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- 1. Chapter 4 Discrete Probability Distributions Larson/Farber 4th ed 1
- 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 Larson/Farber 4th ed 3
- 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. Larson/Farber 4th ed 5
- 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. x 0 Larson/Farber 4th ed 1 2 3 4 5 6
- 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. x 0 Larson/Farber 4th ed 1 2 3 … 24 7
- 8. Example: Random Variables Decide whether the random variable x is discrete or continuous. 1. x = The number of stocks in the Dow Jones Industrial Average that have share price increases on a given day. Solution: Discrete random variable (The number of stocks whose share price increases can be counted.) x 0 Larson/Farber 4th ed 1 2 3 … 30 8
- 9. Example: Random Variables Decide whether the random variable x is discrete or continuous. 2. x = The volume of water in a 32-ounce container. Solution: Continuous random variable (The amount of water can be any volume between 0 ounces and 32 ounces) x 0 Larson/Farber 4th ed 1 2 3 … 32 9
- 10. Textbook Exercises. Page 197 • Distinguish between Discrete and Continuous random variable 14. 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. 16. x represents number of tornadoes in the month of June in Oklahoma. x is a discrete random variable because number of tornadoes 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: In Words In Symbols 1. The probability of each value of the discrete random variable is between 0 and 1, inclusive. 0 ≤ P (x) ≤ 1 2. The sum of all the probabilities is 1. ΣP (x) = 1 Larson/Farber 4th ed 11
- 12. What is a PROBABILITY? 0% 25% 0 Impossible Certain 50% 75% ¼ or .25 Not Very Likely Larson/Farber 4th ed 100% ½ 0r .5 ¾ or .75 Equally Likely 1 Somewhat Likely 12
- 13. Constructing a Discrete Probability Distribution Let x be a discrete random variable with possible outcomes x1, x2, … , xn. 1. Make a frequency distribution for the possible outcomes. 2. Find the sum of the frequencies. 3. Find the probability of each possible outcome by dividing its frequency by the sum of the frequencies. 4. Check that each probability is between 0 and 1 and that the sum is 1. Larson/Farber 4th ed 13
- 14. 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 aggressive. A score of 3 indicated Score, x Frequency, f neither trait. Construct a 1 24 probability distribution for the 2 33 random variable x. Then graph the 3 42 4 30 distribution using a histogram. 5 Larson/Farber 4th ed 21 14
- 15. 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. P (1) = 24 = 0.16 150 30 P (4) = = 0.20 150 P (2) = 33 = 0.22 150 42 P (3) = = 0.28 150 21 P (5) = = 0.14 150 • Discrete probability distribution: x 1 2 3 4 5 P(x) 0.16 0.22 0.28 0.20 0.14 Larson/Farber 4th ed 15
- 16. Solution: Constructing a Discrete Probability Distribution x 1 2 3 4 5 P(x) 0.16 0.22 0.28 0.20 0.14 This is a valid discrete probability distribution since 1. Each probability is between 0 and 1, inclusive, 0 ≤ P(x) ≤ 1. 2. The sum of the probabilities equals 1, ΣP(x) = 0.16 + 0.22 + 0.28 + 0.20 + 0.14 = 1. Larson/Farber 4th ed 16
- 17. Solution: Constructing a Discrete Probability Distribution • Histogram Because the width of each bar is one, the area of each bar is equal to the probability of a particular outcome. Larson/Farber 4th ed 17
- 18. Textbook Exercises. Page 198 • Problem 22 Blood Donations x being number of donations is countable and hence a discrete random variable. Also, number of donations are mutually exclusive. a. 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 b. P(X < 3) • Problem 24 x 0 P(x) 0.05 = P(x = 2) + P(x = 1) + P(x = 0) = 0.25 + 0.25 + 0.30 = 0.80 Missing Probability 1 2 3 ? 0.23 0.21 4 5 6 0.17 0.11 0.08 Since it is a probability distribution sum of all the probabilities is equal to one. missing value = 1 – sum of rest of the probabilities Larson/Farber 4th ed = 1 – 0.85 = 0.15 18
- 19. Mean Mean of a discrete probability distribution • μ = ΣxP(x) • Each value of x is multiplied by its corresponding probability and the products are added. Larson/Farber 4th ed 19
- 20. Example: Finding the Mean The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the mean. x P(x) xP(x) 1 0.16 1(0.16) = 0.16 2 0.22 2(0.22) = 0.44 3 0.28 3(0.28) = 0.84 4 0.20 4(0.20) = 0.80 5 Solution: 0.14 5(0.14) = 0.70 μ = ΣxP(x) = 2.94 Larson/Farber 4th ed 20
- 21. Variance and Standard Deviation Variance of a discrete probability distribution • σ2 = Σ(x – μ)2P(x) Standard deviation of a discrete probability distribution • σ = σ 2 = Σ( x − µ ) 2 P ( x) Larson/Farber 4th ed 21
- 22. 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) x 1 0.16 2 0.22 3 0.28 4 0.20 5 Larson/Farber 4th ed P(x) 0.14 22
- 23. Solution: Finding the Variance and Standard Deviation Recall μ = 2.94 x P(x) x–μ (x – μ)2 (x – μ)2P(x) 1 0.16 1 – 2.94 = –1.94 (–1.94)2 = 3.764 3.764(0.16) = 0.602 2 0.22 2 – 2.94 = –0.94 (–0.94)2 = 0.884 0.884(0.22) = 0.194 3 0.28 3 – 2.94 = 0.06 (0.06)2 = 0.004 0.004(0.28) = 0.001 4 0.20 4 – 2.94 = 1.06 (1.06)2 = 1.124 1.124(0.20) = 0.225 5 0.14 5 – 2.94 = 2.06 (2.06)2 = 4.244 4.244(0.14) = 0.594 σ Variance:2 = Σ(x – μ)2P(x) = 1.616 σ = σ 2 = 1.616 ≈ 1.3 Standard Deviation: Larson/Farber 4th ed 23
- 24. Textbook Exercises. Page 199 • Problem 30 Camping Chairs Construct the probability distribution and determine : a. Mean b. Standard Deviation c. Variance Defects, x Batches, f P(x) 0 95 95 ÷ 380 = 0.250 1 113 113 ÷ 380 = 0.297 2 87 0.229 3 64 0.168 4 13 0.034 5 8 0.021 n = 380 ∑ P(x) = 1 sample mean x = 1.5 sample standard deviation ≈ 1.24 sample variance σ 2 ≈ 1.54 Larson/Farber 4th ed To determine the mean, standard deviation and variance we will use TI 83/84. First store the x-values and the P(x) values in two of the lists, say L1 and L2. Make sure that x goes into L1 and P(x) goes into L2. Do not reverse that order. Then using 1-var stats from CALC menu enter L1 comma L2 and hit the ENTER key. You may notice that the sample standard deviation is not shown but you may use population standard deviation to get an estimate. Double check your results with hand calculations. 24
- 25. Expected Value Expected value of a discrete random variable • Equal to the mean of the random variable. • E(x) = μ = ΣxP(x) Larson/Farber 4th ed 25
- 26. 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? Larson/Farber 4th ed 26
- 27. 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 Larson/Farber 4th ed 27
- 28. Solution: Finding an Expected Value • Probability distribution for the possible gains (outcomes) Gain, x $498 $248 $148 $73 –$2 P(x) 1 1500 1 1500 1 1500 1 1500 1496 1500 E (x ) = ΣxP (x ) 1 1 1 1 1496 = $498 × + $248 × + $148 × + $73 × + (−$2) × 1500 1500 1500 1500 1500 = −$1.35 You can expect to lose an average of $1.35 for each ticket you buy. Larson/Farber 4th ed 28
- 29. 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 29
- 30. Section 4.2 Binomial Distributions Larson/Farber 4th ed 30
- 31. 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 31
- 32. Binomial Experiments 1. The experiment is repeated for a fixed number of trials, where each trial is independent of other trials. 2. 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). 3. The probability of a success P(S) is the same for each trial. 4. The random variable x counts the number of successful trials. Larson/Farber 4th ed 32
- 33. Notation for Binomial Experiments Symbol n Description The number of times a trial is repeated p = P(s) q = P(F) The probability of success in a single trial The probability of failure in a single trial (q = 1 – p) x The random variable represents a count of the number of successes in n trials: x = 0, 1, 2, 3, … , n. Larson/Farber 4th ed 33
- 34. 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. 1. A certain surgical procedure has an 85% chance of success. A doctor performs the procedure on eight patients. The random variable represents the number of successful surgeries. Larson/Farber 4th ed 34
- 35. Solution: Binomial Experiments Binomial Experiment 1. Each surgery represents a trial. There are eight surgeries, and each one is independent of the others. 2. There are only two possible outcomes of interest for each surgery: a success (S) or a failure (F). 3. The probability of a success, P(S), is 0.85 for each surgery. 4. The random variable x counts the number of successful surgeries. Larson/Farber 4th ed 35
- 36. 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 36
- 37. 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. 2. A jar contains five red marbles, nine blue marbles, and six green marbles. You randomly select three marbles from the jar, without replacement. The random variable represents the number of red marbles. Larson/Farber 4th ed 37
- 38. 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 38
- 39. Textbook Exercises. Pages 211 - 212 • Problem 4 Graphical Analysis a) p = 0.75 (graph is skewed to left implies p is greater than 0.5 ) b) p = 0.50 (graph is symmetric) c) p = 0.25 (graph is skewed to right implies p is smaller than 0.5) • Problem 10 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 12 Lottery It is not a binomial experiment because probability of success is not same for each trial. Larson/Farber 4th ed 39
- 40. Binomial Probability Formula Binomial Probability Formula • The probability of exactly x successes in n trials is P( x) = n Cx p q x • • • • n− x n! = p x q n− x (n − x)! x ! n = number of trials p = probability of success q = 1 – p probability of failure x = number of successes in n trials Larson/Farber 4th ed 40
- 41. 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 41
- 42. Solution: Finding Binomial Probabilities Method 1: Draw a tree diagram and use the Multiplication Rule 9 P (2 successful surgeries ) = 3 ÷ ≈ 0.422 64 Larson/Farber 4th ed 42
- 43. Solution: Finding Binomial Probabilities Method 2: Binomial Probability Formula n = 3, 3 1 p = , q = 1− p = , x = 2 4 4 2 3 P (2 successful surgeries ) = 3 C2 ÷ 4 3− 2 1 ÷ 4 2 3! 3 = ÷ (3 − 2)!2! 4 1 1 ÷ 4 9 1 27 = 3 ÷ ÷ = ≈ 0.422 16 4 64 Larson/Farber 4th ed 43
- 44. Binomial Probability Distribution Binomial Probability Distribution • List the possible values of x with the corresponding probability of each. • Example: Binomial probability distribution for 3 Microfacture knee surgery: n = 3, p = 4 x 0 1 2 3 P(x) 0.016 0.141 0.422 0.422 Use binomial probability formula to find probabilities. Larson/Farber 4th ed 44
- 45. 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 Security for retirement income. Create a binomial probability distribution for the number of workers who respond yes. Larson/Farber 4th ed 45
- 46. 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 P(x = 0) = 7C0(0.25)0(0.75)7 = 1(0.25)0(0.75)7 ≈ 0.1335 P(x = 1) = 7C1(0.25)1(0.75)6 = 7(0.25)1(0.75)6 ≈ 0.3115 P(x = 2) = 7C2(0.25)2(0.75)5 = 21(0.25)2(0.75)5 ≈ 0.3115 P(x = 3) = 7C3(0.25)3(0.75)4 = 35(0.25)3(0.75)4 ≈ 0.1730 P(x = 4) = 7C4(0.25)4(0.75)3 = 35(0.25)4(0.75)3 ≈ 0.0577 P(x = 5) = 7C5(0.25)5(0.75)2 = 21(0.25)5(0.75)2 ≈ 0.0115 P(x = 6) = 7C6(0.25)6(0.75)1 = 7(0.25)6(0.75)1 ≈ 0.0013 P(x = 7) = 7C7(0.25)7(0.75)0 = 1(0.25)7(0.75)0 ≈ 0.0001 Larson/Farber 4th ed 46
- 47. Solution: Constructing a Binomial Distribution x P(x) 0 0.1335 1 0.3115 2 0.3115 3 0.1730 4 0.0577 5 0.0115 6 0.0013 7 0.0001 Larson/Farber 4th ed All of the probabilities are between 0 and 1 and the sum of the probabilities is 1.00001 ≈ 1. 47
- 48. 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. Solution: • n = 4, p = 0.41, q = 0.59 • At least two means two or more. • Find the sum of P(2), P(3), and P(4). Larson/Farber 4th ed 48
- 49. Solution: Finding Binomial Probabilities P(x = 2) = 4C2(0.41)2(0.59)2 = 6(0.41)2(0.59)2 ≈ 0.351094 P(x = 3) = 4C3(0.41)3(0.59)1 = 4(0.41)3(0.59)1 ≈ 0.162654 P(x = 4) = 4C4(0.41)4(0.59)0 = 1(0.41)4(0.59)0 ≈ 0.028258 P(x ≥ 2) = P(2) + P(3) + P(4) ≈ 0.351094 + 0.162654 + 0.028258 ≈ 0.542 Larson/Farber 4th ed 49
- 50. 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) Solution: • Binomial with n = 100, p = 0.59, x = 65 Larson/Farber 4th ed 50
- 51. Solution: Finding Binomial Probabilities Using Technology From the displays, you can see that the probability that exactly 65 households use a gas grill is about 0.04. Larson/Farber 4th ed 51
- 52. 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) Solution: • Binomial with n = 6, p = 0.30, x = 3 Larson/Farber 4th ed 52
- 53. Solution: Finding Binomial Probabilities Using a Table • A portion of Table 2 is shown The probability that exactly three of the six workers spend less than 15 minutes each way commuting to work is 0.185. Larson/Farber 4th ed 53
- 54. 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) Solution: • n = 6, p = 0.59, q = 0.41 • Find the probability for each value of x Larson/Farber 4th ed 54
- 55. Solution: Graphing a Binomial Distribution x 0 1 2 3 4 5 6 P(x) 0.005 0.041 0.148 0.283 0.306 0.176 0.042 Histogram: Larson/Farber 4th ed 55
- 56. Textbook Exercises. Page 213 • Problem 22 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 a. 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 b. 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 c. P (x ≤ 1) = 1 – P(x > 1) (using complement rule) = 1 – 0.9999 ≈0 Larson/Farber 4th ed 56
- 57. Textbook Exercises. Page 213 • Problem 26 Movies on Phone n = 12 adults p = 0.25 (probability of success) Follow the procedure described bellow to find the probabilities using TI 83/84 a. 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, 0.25, 4) = 0.194 b. P(x > 4) = 1 – P(x ≤ 4) (using complement rule) = 1 – [P(0) + P(1) + P(2) + P(3) + P(4)] = 1 – [0.032 + 0.127 + 0.232 + 0.258 + 0.194 ] = 0.157 c. P (4 ≤ x ≤ 8) = P(4) + P(5) + P(6) + P(7) + P(8) = 0.194 + 0.103+ 0.04 + 0.011 + 0.002 = 0.35 Larson/Farber 4th ed 57
- 58. Mean, Variance, and Standard Deviation • Mean: μ = np • Variance: σ2 = npq • Standard Deviation: σ = npq Larson/Farber 4th ed 58
- 59. 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) Solution: n = 30, p = 0.56, q = 0.44 Mean: μ = np = 30∙0.56 = 16.8 Variance: σ2 = npq = 30∙0.56∙0.44 ≈ 7.4 Standard Deviation: σ = npq = 30 ×0.56 ×0.44 ≈ 2.7 Larson/Farber 4th ed 59
- 60. Solution: Finding the Mean, Variance, and Standard Deviation μ = 16.8 σ2 ≈ 7.4 σ ≈ 2.7 • On average, there are 16.8 cloudy days during the month of June. • The standard deviation is about 2.7 days. • Values that are more than two standard deviations from the mean are considered unusual. 16.8 – 2(2.7) =11.4, A June with 11 cloudy days would be unusual. 16.8 + 2(2.7) = 22.2, A June with 23 cloudy days would also be unusual. Larson/Farber 4th ed 60
- 61. 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 61

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