Probability Distribution Probability Distribution A variable is a symbol (A, B, x, y, etc.) that can take on any of a specified set of values When the value of a variable is the outcome of a statistical experiment, that variable is a random variable. Generally, statisticians use a capital letter to represent a random variable and a lower-case letter, to represent one of its values. For example, X represents the random variable X. P(X) represents the probability of X. P(X = x) refers to the probability that the random variable X is equal to a particular value, denoted by x. As an example, P(X = 1) refers to the probability that the random variable X is equal to 1. Probability Distribution An example will make clear the relationship between random variables and probability distributions. Suppose you flip a coin two times. This simple statistical experiment can have four possible outcomes: HH, HT, TH, and TT. Now, let the variable X represent the number of Heads that result from this experiment. The variable X can take on the values 0, 1, or 2. In this example, X is a random variable; because its value is determined by the outcome of a statistical experiment. A probability distribution is a table or an equation that links each outcome of a statistical experiment with its probability of occurrence. Consider the coin flip experiment described above. The table which associates each outcome with its probability, is an example of a probability distribution. It represents the probability distribution of random variable x. Discrete & Continuous Probability Distribution If a variable can take on any value between two specified values, it is called a continuous variable; otherwise, it is called a discrete variable. Some examples will clarify the difference between discrete and continuous variables. Suppose the fire department mandates that all fire fighters must weigh between 150 and 250 pounds. The weight of a fire fighter would be an example of a continuous variable; since a fire fighter's weight could take on any value between 150 and 250 pounds. Suppose we flip a coin and count the number of heads. The number of heads could be any integer value between 0 and plus infinity. However, it could not be any number between 0 and plus infinity. We could not, for example, get 2.5 heads. Therefore, the number of heads must be a discrete variable. Just like variables, Probability Distributions can be classified as discrete or continuous. Discrete Probability Distribution If a random variable is a discrete variable, its probability distribution is called a discrete probability distribution. An example will make this clear. Suppose you flip a coin two times. This simple statistical experiment can have four possible outcomes: HH, HT, TH, and TT. Now, let the random variable X represent the number of Heads that result from this experiment. The random variable X can only take on the values 0, 1, or 2, so it is a discrete random variable. The probability distribution for this statistical experiment appears below. table represents a discrete probability distribution because it relates each value of a discrete random variable with its probability of occurrence Discrete Probability Distribution Most well-known discrete probability distributions that are used for statistical modeling are Binomial Probability Distribution Hypergeometric Probability Distribution Poisson Probability Distribution Note: With a discrete probability distribution, each possible value of the discrete random variable can be associated with a non-zero probability. Thus, a discrete probability distribution can always be presented in tabular form. Binomial Probability Distribution Binomial Experiment A binomial experiment (also known as a Bernoulli trial) is a statistical experiment that has the following properties: The experiment consists of n repeated trials. Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure. The probability of success, denoted by P, is the same on every trial. The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials. Consider the following statistical experiment. You flip a coin 2 times and count the number of times the coin lands on heads. This is a binomial experiment because: The experiment consists of repeated trials. We flip a coin 2 times. Each trial can result in just two possible outcomes - heads or tails. The probability of success is constant - 0.5 on every trial. The trials are independent; that is, getting heads on one trial does not affect whether we get heads on other trials. Binomial Probability Distribution Binomial Distribution Possible Binomial Distribution Settings: A manufacturing plant labels items as either defective or acceptable A firm bidding for contracts will either get a contract or not A marketing research firm receives survey responses of “yes I will buy” or “no I will not” New job applicants either accept the offer or reject it Binomial Probability Distribution Notation The following notation is helpful, when we talk about binomial probability. x: The number of successes that result from the binomial experiment. n: The number of trials in the binomial experiment. P: The probability of success on an individual trial. Q: The probability of failure on an individual trial. (This is equal to 1 - P) b(x; n, P): Binomial probability - the probability that an n-trial binomial experiment results in exactly x successes, when the probability of success on an individual trial is P. nCr : The number of combinations of n things, taken r at a time. Binomial Probability Distribution Binomial Distribution A binomial random variable is the number of successes x in n repeated trials of a binomial experiment. The Probability Distribution of a binomial random variable is called a binomial distribution (also known as a Bernoulli distribution). Suppose we flip a coin two times and count the number of heads (successes). The binomial random variable is the number of heads, which can take on values of 0, 1, or 2. The binomial distribution is presented below. The binomial distribution has the following properties: The mean of the distribution (μx) is equal to n * P . The variance (σ2x) is n * P * ( 1 - P ). The Standard Deviation (σx) is sqrt[ n * P * ( 1 - P ) ]. Binomial Probability Distribution Binomial Probability refers to the probability that a binomial experiment results in exactly x successes. For example, in the above table, we see that the binomial probability of getting exactly one head in two coin flips is 0.50. Suppose we flip a coin two times and count the number of heads (successes). The binomial random variable is the number of heads, which can take on values of 0, 1, or 2. The binomial distribution is presented on the previous slide. Binomial Probability Distribution Example 1: Suppose a die is tossed 5 times. What is the probability of getting exactly 2 fours? Solution: This is a binomial experiment in which the number of trials is equal to 5, the number of successes is equal to 2, and the probability of success on a single trial is 1/6 or about 0.167. Therefore, the binomial probability is: b(2; 5, 0.167) = 5C2 * (0.167)2 * (0.833)3 b(2; 5, 0.167) = 0.161 Binomial Probability Distribution Example 2: Evans is concerned about a low retention rate for employees. In recent years, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employee chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year. Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company this year? Solution: This is a binomial experiment in which the number of trials is equal to 3, the number of successes is equal to 1, and the probability of success on a single trial is 10% or 0.10. Therefore, the binomial probability is: b(1; 3, 0.10) = 3C1 * (0.10)1 * (0.9)2 B(1; 3, 0.10) = 0.243 Binomial Probability Distribution Example 3: You are performing a cohort study. If the probability of developing disease in the exposed group is .05 for the study duration, then if you sample (randomly) 500 exposed people, how many do you expect to develop the disease? Give a margin of error (+/- 1 standard deviation) for your estimate. Solution: B(x; 500, 0.05) E(x) or μx is equal to n * P = 500 * 0.05 = 25 Var(x) or σ2x is n * P * ( 1 - P ) = 500 * 0.05 * 0.95 = 23.75 StdDev(x) or σx is sqrt[ n * P * ( 1 - P ) ] = sqrt[23.75] = 4.87 Therefore: 25 4.87 Binomial Probability Distribution Cumulative Binomial Probability A cumulative binomial probability refers to the probability that the binomial random variable falls within a specified range (e.g., is greater than or equal to a stated lower limit and less than or equal to a stated upper limit). For example, we might be interested in the cumulative binomial probability of obtaining 45 or fewer heads in 100 tosses of a coin (see Example 1 below). This would be the sum of all these individual binomial probabilities. b(x < 45; 100, 0.5) = b(x = 0; 100, 0.5) + b(x = 1; 100, 0.5) + ... + b(x = 44; 100, 0.5) + b(x = 45; 100, 0.5) Binomial Probability Distribution Example 4: The probability that a student is accepted to a prestigious college is 0.3. If 5 students from the same school apply, what is the probability that at most 2 are accepted? Solution: To solve this problem, we compute 3 individual probabilities, using the binomial formula. The sum of all these probabilities is the answer we seek. Thus, b(x < 2; 5, 0.3) = b(x = 0; 5, 0.3) + b(x = 1; 5, 0.3) + b(x = 2; 5, 0.3) b(x < 2; 5, 0.3) = 0.1681 + 0.3601 + 0.3087 b(x < 2; 5, 0.3) = 0.8369 Binomial Probability Distribution Example 5: If the probability of being a smoker among a group of cases with lung cancer is .6, what’s the probability that in a group of 8 cases you have less than 2 smokers? More than 5? What are the expected value and variance of the number of smokers? Solution: 1 11 121 1331 14641 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 Binomial Probability Distribution Example 5: If the probability of being a smoker among a group of cases with lung cancer is 0.6, what’s the probability that in a group of 8 cases you have less than 2 smokers? More than 5? What are the expected value and variance of the number of smokers? Solution: P(<2)=.00065 + .008 = .00865 P(>5)=.21+.09+.0168 = .3168 0 1 2 3 4 5 6 7 8 E(X) = 8 (.6) = 4.8 Var(X) = 8 (.6) (.4) =1.92 StdDev(X) = 1.38 Binomial Probability Distribution Using Tables of Binomial Probabilities Hypergeometric Probability Distribution Hypergeometric Experiment A hypergeometric experiment is a statistical experiment that has the following properties: “n” trials in a sample taken from a finite population of size N Sample taken without replacement Outcomes of trials are dependent Concerned with finding the probability of “X” successes in the sample where there are “A” successes in the population Consider the following statistical experiment. You have a sum of 10 marbles - 5 red and 5 green. You randomly select 2 marbles without replacement and count the number of red marbles you have selected. This would be a hypergeometric experiment. Hypergeometric Probability Distribution Hypergeometric Experiment If you select a red marble on the first trial, the probability of selecting a red marble on the second trial is 4/9. And if you select a green marble on the first trial, the probability of selecting a red marble on the second trial is 5/9. Note further that if you selected the marbles with replacement, the probability of success would not change. It would be 5/10 on every trial. Then, this would be a binomial experiment. Hypergeometric Probability Distribution Notation The following notation is helpful, when we talk about hypergeometric probability. N: The number of items in the population k: The number of items in the population that are classified as successes. n: The number of items in the sample x: The number of items in the sample that are classified as successes. kCx: The number of k things, taken x at a time. h(x; N, n, k) : hypergeometric probability - the probability that an ntrial hypergeometric experiment results in exactly x successes, when the population consists of N items, k of which are classified as successes. Hypergeometric Probability Distribution Hypergeometric Distribution A hypergeometric random variable is the number of successes that result from a hypergeometric experiment. The probability distribution of a hypergeometric random variable is called a hypergeometric distribution. Given x, N, n, and k, we can compute the hypergeometric probability based on the following formula: The hypergeometric distribution has the following properties: The mean of the distribution is equal to n * k / N . The variance is n * k * ( N - k ) * ( N - n ) / [ N2 * ( N - 1 ) ] . Hypergeometric Probability Distribution Example 1: Suppose we randomly select 5 cards without replacement from an ordinary deck of playing cards. What is the probability of getting exactly 2 red cards (i.e., hearts or diamonds)? Solution: This is a hypergeometric experiment in which we know the following: N = 52; since there are 52 cards in a deck. k = 26; since there are 26 red cards in a deck. n = 5; since we randomly select 5 cards from the deck. x = 2; since 2 of the cards we select are red. Hypergeometric Probability Distribution Example 1: continuation… Suppose we randomly select 5 cards without replacement from an ordinary deck of playing cards. What is the probability of getting exactly 2 red cards (i.e., hearts or diamonds)? Solution: We plug these values into the hypergeometric formula as follows: h(x; N, n, k) = [ kCx ] [ N-kCn-x ] / [ NCn ] h(2; 52, 5, 26) = [ 26C2 ] [ 26C3 ] / [ 52C5 ] h(2; 52, 5, 26) = [ 325 ] [ 2600 ] / [ 2,598,960 ] = 0.32513 Thus, the probability of randomly selecting 2 red cards is 0.32513. Hypergeometric Probability Distribution Example 2: 3 different computers are checked from 10 in the department. 4 of the 10 computers have illegal software loaded. What is the probability that 2 of the 3 selected computers have illegal software loaded? Solution: This is a hypergeometric experiment in which we know the following: N = 10; since there are 10 computers. k = 4; since there are 4computers with illegal software. n = 3; since we randomly select 3 computers with illegal software. x = 2; since 2 of the computers were selected. Hypergeometric Probability Distribution Example 2: continuation… 3 different computers are checked from 10 in the department. 4 of the 10 computers have illegal software loaded. What is the probability that 2 of the 3 selected computers have illegal software loaded? Solution: We plug these values into the hypergeometric formula as follows: h(x; N, n, k) = [ kCx ] [ N-kCn-x ] / [ NCn ] h(2; 10, 3, 4) = [ 4C2 ] [ 6C1 ] / [ 10C3 ] h(2; 10, 3, 4) = [ 6 ] [ 6 ] / [ 120 ] = 0.3 The probability that 2 of the 3 selected computers have illegal software loaded is 0.30, or 30%. Hypergeometric Probability Distribution Cumulative Hypergeometric Probability A cumulative hypergeometric probability refers to the probability that the hypergeometric random variable is greater than or equal to some specified lower limit and less than or equal to some specified upper limit. For example, suppose we randomly select five cards from an ordinary deck of playing cards. We might be interested in the cumulative hypergeometric probability of obtaining 2 or fewer hearts. This would be the probability of obtaining 0 hearts plus the probability of obtaining 1 heart plus the probability of obtaining 2 hearts, as shown in the example below. Hypergeometric Probability Distribution Example 3 Suppose we select 5 cards from an ordinary deck of playing cards. What is the probability of obtaining 2 or fewer hearts? Solution: This is a hypergeometric experiment in which we know the following: N = 52; since there are 52 cards in a deck. k = 13; since there are 13 hearts in a deck. n = 5; since we randomly select 5 cards from the deck. x = 0 to 2; since our selection includes 0, 1, or 2 hearts. Hypergeometric Probability Distribution Example 3 continuation… Suppose we select 5 cards from an ordinary deck of playing cards. What is the probability of obtaining 2 or fewer hearts? Solution: We plug these values into the hypergeometric formula as follows: h(x < x; N, n, k) = h(x < 2; 52, 5, 13) h(x < 2; 52, 5, 13) = h(x=0; 52, 5, 13) + h(x=1; 52, 5, 13) + h(x=2; 52, 5, 13) h(x < 2; 52, 5, 13) = [ (13C0) (39C5) / (52C5) ] + [ (13C1) (39C4) / (52C5) ] + [ (13C2) (39C3) / (52C5) ] h(x < 2; 52, 5, 13) = [ (1)(575,757)/(2,598,960) ] + [ (13)(82,251)/(270,725) ] + [ (78)(9139)/(22,100) ] h(x < 2; 52, 5, 13) = [ 0.2215 ] + [ 0.4114 ] + [ 0.2743 ] h(x < 2; 52, 5, 13) = 0.9072 Thus, the probability of randomly selecting at most 2 hearts is 0.9072. Poisson Probability Distribution Poisson Experiment A poisson experiment is a statistical experiment that has the following properties: The experiment results in outcomes that can be classified as successes or failures. The average number of successes (μ) that occurs in a specified region is known. The probability that a success will occur is proportional to the size of the region. The probability that a success will occur in an extremely small region is virtually zero. Note that the specified region could take many forms. For instance, it could be a length, an area, a volume, a period of time, etc. Poisson Probability Distribution Notation The following notation is helpful, when we talk about binomial probability. e: A constant equal to approximately 2.71828. (Actually, e is the base of the natural logarithm system.) μ: The mean number of successes that occur in a specified region. x: The actual number of successes that occur in a specified region. P(x; μ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is μ. Poisson Probability Distribution Poisson Distribution A Poisson random variable is the number of successes that result from a Poisson experiment. The Probability Distribution of a Poisson random variable is called a Poisson distribution. Given the mean number of successes (μ) that occur in a specified region, we can compute the Poisson probability based on the following formula: The poisson distribution has the following properties: The mean of the distribution is equal to μ which is equal to n*P. The variance is also equal to μ . Poisson Probability Distribution Example 1 The average number of homes sold by the Acme Realty company is 2 homes per day. What is the probability that exactly 3 homes will be sold tomorrow? Solution: This is a poisson experiment in which we know the following: μ = 2; since 2 homes are sold per day, on average. x = 3; since we want to find the likelihood that 3 homes will be sold tomorrow. e = 2.71828; since e is a constant equal to approximately 2.71828. Poisson Probability Distribution Example 1 continuation… The average number of homes sold by the Acme Realty company is 2 homes per day. What is the probability that exactly 3 homes will be sold tomorrow? Solution: We plug these values into the poisson formula as follows: P(x; μ) = (e-μ) (μx) / x! P(3; 2) = (2.71828-2) (23) / 3! P(3; 2) = (0.13534) (8) / 6 P(3; 2) = 0.180 Thus, the probability of selling 3 homes tomorrow is 0.180 . Poisson Probability Distribution Example 2 The number of typographical errors in new editions of textbooks varies considerably from book to book. After some analysis he concludes that the number of errors is Poisson distributed with a mean of 1.5 typos per 100 pages. The instructor randomly selects 100 pages of a new book. What is the probability that there are no typos? Solution: We plug these values into the poisson formula as follows: P(x; μ) = (e-μ) (μx) / x! P(0; 1.5) = (2.71828-1.5) (1.50) / 0! P(0; 1.5) = 0.2231 There is about a 22% chance of finding zero errors Poisson Probability Distribution Example 3 Suppose that a rare disease has an incidence of 1 in 1000 personyears. Assuming that members of the population are affected independently, find the probability of k cases in a population of 10,000 (followed over 1 year) for k=0,1,2. Solution: The expected value (mean) = μ = N*P = 10,000 * 0.001 = 10 10 new cases expected in this population per year P(x; μ) = (e-μ) (μx) / x! P(0; 10) = (2.71828-10) (100) / 0! = 0.0000454 P(1; 10) = (2.71828-10) (101) / 1! = 0.000454 P(2; 10) = (2.71828-10) (102) / 2! = 0.00227 Poisson Probability Distribution Poisson Distribution “Poisson Process” (rates) Note that the Poisson parameter can be given as the mean number of events that occur in a defined time period OR, equivalently, can be given as a rate, such as =2/month (2 events per 1 month) that must be multiplied by t=time (called a “Poisson Process”) X ~ Poisson () P(x=K, t) = (e- t) ( t)x / x! E(X) = t Var(X) = t Poisson Probability Distribution Example 4 If new cases of West Nile in New England are occurring at a rate of about 2 per month, then what’s the probability that exactly 4 cases will occur in the next 3 months? Exactly 6 cases? Solution: X ~ Poisson (=2/month) X = 4 cases, twice in 3 months P(x=K, t) = (e-t) ( t)x / x! P(x=4, 2*3) = (e-2*3) (2*3)4 / 4! = 0.1339 X = 6 cases, twice in 3 months P(x=K, t) = (e-t) ( t)x / x! P(x=6, 2*3) = (e-2*3) (2*3) 6 / 6! = 0.1606 Poisson Probability Distribution Example 5 If calls to your cell phone are a Poisson process with a constant rate =2 calls per hour, what’s the probability that, if you forget to turn your phone off in a 1.5 hour movie, your phone rings during that time? How many phone calls do you expect to get during the movie? Solution: Probability that your phone rings in a 1.5hr movie X ~ Poisson (=2 calls/hour) P(X≥1)=1 – P(X=0) X = 0 calls, 2 calls /hr in 1.5hrs P(x=K, t) = (e-t) ( t)x / x! P(x=0, 2*1.5) = (e-2*1.5) (2*1.5)0 / 0! = 0.05% Therefore: P(X≥1)=1 – 0.05 = 95% chance Poisson Probability Distribution Example 5 If calls to your cell phone are a Poisson process with a constant rate =2 calls per hour, what’s the probability that, if you forget to turn your phone off in a 1.5 hour movie, your phone rings during that time? How many phone calls do you expect to get during the movie? Solution: Number of phone calls during a 1.5hr movie X ~ Poisson (=2 calls/hour) 1.5hr movie E(X) = t = 2(1.5) = 3 calls Poisson Probability Distribution Cumulative Poisson Probability A cumulative Poisson probability refers to the probability that the Poisson random variable is greater than some specified lower limit and less than some specified upper limit. Example 6 Suppose the average number of lions seen on a 1-day safari is 5. What is the probability that tourists will see fewer than four lions on the next 1-day safari? Poisson Probability Distribution Solution: This is a poisson experiment in which we know the following: μ = 5; since 5 lions are seen per safari, on average. x = 0, 1, 2, or 3; since we want to find the likelihood that tourists will see fewer than 4 lions; that is, we want the probability that they will see 0, 1, 2, or 3 lions. e = 2.71828; since e is a constant equal to approximately 2.71828. To solve this problem, we need to find the probability that tourists will see 0, 1, 2, or 3 lions. Thus, we need to calculate the sum of four probabilities: P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5). Poisson Probability Distribution Example 6 continuation... To compute this sum, we use the Poisson formula: P(x < 3, 5) = P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5) P(x < 3, 5) = [ (e-5)(50) / 0! ] + [ (e-5)(51) / 1! ] + [ (e-5)(52) / 2! ] + [ (e-5)(53) / 3! ] P(x < 3, 5) = [ (0.006738)(1) / 1 ] + [ (0.006738)(5) / 1 ] + [ (0.006738)(25) / 2 ] + [ (0.006738)(125) / 6 ] P(x < 3, 5) = [ 0.0067 ] + [ 0.03369 ] + [ 0.084224 ] + [ 0.140375 ] P(x < 3, 5) = 0.2650 Thus, the probability of seeing at no more than 3 lions is 0.2650. Discrete Probability Distribution Binomial Probability Distribution b(x; n, p) = nCx * (p)x * (n – p)n-x **The mean of the distribution (μx) is equal to n * P . **The variance (σ2x) is n * P * ( 1 - P ). **The Standard Deviation (σx) is sqrt[ n * P * ( 1 - P ) ]. Hypergeometric Probability Distribution h(x; N, n, k) = [ kCx ] [ N-kCn-x ] / [ NCn ] **The mean of the distribution is equal to n * k / N . **The variance is n * k * ( N - k ) * ( N - n ) / [ N2 * ( N - 1 ) ] Poisson Probability Distribution P(x; μ) = (e-μ) (μx) / x! E(X) = t Var(X) = t **The mean of the distribution is equal to μ which is equal to n*P. **The variance is also equal to μ