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Discrete Probability Distributions

Chapter Outline. 4.1 Probability Distributions4.2 Binomial Distributions4.3 More Discrete Probability Distributions (not in syllabus). Larson/Farber 4th ed. 2. Section 4.1. Probability Distributions. 3. Larson/Farber 4th ed. Section 4.1 Objectives. Distinguish between discrete random variables and

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Discrete Probability Distributions

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    1. Chapter 4 Discrete Probability Distributions 1 Larson/Farber 4th ed

    2. Chapter Outline 4.1 Probability Distributions 4.2 Binomial Distributions 4.3 More Discrete Probability Distributions (not in syllabus) Larson/Farber 4th ed 2

    3. Section 4.1 Probability Distributions 3 Larson/Farber 4th ed

    4. Section 4.1 Objectives Distinguish between discrete random variables and continuous random variables Construct a discrete probability distribution and its graph Determine if a distribution is a probability distribution Find the mean, variance, and standard deviation of a discrete probability distribution Find the expected value of a discrete probability distribution Larson/Farber 4th ed 4

    5. Random Variables Random Variable Represents a numerical value associated with each outcome of a probability distribution. Denoted by x Examples x = Number of sales calls a salesperson makes in one day. x = Hours spent on sales calls in one day. 5 Larson/Farber 4th ed

    6. Random Variables Discrete Random Variable Has a finite or countable number of possible outcomes that can be listed. Example x = Number of sales calls a salesperson makes in one day. 6 Larson/Farber 4th ed

    7. Random Variables Continuous Random Variable Has an uncountable number of possible outcomes, represented by an interval on the number line. Example x = Hours spent on sales calls in one day. 7 Larson/Farber 4th ed

    8. Example: Random Variables Decide whether the random variable x is discrete or continuous. 8 Larson/Farber 4th ed

    9. Example: Random Variables Decide whether the random variable x is discrete or continuous. 9 Larson/Farber 4th ed

    10. Textbook Exercises. Page 201 Distinguish between Discrete and Continuous random variable x represents the length of time it takes to go to work. x is a continuous random variable because length of time is a measurement and not a count. Measurements are continuous. x represents number of rainy days in the month of July in Orlando, Florida. x is a discrete random variable because number of rainy days is a count. Counts are discrete. Larson/Farber 4th ed 10

    11. Discrete Probability Distributions Discrete probability distribution Lists each possible value the random variable can assume, together with its probability. Must satisfy the following conditions: 11 Larson/Farber 4th ed

    12. Constructing a Discrete Probability Distribution Make a frequency distribution for the possible outcomes. Find the sum of the frequencies. Find the probability of each possible outcome by dividing its frequency by the sum of the frequencies. Check that each probability is between 0 and 1 and that the sum is 1. 12 Larson/Farber 4th ed

    13. Example: Constructing a Discrete Probability Distribution An industrial psychologist administered a personality inventory test for passive-aggressive traits to 150 employees. Individuals were given a score from 1 to 5, where 1 was extremely passive and 5 extremely 13 Larson/Farber 4th ed

    14. Solution: Constructing a Discrete Probability Distribution Divide the frequency of each score by the total number of individuals in the study to find the probability for each value of the random variable. 14 Larson/Farber 4th ed

    15. Solution: Constructing a Discrete Probability Distribution This is a valid discrete probability distribution since Each probability is between 0 and 1, inclusive, 0 = P(x) = 1. The sum of the probabilities equals 1, SP(x) = 0.16 + 0.22 + 0.28 + 0.20 + 0.14 = 1. 15 Larson/Farber 4th ed

    16. Solution: Constructing a Discrete Probability Distribution Histogram 16 Larson/Farber 4th ed

    17. Textbook Exercises. Page 202 Problem 22 Blood Donations x being number of donations is countable and hence a discrete random variable. Also, number of donations are mutually exclusive. P(X>1) = P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) = 0.25 + 0.10 + 0.05 + 0.03 + 0.02 = 0.45 P(X < 3) = P(x = 2) + P(x = 1) + P(x = 0) = 0.25 + 0.25 + 0.30 = 0.80 Problem 24 Dependent Children Since it is a probability distribution sum of all the probabilities is equal to one. missing value = 1 – sum of rest of the probabilities = 1 – 0.85 = 0.15 Larson/Farber 4th ed 17

    18. Mean Mean of a discrete probability distribution µ = SxP(x) Each value of x is multiplied by its corresponding probability and the products are added. 18 Larson/Farber 4th ed

    19. Example: Finding the Mean The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the mean. 19 Larson/Farber 4th ed

    20. Variance and Standard Deviation Variance of a discrete probability distribution s2 = S(x – µ)2P(x) Standard deviation of a discrete probability distribution 20 Larson/Farber 4th ed

    21. Example: Finding the Variance and Standard Deviation The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the variance and standard deviation. ( µ = 2.94) 21 Larson/Farber 4th ed

    22. Solution: Finding the Variance and Standard Deviation Recall µ = 2.94 22 Larson/Farber 4th ed

    23. Textbook Exercises. Page 203 Problem 32 DVDs Construct the probability distribution and determine : a. Mean b. Standard Deviation c. Variance Larson/Farber 4th ed 23

    24. Expected Value Expected value of a discrete random variable Equal to the mean of the random variable. E(x) = µ = SxP(x) 24 Larson/Farber 4th ed

    25. Example: Finding an Expected Value At a raffle, 1500 tickets are sold at $2 each for four prizes of $500, $250, $150, and $75. You buy one ticket. What is the expected value of your gain? 25 Larson/Farber 4th ed

    26. Solution: Finding an Expected Value To find the gain for each prize, subtract the price of the ticket from the prize: Your gain for the $500 prize is $500 – $2 = $498 Your gain for the $250 prize is $250 – $2 = $248 Your gain for the $150 prize is $150 – $2 = $148 Your gain for the $75 prize is $75 – $2 = $73 If you do not win a prize, your gain is $0 – $2 = –$2 26 Larson/Farber 4th ed

    27. Solution: Finding an Expected Value Probability distribution for the possible gains (outcomes) 27 Larson/Farber 4th ed

    28. Section 4.1 Summary Distinguished between discrete random variables and continuous random variables Constructed a discrete probability distribution and its graph Determined if a distribution is a probability distribution Found the mean, variance, and standard deviation of a discrete probability distribution Found the expected value of a discrete probability distribution Larson/Farber 4th ed 28

    29. Section 4.2 Binomial Distributions Larson/Farber 4th ed 29

    30. Section 4.2 Objectives Determine if a probability experiment is a binomial experiment Find binomial probabilities using the binomial probability formula Find binomial probabilities using technology and a binomial table Graph a binomial distribution Find the mean, variance, and standard deviation of a binomial probability distribution Larson/Farber 4th ed 30

    31. Binomial Experiments The experiment is repeated for a fixed number of trials, where each trial is independent of other trials. There are only two possible outcomes of interest for each trial. The outcomes can be classified as a success (S) or as a failure (F). The probability of a success P(S) is the same for each trial. The random variable x counts the number of successful trials. Larson/Farber 4th ed 31

    32. Notation for Binomial Experiments Larson/Farber 4th ed 32

    33. Example: Binomial Experiments Decide whether the experiment is a binomial experiment. If it is, specify the values of n, p, and q, and list the possible values of the random variable x. Larson/Farber 4th ed 33

    34. Solution: Binomial Experiments Binomial Experiment Each surgery represents a trial. There are eight surgeries, and each one is independent of the others. There are only two possible outcomes of interest for each surgery: a success (S) or a failure (F). The probability of a success, P(S), is 0.85 for each surgery. The random variable x counts the number of successful surgeries. Larson/Farber 4th ed 34

    35. Solution: Binomial Experiments Binomial Experiment n = 8 (number of trials) p = 0.85 (probability of success) q = 1 – p = 1 – 0.85 = 0.15 (probability of failure) x = 0, 1, 2, 3, 4, 5, 6, 7, 8 (number of successful surgeries) Larson/Farber 4th ed 35

    36. Example: Binomial Experiments Decide whether the experiment is a binomial experiment. If it is, specify the values of n, p, and q, and list the possible values of the random variable x. Larson/Farber 4th ed 36

    37. Solution: Binomial Experiments Not a Binomial Experiment The probability of selecting a red marble on the first trial is 5/20. Because the marble is not replaced, the probability of success (red) for subsequent trials is no longer 5/20. The trials are not independent and the probability of a success is not the same for each trial. Larson/Farber 4th ed 37

    38. Textbook Exercises. Pages 215 - 216 Problem 2 Graphical Analysis p = 0.75 (graph is skewed to left implies p is greater than 0.5 ) p = 0.50 (graph is symmetric) p = 0.25 (graph is skewed to right implies p is smaller than 0.5) Problem 8 Clothing store purchases Since it satisfies all the requirements , it is a binomial experiment. where, Success is: person does not make a purchase n = 18, p = 0.74, q = 0.26 and x = 0, 1,2, …, 18 Problem 10 Lottery It is not a binomial experiment because probability of success is not same for each trial. Larson/Farber 4th ed 38

    39. Binomial Probability Formula 39 Larson/Farber 4th ed Binomial Probability Formula The probability of exactly x successes in n trials is

    40. Example: Finding Binomial Probabilities Microfracture knee surgery has a 75% chance of success on patients with degenerative knees. The surgery is performed on three patients. Find the probability of the surgery being successful on exactly two patients. Larson/Farber 4th ed 40

    41. Solution: Finding Binomial Probabilities Method 1: Draw a tree diagram and use the Multiplication Rule Larson/Farber 4th ed 41

    42. Solution: Finding Binomial Probabilities Method 2: Binomial Probability Formula Larson/Farber 4th ed 42

    43. Binomial Probability Distribution Binomial Probability Distribution List the possible values of x with the corresponding probability of each. Example: Binomial probability distribution for Microfacture knee surgery: n = 3, p = Use binomial probability formula to find probabilities. Larson/Farber 4th ed 43

    44. Example: Constructing a Binomial Distribution In a survey, workers in the U.S. were asked to name their expected sources of retirement income. Seven workers who participated in the survey are randomly selected and asked whether they expect to rely on Social Larson/Farber 4th ed 44

    45. Solution: Constructing a Binomial Distribution 25% of working Americans expect to rely on Social Security for retirement income. n = 7, p = 0.25, q = 0.75, x = 0, 1, 2, 3, 4, 5, 6, 7 Larson/Farber 4th ed 45

    46. Solution: Constructing a Binomial Distribution Larson/Farber 4th ed 46

    47. Example: Finding Binomial Probabilities A survey indicates that 41% of women in the U.S. consider reading their favorite leisure-time activity. You randomly select four U.S. women and ask them if reading is their favorite leisure-time activity. Find the probability that at least two of them respond yes. Larson/Farber 4th ed 47

    48. Solution: Finding Binomial Probabilities Larson/Farber 4th ed 48

    49. Example: Finding Binomial Probabilities Using Technology The results of a recent survey indicate that when grilling, 59% of households in the United States use a gas grill. If you randomly select 100 households, what is the probability that exactly 65 households use a gas grill? Use a technology tool to find the probability. (Source: Greenfield Online for Weber-Stephens Products Company) Larson/Farber 4th ed 49

    50. Solution: Finding Binomial Probabilities Using Technology Larson/Farber 4th ed 50

    51. Example: Finding Binomial Probabilities Using a Table About thirty percent of working adults spend less than 15 minutes each way commuting to their jobs. You randomly select six working adults. What is the probability that exactly three of them spend less than 15 minutes each way commuting to work? Use a table to find the probability. (Source: U.S. Census Bureau) Larson/Farber 4th ed 51

    52. Solution: Finding Binomial Probabilities Using a Table A portion of Table 2 is shown Larson/Farber 4th ed 52

    53. Example: Graphing a Binomial Distribution Fifty-nine percent of households in the U.S. subscribe to cable TV. You randomly select six households and ask each if they subscribe to cable TV. Construct a probability distribution for the random variable x. Then graph the distribution. (Source: Kagan Research, LLC) Larson/Farber 4th ed 53

    54. Solution: Graphing a Binomial Distribution Larson/Farber 4th ed 54

    55. Textbook Exercises. Page 217 Problem 20 Honeymoon Financing n = 20 married couples p = 0.70 (probability of success) Follow the procedure described bellow to find the probabilities using TI 83/84 P(x = 1) Press 2nd key and DISTR to see the list of different distributions. Scroll down until you see binompdf(. Choose that option and then enter the values of n comma p comma x in this order and close the parenthesis. Hit ENTER. binompdf(20, 0.7, 1) ? 1.627 ? 10 -9 P(x ? 1) = 1 – [P(x = 0) + P( x = 1)] (using complement rule) = 1 – [ binompdf(20, 0.7,0) + binompdf(20 , 0.7, 1)] = 1 – [ 3.487 ? 10-11 + 1.627 ? 10-9 ] = 0.99999 ? 1 P (x ? 1) = 1 – P(x ? 1) (using complement rule) = 1 – 0.9999 ? 0 Larson/Farber 4th ed 55

    56. Textbook Exercises. Page 217 Problem 24 Career Advancement n = 12 executives p = 0.24 (probability of success) Follow the procedure described bellow to find the probabilities using TI 83/84 P(x = 4) Press 2nd key and DISTR to see the list of different distributions. Scroll down until you see binompdf(. Choose that option and then enter the values of n comma p comma x in this order and close the parenthesis. Hit ENTER. binompdf(12, 0.24, 4) = 0.183 P(x ? 4) = 1 – P(x ? 4) (using complement rule) = 1 – [P(0) + P(1) + P(2) + P(3) + P(4)] = 1 – [0.037 + 0.141 + 0.244 + 0.257 + 0.183 ] = 0.138 P (4 ? x ? 8) = P(4) + P(5) + P(6) + P(7) + P(8) = 0.183 + 0.092 + 0.034 + 0.009 + 0.002 = 0.32 Larson/Farber 4th ed 56

    57. Mean, Variance, and Standard Deviation Mean: µ = np Variance: s2 = npq Standard Deviation: Larson/Farber 4th ed 57

    58. Example: Finding the Mean, Variance, and Standard Deviation In Pittsburgh, Pennsylvania, about 56% of the days in a year are cloudy. Find the mean, variance, and standard deviation for the number of cloudy days during the month of June. Interpret the results and determine any unusual values. (Source: National Climatic Data Center) Larson/Farber 4th ed 58

    59. Solution: Finding the Mean, Variance, and Standard Deviation Larson/Farber 4th ed 59

    60. Section 4.2 Summary Determined if a probability experiment is a binomial experiment Found binomial probabilities using the binomial probability formula Found binomial probabilities using technology and a binomial table Graphed a binomial distribution Found the mean, variance, and standard deviation of a binomial probability distribution Larson/Farber 4th ed 60

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