)  µ2 = µ = expected number of occurrences. Variance Example 6.4.1: A large bakery determined that the expected number of Example delivery truck breakdowns per day is 1.5. Assume that the number of breakdowns is independent from day to day. breakdowns 32 What is the probability that there will be exactly two breakdowns tomorrow? µ = 1.5, P(X = 2) = 0.251. What is the probability that there will be exactly two breakdowns during next two What days? days? µ = 2(1.5) = 3, P(X = 2) = 0.224. Using Excel’s POISSON The Excel function here is named POISSON, which is in the list under the Statistical category You can obtain cumulative Poisson probabilities using Excel The following slide contains the window to obtain the “=“ probabilities for this example 33 Using Excel’s POISSON 34 After clicking on OK and dragging, you get the probabilities on the next slide Using Excel’s POISSON 35 ...
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This note was uploaded on 04/15/2011 for the course DSCI 2710 taught by Professor Hossain during the Spring '08 term at North Texas.
 Spring '08
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