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Econ10/Mgt 10 Stuffler 1 Discrete Random Variables Discrete Probability Distributions Binomial Distribution Poisson Distribution Hypergeometric Distribution

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Econ10/Mgt 10 Stuffler 2 Discrete Probability Distributions Discrete Random Variables – Sample Space includes all mutually exclusive outcomes Probabilities – from subjective, frequency or subjective methods Two conditions apply:

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Econ10/Mgt 10 Stuffler 3 Probability distributions can be estimated from relative frequencies. Consider the discrete (countable) number of televisions per household (millions) from US survey data: Discrete Probability Distributions

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Econ10/Mgt 10 Stuffler 4 Discrete Probability Distributions Probability distributions can be estimated from relative frequencies. Consider the discrete (countable) number of televisions per household from US survey data: 1,218 ÷ 101,501 = 0.012

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Econ10/Mgt 10 Stuffler 5 Discrete Probability Distributions Probability distributions can be estimated from relative frequencies. Consider the discrete (countable) number of televisions per household from US survey data: EX: P( X =4) = P(4) = = 7.6%

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Econ10/Mgt 10 Stuffler 6 Discrete Probability Distributions What is the probability there is at least one television but no more than three in any given household?

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Econ10/Mgt 10 Stuffler 7 Discrete Probability Distributions What is the probability there is at least one television but no more than three in any given household?

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Econ10/Mgt 10 Stuffler 8 Discrete Probability Distributions What is the probability there is at least one television but no more than three in any given household? at least one television but no more than three P(1 X 3) = P(1) + P(2) + P(3) = =.884

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Econ10/Mgt 10 Stuffler 9 Discrete Probability Distributions Assume a mutual fund salesman knows that there is 20% chance of closing a sale on each call he makes.

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Econ10/Mgt 10 Stuffler 10 Discrete Probability Distributions What is the probability distribution of the number of sales if he plans to call three customers? Let S denote success, making a sale: P(S) =.20, then S ʹ, not making a sale: P(S ʹ ) =.80

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Econ10/Mgt 10 Stuffler 11 Discrete Probability Distributions Developing a Probability Distribution Tree P(S)=.2 P(S)=.8 Sales Call 1

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Econ10/Mgt 10 Stuffler 12 Discrete Probability Distributions Developing a Probability Distribution Tree P(S)=.2 P(S)=.8 P(S)=.2 P(S)=.8 P(S)=.2 Sales Call 1Sales Call 2

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Econ10/Mgt 10 Stuffler 13 Discrete Probability Distributions Developing a Probability Distribution Tree P(S)=.2 P(S)=.8 P(S)=.2 P(S)=.8 S S S P(S)=.2 P(S)=.8 P(S)=.2 Sales Call 1Sales Call 2Sales Call 3

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Econ10/Mgt 10 Stuffler 14 Discrete Probability Distributions Developing a Probability Distribution Tree P(S)=.2 P(S)=.8 P(S)=.2 P(S)=.8 S S S P(S)=.2 P(S)=.8 P(S)=.2 Sales Call 1Sales Call 2Sales Call 3

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Econ10/Mgt 10 Stuffler 15 Discrete Probability Distributions Developing a Probability Distribution P(S)=.2 P(S)=.8 P(S)=.2 P(S)=.8 S S S P(S)=.2 )=.8 P(S)=.8 P(S)=.2 XP(x) = (.032)= (.128)= =.512 Sales Call 1Sales Call 2Sales Call 3

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Econ10/Mgt 10 Stuffler 16 Discrete Probability Distributions Explain how to derive.032 and.128 P(S)=.2 P(S C )=.8 P(S)=.2 P(S C )=.8 S S S S S S C S S C S S S C S C S C S S S C S S C S C S C S S C S C S C P(S)=.2 P(S C )=.8 P(S)=.2 XP(x) = (.032)= (.128)= =.512 Sales Call 1Sales Call 2Sales Call 3

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Econ10/Mgt 10 Stuffler 17 Discrete Probability Distributions The P(X=2) is: P(S)=.2 P(S C )=.8 P(S)=.2 P(S C )=.8 S S S S S S C S S C S S S C S C S C S S S C S S C S C S C S S C S C S C P(S)=.2 P(S C )=.8 P(S)=.2 XP(x) = (.032)= (.128)= =.512 (.2)(.2)(.8)=.032 Sales Call 1Sales Call 2Sales Call 3

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Econ10/Mgt 10 Stuffler 18 Discrete Probability Distributions A discrete probability distribution represents a population Example: Population of number of TVs per household Example: Population of sales call outcomes Since we have populations, we can describe them by computing various parameters: Population Mean and Population Variance

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Econ10/Mgt 10 Stuffler 19 Discrete Probability Distributions Population Mean (Expected Value) - W eighted average of all values with the weights being the probabilities - Expected value of X, E(X)

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Econ10/Mgt 10 Stuffler 20 Discrete Probability Distributions Population variance - weighted average of the squared deviations from the mean. Short cut formula for the variance Standard deviation formula

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Econ10/Mgt 10 Stuffler 21 Discrete Probability Distributions Find the mean, variance, and standard deviation for the population of the number of color televisions per household. = 0(.012) + 1(.319) + 2(.374) + 3(.191) + 4(.076) + 5(.028) = 2.084

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Econ10/Mgt 10 Stuffler 22 Discrete Probability Distributions Find the mean, variance, and standard deviation for the population of the number of color televisions per household. = (0 – 2.084) 2 (.012) + (1 – 2.084) 2 (.319)+…+(5 – 2.084) 2 (.028) = 1.107

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Econ10/Mgt 10 Stuffler 23 Discrete Probability Distributions Find the mean, variance, and standard deviation for the population of the number of color televisions per household. = 1.052

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Econ10/Mgt 10 Stuffler 24 Discrete Probability Distributions Special application of Expected Value Suppose the probability that an insurance agent makes a sale is.20 and after costs earns a commission of $525. If he/she does not make a sale, they must pay $75 in costs. What is their expected value from a sales call? Does the benefit exceed the cost or is the opposite true?

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Econ10/Mgt 10 Stuffler 25 Discrete Probability Distributions Special application of Expected Value Let X be the discrete random variable of making a sale call x P(x) xP(x) Sale $ $ 105 No Sale - $ $ 60 E(x) = μ = xP(x) = $ 45

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Econ10/Mgt 10 Stuffler 26 Binomial Distribution Binomial distribution is the probability distribution that results from doing a binomial experiment which have the properties: 1.Fixed number of identical trials, represented as n. 2.Each trial has two possible outcomes: a success or failure 3.For all trials, the probability of success, P(success)=p, and the probability of failure, P(failure)=1–p=q, are constant. 4.The trials are independent

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Econ10/Mgt 10 Stuffler 27 Several Binomial Distributions

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Econ10/Mgt 10 Stuffler 28 Binomial Distribution Success and failure: labels for binomial experiment outcomes, no value judgment is implied. EX: Coin flip results in either heads or tails. If we define heads as success, then tails is considered a failure. Other binomial examples: An election candidate wins or loses An employee is male or female A worker is employed or unemployed

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Econ10/Mgt 10 Stuffler 29 Binomial Distribution where x = number of successes in n trials, n – x = number of failures in n trials, p x = the probability of success raised to the number of successes, and q n-x = probability of failure raised to the number of failures Binomial Formula:

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Econ10/Mgt 10 Stuffler 30 Binomial Distribution The random variable of a binomial experiment is defined as the number of successes in the n trials, and is called the binomial random variable. EX: Flip a fair coin 10 times –1) Fixed number of trials n=10 –2) Each trial has two possible outcomes {heads (success), tails (failure)} –3) P(success)= 0.50; P(failure)=1–0.50 = 0.50 –4) The trials are independent (i.e. the outcome of heads on the first flip will have no impact on subsequent coin flips). Flipping a coin ten times is a binomial experiment since all conditions are met.

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Econ10/Mgt 10 Stuffler 31 Binomial Distribution Another sales call example: Assume a mutual fund salesman knows that there is 20% chance of closing a sale on each call he makes. We want to determine the probability of making two sales in three calls: P(sale) =.2 P(no sale) =.8 P(X=2) = 3! =.096 2!(3-2)!

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Econ10/Mgt 10 Stuffler 32 Binomial Distribution Another sales call example: Assume a mutual fund salesman knows that there is 20% chance of closing a sale on each call he makes We want probability of making two sales in three calls P(sale) =.2 P(no sale) =.8 GOOD NEWS! MegastatProbabilityDiscrete DistributionsBinomial

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Econ10/Mgt 10 Stuffler 33 Binomial Distribution

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Econ10/Mgt 10 Stuffler 34 Pat Statsly Pat Statsly is a student (not a good student) taking a statistics course. Pats exam strategy is to rely on luck for the first test. The test consists of 10 multiple-choice questions. Each question has five possible answers, only one of which is correct. Pat plans to guess the answer to each question. What is the probability that Pat gets no answers correct? What is the probability that Pat gets two answers correct?

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Econ10/Mgt 10 Stuffler 35 Pat Statsly n=10 P(correct) = p = 1/5 =.20 P(wrong) = q =.80 Is this a binomial experiment? Check the conditions

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Econ10/Mgt 10 Stuffler 36 Pat Statsly n=10 P(correct) = p = 1/5 =.20 P(wrong) = q =.80 Is this a binomial experiment? Check the conditions: There is a fixed finite number of trials (n=10). An answer can be either correct or incorrect. The probability of a correct answer (P(success)=.20) does not change from question to question. Each answer is independent of the others.

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Econ10/Mgt 10 Stuffler 37 Pat Statsly n=10, and P(success) =.20 What is the probability that Pat gets no answers correct? EX: P(x=0) Whats the interpretation of this result?

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Econ10/Mgt 10 Stuffler 38 Pat Statsly n=10, and P(success) =.20 What is the probability that Pat gets no answers correct? EX: P(x=0) Pat has about an 11% chance of getting no answers correct using the guessing strategy.

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Econ10/Mgt 10 Stuffler 39 Pat Statsly

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Econ10/Mgt 10 Stuffler 40 Pat Statsly n=10, and P(success) =.20 What is the probability that Pat gets two answers correct? EX: P(x=2)

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Econ10/Mgt 10 Stuffler 41 Pat Statsly We have been using the binomial probability distribution to find probabilities for individual values of x. To answer the question: Find the probability that Pat fails the quiz requires a cumulative probability, that is, P(X x) If a grade on the quiz is less than 50% (i.e. 5 questions out of 10), thats considered a failed quiz.

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Econ10/Mgt 10 Stuffler 42 Pat Statsly We have been using the binomial probability distribution to find probabilities for individual values of x. To answer the question: Find the probability that Pat fails the test requires a cumulative probability, that is, P(X x) If a grade on the test is less than 50% (i.e., 5 questions out of 10), thats considered a failed test. We want to know what is: P(X 4) to answer

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Econ10/Mgt 10 Stuffler 43 Pat Statsly P(X 4) = P(0) + P(1) + P(2) + P(3) + P(4) =.9672 What is the interpretation of this result?

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Econ10/Mgt 10 Stuffler 44 Pat Statsly Its about 97% probable that Pat will fail the test using the luck strategy and guessing at answers.

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Econ10/Mgt 10 Stuffler 45 Binomial Table Calculating binomial probabilities by hand is tedious and error prone. There is an easier way. Refer to Table E-6 in the Appendices. For the Pat Statsly example, n=10, so go to the n=10 table: Look in the Column, p=.20 and substitute into: P(X 4) = P(0) + P(1) + P(2) + P(3) + P(4) P(X 4) = =.9672 The probability of Pat failing the test is 96.72%

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Econ10/Mgt 10 Stuffler 46 Poisson Distribution Named for Simeon Poisson, Poisson distribution - Discrete probability distribution There there are no trials Number of independent events (successes) occurring in a fixed time period or region of space that occur with a known average rate such as, arrivals, departures, or accidents, number of baskets in a quarter, etc.

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Econ10/Mgt 10 Stuffler 47 Poisson Distribution For example: The number of cars arriving at a service station in 1 hour. (The interval of time is 1 hour.) The number of flaws in a bolt of cloth. (The specific region is a bolt of cloth.) The number of accidents in 1 day on a particular stretch of highway. (The interval is defined by both time, 1 day, and space, the particular stretch of highway.)

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Econ10/Mgt 10 Stuffler 48 Poisson Distribution Poisson random variable - number of successes that occur in a period of time or an interval of space EX: On average, 96 trucks arrive at a border crossing every hour. EX: Number of typographical errors in a new textbook edition averages 1.5 per 100 pages.

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Econ10/Mgt 10 Stuffler 49 Poisson Distribution… The Poisson random variable is the number of successes that occur in a period of time or an interval of space in a Poisson experiment. EX: On average, 96 trucks arrive at a border crossing every hour. E.g. The number of typographic errors in a new textbook edition averages 1.5 per 100 pages. successes time period successes (?!)interval

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Econ10/Mgt 10 Stuffler 50 Poisson Distribution Poisson experiment has four defining characteristics or properties: 1.The number of successes that occur in any interval is independent of the number of successes that occur in any other interval 2.The probability of a success in an interval is the same for all equal-size intervals 3.The probability of a success is proportional to the size of the interval 4.Only one value is required to determine the probability of a designated number of events occurring during an interval

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Econ10/Mgt 10 Stuffler 51 Poisson Distribution The probability that a Poisson random variable assumes a value of x is given by: µ = n(p) = number of successes times the probability of success The unit of time or the interval should be short enough so that the mean rate is not large (µ < 20) YOUR TEXTBOOK USES λ (lambda) REPRESENT THE MEAN NUMBER OF SUCCESSES IN THE INTERVAL

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Econ10/Mgt 10 Stuffler 52 Poisson Distribution EXAMPLE The number of typographical errors in new editions of textbooks varies considerably from book to book. After some analysis she concludes that the number of errors is Poisson distributed with a mean of 1.5 per 100 pages. The instructor randomly selects 100 pages of a new book. What is the probability that there are no typos? MEGASTATPROBABILITYDISCRETE DISTRIBUTIONSPOISSON AND ENTER THE MEAN VALUE: 1.5, CLICK OK

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Econ10/Mgt 10 Stuffler 53 Poisson Distribution

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Econ10/Mgt 10 Stuffler 54 Mean and Variance of a Poisson Random Variable If x is a Poisson random variable with parameter, then Mean number of events per unit of time or space = µ Variance = V(X) = σ 2 = μ

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Econ10/Mgt 10 Stuffler 55 Poisson Distribution As mentioned on the first Poisson Distribution slide: The probability of a success is proportional to the size of the interval Thus, knowing an error rate of 1.5 typos per 100 pages, we can determine a mean value for a 400 page book as: μ = n(p) = 1.5(4) = 6 typos / 400 pages.

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Econ10/Mgt 10 Stuffler 56 Poisson Distribution For a 400 page book, what is the probability that there are no typos? P(X=0) = Interpretation?

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Econ10/Mgt 10 Stuffler 57 GOOD NEWS! Go to Excel, Megastat, Probability, Discrete, Poisson. Type in the mean, μ, 6, then OK. For a mean of 6,the Excel result is

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Econ10/Mgt 10 Stuffler 58 Poisson Distribution For a 400 page book, what is the probability that there are five or less typos? P(X5) = P(0) + P(1) + … + P(5) Another alternative is to refer to Table E-7 in the Appendices For x = 5, μ =6, and P(X 5) = P(0) + … + P(5) =.4456 There is about a 45% chance that there will be 5 or less typos

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Econ10/Mgt 10 Stuffler 59 Hypergeometric Distribution If a sample is taken from a finite population without replacement, the probability of X number of successes, follows the hypergeometric distribution. NOTE: When you see without replacement, this indicates that hypergeometric trials are dependent.

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Econ10/Mgt 10 Stuffler 60 Hypergeometric Distribution For both the Binomial Distribution and the Hypergeometric Distribution, there are two possible outcomes for the random discrete variable: Success or Failure

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Econ10/Mgt 10 Stuffler 61 Hypergeometric Distribution The difference is that for Hypergeometric each trial is without replacement, so the probability changes from one draw to the next For Binomial, the probability of success and failure remain constant since each trial or draw is independent

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Econ10/Mgt 10 Stuffler 62 Hypergeometric Distribution 1. Trials are NOT independent 2. Finite population 3. Sample without replacement 4.Probability of success and failure changes from trial to trial

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Econ10/Mgt 10 Stuffler 63 Hypergeometric Distribution r N – r μ = n r P(x) = x n – x N N n σ 2 = r (N-r) n (N-n) N 2 (N-1) where N = Total number in the population n = number in the sample, x = number selected or drawn, r = number of successes in N

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Econ10/Mgt 10 Stuffler 64 Hypergeometric Distribution EXAMPLE: Assume that 3 stocks are chosen randomly from a list of 10 stocks and that of the 10 stocks 4 stocks pay dividends. The number (x) of the three selected stocks that pay a dividend is a hypergeometric random variable.

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Econ10/Mgt 10 Stuffler 65 Hypergeometric Distribution N=10, n=3, r = 4 and x = number of the 3 stocks selected that pay a dividend Calculate the mean, the variance and the probability that none of the 3 stocks pay dividends?

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Econ10/Mgt 10 Stuffler 66 Hypergeometric Distribution 4 10 – 4 4! 6! P(0) = = 0!(4-0)! 3!(6-3)! = ! 6 3 3!(10-3)! μ = (3)(4) = σ 2 = 4(10-4)3(10-3) =.56, σ =.75 (10) 2 (10-1)

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Econ10/Mgt 10 Stuffler 67 Hypergeometric Distribution GOOD NEWS! Go to ExcelMegastatProbability Discrete Probability Distributions Hypergeometric, enter the required data

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Econ10/Mgt 10 Stuffler 68 Hypergeometric Distribution

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Econ10/Mgt 10 Stuffler 69 Discrete Probability WHICH DISTRIBUTION? Sony issued a recall for their plasma screen TVs. They choose 25 TVs and 7 do not operate. If the inspector chooses 3 at random from the 25, what is the probability that 2 of the 3 do not operate?

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Econ10/Mgt 10 Stuffler 70 Discrete Probability WHICH DISTRIBUTION? When Wet Water drills a well, their success rate is.55 and their profit is $2,550. If they do not find water, they lose $1,500.

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Econ10/Mgt 10 Stuffler 71 Discrete Probability WHICH DISTRIBUTION? The US unemployment rate is calculated from survey data of 60,000 households. What is the probability that no more than 5,000 respondents are unemployed?

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Econ10/Mgt 10 Stuffler 72 Discrete Probability WHICH DISTRIBUTION? US Transportation and Safety Board reports the annual average number of plane crashes is For a given month, what is the probability that at 2 crashes will occur? That 3 crashes will occur?

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Econ10/Mgt 10 Stuffler 73 Discrete Probability WHICH DISTRIBUTION? The US Statistical Abstract of the US tracks many data items such as, the average number of automobiles per household. The average for 2009 is 2.5 autos. What is the probability that the 2010 average is at least 3 autos?

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Econ10/Mgt 10 Stuffler 74 Discrete Random Variables Two Types of Random Variables Discrete Probability Distributions The Binomial Distribution The Poisson Distribution The Hypergeometric Distribution Summary:

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