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B|B) here means P(B on 2nd|B on 1st) 3.25 In a standard deck, there are 52 cards. Twelve cards are face cards (F) and 40 cards are not face cards (N). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities. 216 Chapter 3 | Probability Topics Figure 3.14 a. Find P(FN OR NF). b. Find P(N|F). c. Find P(at most one face card). Hint: At most one face card means zero or one face card. d. Find P(at least one face card). Hint: At least one face card means one or two face cards. Example 3.26 A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 217 a. Which shows the probability that both kittens are tabby? ⎛ a. What is the probability that one kitten of each coloring is selected? ⎛ a. What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first? d. What is the probability of choosing two kittens of the same color? Solution 3.26 ⎠, b. ⎛ a. 4 8, d. 32 72 3.26 Suppose there are four red balls and three yellow balls in a box. Three balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected? Venn Diagram A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. Example 3.27 Suppose an experiment has the outcomes 1, 2, 3,..., 12 where each outcome has an equal chance of occurring. Let event A = {1, 2, 3, 4, 5, 6} and event B = {6, 7, 8, 9}. Then A AND B = {6} and A OR B = {1, 2, 3, 4, 5, 6, 7, 218 Chapter 3 | Probability Topics 8, 9}. The V |
enn diagram is as follows: Figure 3.15 3.27 Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event C = {green, blue, purple} and event P = {red, yellow, blue}. Then C AND P = {blue} and C OR P = {green, blue, purple, red, yellow}. Draw a Venn diagram representing this situation. Example 3.28 Flip two fair coins. Let A = tails on the first coin. Let B = tails on the second coin. Then A = {TT, TH} and B = {TT, HT}. Therefore, A AND B = {TT}. A OR B = {TH, TT, HT}. The sample space when you flip two fair coins is X = {HH, HT, TH, TT}. The outcome HH is in NEITHER A NOR B. The Venn diagram is as follows: Figure 3.16 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 219 3.28 Roll a fair, six-sided die. Let A = a prime number of dots is rolled. Let B = an odd number of dots is rolled. Then A = {2, 3, 5} and B = {1, 3, 5}. Therefore, A AND B = {3, 5}. A OR B = {1, 2, 3, 5}. The sample space for rolling a fair die is S = {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation. Example 3.29 Forty percent of the students at a local college belong to a club and 50 percent work part time. Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let C = student belongs to a club and PT = student works part time. Start by drawing a rectangle to represent the sample space. Then draw two circles or ovals inside the rectangle to represent the events of interest: belonging to a club (C) and working part time (PT). Always draw overlapping shapes to represent outcomes that are in both events. Figure 3.17 Label each piece of the diagram clearly and note the probability or frequency of each part. Start by labeling the overlapping section first. Note that the probabilities in C total 0.40 and the |
sum of the probabilities in PT is 0.50. The total of all probabilities displayed must be 1, representing 100 percent of the sample space. If a student is selected at random, find the following: a. b. c. d. e. the probability that the student belongs to a club. the probability that the student works part time. the probability that the student belongs to a club AND works part time. the probability that the student belongs to a club given that the student works part time. the probability that the student belongs to a club OR works part time. Solution 3.29 P(C) =.40 220 Chapter 3 | Probability Topics P(PT) =.50 P(C AND PT) =.05 P(C|PT) = P(C AND PT) P(PT) =.05.50 =.1 P(C OR PT) = P(C) + P(PT) − P(C AND PT) =.40 +.50 −.05 =.85 3.29 Fifty percent of the workers at a factory work a second job, 25 percent have a spouse who also works, and 5 percent work a second job and have a spouse who also works. Draw a Venn diagram showing the relationships. Let W = works a second job and S = spouse also works. Example 3.30 A person with type O blood and a negative Rh factor (Rh–) can donate blood to any person with any blood type. Four percent of African Americans have type O blood and a negative Rh factor, 5−10 percent of African Americans have the Rh– factor, and 51 percent have type O blood. Figure 3.18 The “O” circle represents the African Americans with type O blood. The “Rh––" oval represents the African Americans with the Rh– –factor. We will use the average of 5 percent and 10 percent, 7.5 percent, as the percentage of African Americans who have the Rh–– factor. Let O = African American with Type O blood and R = African American with Rh– –factor. a. P(O) = ___________ b. P(R) = ___________ c. P(O AND R) = ___________ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 221 d. P(O OR R) = ____________ e. f. In the Venn |
Diagram, describe the overlapping area using a complete sentence. In the Venn Diagram, describe the area in the rectangle but outside both the circle and the oval using a complete sentence. Solution 3.30 a. P(O) =.51 b. P(R) =.075 because an average of 7.5 percent of African Americans have the Rh– –factor. c. P(O AND R) = 0.04 because 4 percent of African Americans have both Type O blood and the Rh– –factor. d. P(O OR R) = P(O) + P(R) - P(O AND R) =.51 +.075 −.04 =.545 e. The area represents the African Americans that have type O blood and the Rh–– factor. f. The area represents the African Americans that have neither type O blood nor the Rh–– factor. 3.30 In a bookstore, the probability that the customer buys a novel is.6, and the probability that the customer buys a nonfiction book is.4. Suppose that the probability that the customer buys both is.2. a. Draw a Venn diagram representing the situation. b. Find the probability that the customer buys either a novel or a nonfiction book. c. In the Venn diagram, describe the overlapping area using a complete sentence. d. Suppose that some customers buy only compact disks. Draw an oval in your Venn diagram representing this event. 3.6 | Probability Topics 222 Chapter 3 | Probability Topics 3.1 Probability Topics Student Learning Outcomes • The student will use theoretical and empirical methods to estimate probabilities. • The student will appraise the differences between the two estimates. • The student will demonstrate an understanding of long-term relative frequencies. Do the Experiment Count out 40 mixed-color candies, which is approximately one small bag’s worth. Record the number of each color in Table 3.12. Use the information from this table to complete Table 3.13. Next, put the candies in a cup. The experiment is to pick two candies, one at a time. Do not look into the cup as you pick them. The first time through, replace the first candy before picking the second one. Record the results in the With Replacement column of Table 3.14. Do this 24 times. The second time through, after picking the first candy, do not replace it before picking the second one. Then, pick the second one. |
Record the results in the Without Replacement column section of Table 3.15. After you record the pick, put both candies back. Do this a total of 24 times, also. Use the data from Table 3.15 to calculate the empirical probability questions. Leave your answers in unreduced fractional form. Do not multiply out any fractions. Color Quantity Yellow (Y) Green (G) Blue (BL) Brown (B) Orange (O) Red (R) Table 3.12 Population With Replacement Without Replacement P(2 reds) P(R1B2 OR B1R2) P(R1 AND G2) P(G2|R1) P(no yellows) P(doubles) P(no doubles) Table 3.13 Theoretical Probabilities This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 223 NOTE G2 = green on second pick, R1 = red on first pick, B1 = brown on first pick, B2 = brown on second pick, doubles = both picks are the same color. With Replacement Without Replacement ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) Table 3.14 Empirical Results With Replacement Without Replacement P(2 reds) P(R1B2 OR B1 |
R2) P(R1 AND G2) P(G2|R1) P(no yellows) P(doubles) P(no doubles) Table 3.15 Empirical Probabilities Discussion Questions 1. Why are the With Replacement and Without Replacement probabilities different? 2. Convert P(no yellows) to decimal format for both Theoretical With Replacement and for Empirical With Replacement. Round to four decimal places. a. Theoretical With Replacement: P(no yellows) = _______ b. Empirical With Replacement: P(no yellows) = _______ c. Are the decimal values close? Did you expect them to be closer together or farther apart? Why? 3. If you increased the number of times you picked two candies to 240 times, why would empirical probability values change? 224 Chapter 3 | Probability Topics 4. Would this change (see Question 3) cause the empirical probabilities and theoretical probabilities to be closer together or farther apart? How do you know? 5. Explain the differences in what P(G1 AND R2) and P(R1|G2) represent. Hint: Think about the sample space for each probability. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 225 KEY TERMS conditional probability the likelihood that an event will occur given that another event has already occurred contingency table the method of displaying a frequency distribution as a table with rows and columns to show how two variables may be dependent (contingent) upon each other; the table provides an easy way to calculate conditional probabilities dependent events if two events are NOT independent, then we say that they are dependent equally likely each outcome of an experiment has the same probability event a subset of the set of all outcomes of an experiment; the set of all outcomes of an experiment is called a sample space and is usually denoted by S. An event is an arbitrary subset in S. It can contain one outcome, two outcomes, no outcomes (empty subset), the entire sample space, and the like. Standard notations for events are capital letters such as A, B, C, and so on experiment a planned activity carried out under controlled conditions independent events The occurrence of one event has no effect on the probability of the occurrence of another event; events A and B are independent if one of the following is true: 1. P(A|B) = P( |
A) 2. P(B|A) = P(B) 3. P(A AND B) = P(A)P(B) mutually exclusive two events are mutually exclusive if the probability that they both happen at the same time is zero; if events A and B are mutually exclusive, then P(A AND B) = 0 outcome a particular result of an experiment probability a number between zero and one, inclusive, that gives the likelihood that a specific event will occur; the foundation of statistics is given by the following three axioms (by A.N. Kolmogorov, 1930s): Let S denote the sample space and A and B are two events in S; then • 0 ≤ P(A) ≤ 1, • If A and B are any two mutually exclusive events, then P(A OR B) = P(A) + P(B), and • P(S) = 1 sample space the set of all possible outcomes of an experiment sampling with replacement if each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once sampling without replacement when sampling is done without replacement, each member of a population may be chosen only once the AND event an outcome is in the event A AND B if the outcome is in both A AND B at the same time the complement event the complement of event A consists of all outcomes that are NOT in A the conditional probability of one event GIVEN another event P(A|B) is the probability that event A will occur given that the event B has already occurred the OR event an outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B the OR of two events an outcome is in the event A OR B if the outcome is in A, is in B, or is in both A and B tree diagram the useful visual representation of a sample space and events in the form of a tree with branches marked by possible outcomes together with associated probabilities (frequencies, relative frequencies) Venn diagram the visual representation of a sample space and events in the form of circles or ovals showing their 226 Chapter 3 | Probability Topics intersections CHAPTER REVIEW 3.1 Terminology In this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, |
inclusive. 3.2 Independent and Mutually Exclusive Events Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other. 3.3 Two Basic Rules of Probability The multiplication rule and the addition rule are used for computing the probability of A and B, as well as the probability of A or B for two given events A, B defined on the sample space. In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events A and B are mutually exclusive events when they do not have any outcomes in common. 3.4 Contingency Tables There are several tools you can use to help organize and sort data when calculating probabilities. Contingency tables, also known as two-way tables, help display data and are particularly useful when calculating probabilites that have multiple dependent variables. 3.5 Tree and Venn Diagrams A tree diagram uses branches to show the different outcomes of experiments and makes complex probability questions easy to visualize. A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. A Venn diagram is especially helpful for visualizing the OR event, the AND event, and the complement of an event and for understanding conditional probabilities. FORMULA REVIEW 3.1 Terminology A and B are events P(S) = 1 where S is the sample space 0 ≤ P(A) ≤ 1 P(A|B) = P(AANDB) P(B) 3.2 Independent and Mutually Exclusive Events If A and B are independent, P(A AND B) = P(A)P(B), This OpenStax book is available for free at |
http://cnx.org/content/col30309/1.8 P(A|B) = P(A), and P(B|A) = P(B). If A and B are mutually exclusive, P(A OR B) = P(A) + P(B) and P(A AND B) = 0. 3.3 Two Basic Rules of Probability The multiplication rule—P(A AND B) = P(A|B)P(B) The addition rule—P(A OR B) = P(A) + P(B) − P(A AND B) Chapter 3 | Probability Topics 227 PRACTICE 3.1 Terminology 1. In a particular college class, there are male and female students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts A through J of this question. Note that you cannot find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols. • Let F be the event that a student is female. • Let M be the event that a student is male. • Let S be the event that a student has short hair. • Let L be the event that a student has long hair. a. The probability that a student does not have long hair. b. The probability that a student is male or has short hair. c. The probability that a student is female and has long hair. d. The probability that a student is male, given that the student has long hair. e. The probability that a student has long hair, given that the student is male. f. Of all female students, the probability that a student has short hair. g. Of all students with long hair, the probability that a student is female. h. The probability that a student is female or has long hair. i. The probability that a randomly selected student is a male student with short hair. j. The probability that a student is female. Use the following information to answer the next four exercises. A box is filled with several party favors. It contains 12 hats, 15 noisemakers, 10 finger traps, and five bags of confetti. Let H = the event of getting a hat. Let N = the event of getting a noisemaker. Let F = the event of getting a finger trap. Let C = the event of getting a bag of confetti. 2. Find P |
(H). 3. Find P(N). 4. Find P(F). 5. Find P(C). Use the following information to answer the next six exercises. A jar of 150 jelly beans contains 22 red jelly beans, 38 yellow, 20 green, 28 purple, 26 blue, and the rest are orange. Let B = the event of getting a blue jelly bean Let G = the event of getting a green jelly bean. Let O = the event of getting an orange jelly bean. Let P = the event of getting a purple jelly bean. Let R = the event of getting a red jelly bean. Let Y = the event of getting a yellow jelly bean. 6. Find P(B). 7. Find P(G). 8. Find P(P). 9. Find P(R). 10. Find P(Y). 11. Find P(O). Use the following information to answer the next six exercises. There are 23 countries in North America, 12 countries in South America, 47 countries in Europe, 44 countries in Asia, 54 countries in Africa, and 14 countries in Oceania (Pacific Ocean region). Let A = the event that a country is in Asia. 228 Chapter 3 | Probability Topics Let E = the event that a country is in Europe. Let F = the event that a country is in Africa. Let N = the event that a country is in North America. Let O = the event that a country is in Oceania. Let S = the event that a country is in South America. 12. Find P(A). 13. Find P(E). 14. Find P(F). 15. Find P(N). 16. Find P(O). 17. Find P(S). 18. What is the probability of drawing a red card in a standard deck of 52 cards? 19. What is the probability of drawing a club in a standard deck of 52 cards? 20. What is the probability of rolling an even number of dots with a fair, six-sided die numbered one through six? 21. What is the probability of rolling a prime number of dots with a fair, six-sided die numbered one through six? Use the following information to answer the next two exercises. You see a game at a local fair. You have to throw a dart at a color wheel. Each section on the color wheel is equal in area. Figure 3.19 Let B = the event of landing on blue. Let R = the event of landing on red. Let |
G = the event of landing on green. Let Y = the event of landing on yellow. 22. If you land on Y, you get the biggest prize. Find P(Y). 23. If you land on red, you don’t get a prize. What is P(R)? Use the following information to answer the next 10 exercises. On a baseball team, there are infielders and outfielders. Some players are great hitters, and some players are not great hitters. Let I = the event that a player in an infielder. Let O = the event that a player is an outfielder. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 229 Let H = the event that a player is a great hitter. Let N = the event that a player is not a great hitter. 24. Write the symbols for the probability that a player is not an outfielder. 25. Write the symbols for the probability that a player is an outfielder or is a great hitter. 26. Write the symbols for the probability that a player is an infielder and is not a great hitter. 27. Write the symbols for the probability that a player is a great hitter, given that the player is an infielder. 28. Write the symbols for the probability that a player is an infielder, given that the player is a great hitter. 29. Write the symbols for the probability that of all the outfielders, a player is not a great hitter. 30. Write the symbols for the probability that of all the great hitters, a player is an outfielder. 31. Write the symbols for the probability that a player is an infielder or is not a great hitter. 32. Write the symbols for the probability that a player is an outfielder and is a great hitter. 33. Write the symbols for the probability that a player is an infielder. 34. What is the word for the set of all possible outcomes? 35. What is conditional probability? 36. A shelf holds 12 books. Eight are fiction and the rest are nonfiction. Each is a different book with a unique title. The fiction books are numbered one to eight. The nonfiction books are numbered one to four. Randomly select one book Let F = event that book is fiction Let N = event that book is nonfiction What is the sample space? 37. What is the sum of the probabilities of an event and its complement? Use the following information to |
answer the next two exercises. You are rolling a fair, six-sided number cube. Let E = the event that it lands on an even number. Let M = the event that it lands on a multiple of three. 38. What does P(E|M) mean in words? 39. What does P(E OR M) mean in words? 3.2 Independent and Mutually Exclusive Events 40. E and F are mutually exclusive events. P(E) =.4; P(F) =.5. Find P(E∣F). 41. J and K are independent events. P(J|K) =.3. Find P(J). 42. U and V are mutually exclusive events. P(U) =.26; P(V) =.37. Find the following: a. P(U AND V) = b. P(U|V) = c. P(U OR V) = 43. Q and R are independent events. P(Q) =.4 and P(Q AND R) =.1. Find P(R). 3.3 Two Basic Rules of Probability Use the following information to answer the next 10 exercises. Forty-eight percent of all voters of a certain state prefer life in prison without parole over the death penalty for a person convicted of first-degree murder. Among Latino registered voters in this state, 55 percent prefer life in prison without parole over the death penalty for a person convicted of first-degree murder. Of all citizens in this state, 37.6 percent are Latino. In this problem, let • C = citizens of a certain state (registered voters) preferring life in prison without parole over the death penalty for a person convicted of first-degree murder. • L = registered voters of the state who are Latino. Suppose that one citizen is randomly selected. 44. Find P(C). Chapter 3 | Probability Topics 230 45. Find P(L). 46. Find P(C|L). 47. In words, what is C|L? 48. Find P(L AND C). 49. In words, what is L AND C? 50. Are L and C independent events? Show why or why not. 51. Find P(L OR C). 52. In words, what is L OR C? 53. Are L and C mutually exclusive events? Show why or why not. 3.4 Contingency Tables Use the following information to answer the next four exercises. |
Table 3.16 shows a random sample of musicians and how they learned to play their instruments. Gender Self-Taught Studied in School Private Instruction Total Female Male Total 12 19 31 Table 3.16 38 24 62 22 15 37 72 58 130 54. Find P(musician is a female). 55. Find P(musician is a male AND had private instruction). 56. Find P(musician is a female OR is self taught). 57. Are the events being a female musician and learning music in school mutually exclusive events? 3.5 Tree and Venn Diagrams 58. The probability that a man develops some form of cancer in his lifetime is 0.4567. The probability that a man has at least one false-positive test result, meaning the test comes back for cancer when the man does not have it, is.51. Let C = a man develops cancer in his lifetime; P = a man has at least one false-positive test. Construct a tree diagram of the situation. BRINGING IT TOGETHER: PRACTICE Use the following information to answer the next seven exercises. An article in the New England Journal of Medicine, reported about a study of people who use a product in California and Hawaii. In one part of the report, the self-reported ethnicity and using the product levels per day were given. Of the people using the product at most 10 times a day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans, and 7,650 whites. Of the people using the product 11 to 20 times per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 whites. Of the people using the product 21 to 30 times per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 Whites. Of the people using the product at least 31 times per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 whites. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 231 59. Complete the table using the data provided. Suppose that one person from the study is randomly selected. |
Find the probability that person used the product 11 to 20 times a day. Product Use (times per day) African Americans Native Hawaiians Latinos Japanese Americans Whites TOTALS 1–10 11–20 21–30 31+ TOTALS Table 3.17 Product Use by Ethnicity 60. Suppose that one person from the study is randomly selected. Find the probability that the person used the product 11 to 20 times per day. 61. Find the probability that the person was Latino. 62. In words, explain what it means to pick one person from the study who is Japanese American AND uses the product 21 to 30 times per day. Also, find the probability. 63. In words, explain what it means to pick one person from the study who is Japanese American OR uses the product 21 to 30 times per day. Also, find the probability. 64. In words, explain what it means to pick one person from the study who is Japanese American GIVEN that the person uses the product 21 to 30 times per day. Also, find the probability. 65. Prove that product use/day and ethnicity are dependent events. HOMEWORK 232 Chapter 3 | Probability Topics 3.1 Terminology 66. Figure 3.20 The graph in Figure 3.20 displays the sample sizes and percentages of people in different age and gender groups who were polled concerning their approval of Mayor Ford’s actions in office. The total number in the sample of all the age groups is 1,045. a. Define three events in the graph. b. Describe in words what the entry 40 means. c. Describe in words the complement of the entry in the previous question. d. Describe in words what the entry 30 means. e. Out of the males and females, what percent are males? f. Out of the females, what percent disapprove of Mayor Ford? g. Out of all the age groups, what percent approve of Mayor Ford? h. Find P(Approve|Male). i. Out of the age groups, what percent are more than 44 years old? j. Find P(Approve|Age < 35). 67. Explain what is wrong with the following statements. Use complete sentences. a. If there is a 60 percent chance of rain on Saturday and a 70 percent chance of rain on Sunday, then there is a 130 percent chance of rain over the weekend. b. The probability that a baseball player hits a home run is greater than the probability that he gets a successful |
hit. 3.2 Independent and Mutually Exclusive Events Use the following information to answer the next 12 exercises. The graph shown is based on more than 170,000 interviews that took place from January through December 2012. The sample consists of employed Americans 18 years of age or older. The Health Index Scores are the sample space. We randomly sample one type of Health Index Score, the emotional wellbeing score. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 233 Figure 3.21 68. Find the probability that a Health Index Score is 82.7. 69. Find the probability that a Health Index Score is 81.0. 70. Find the probability that a Health Index Score is more than 81. 71. Find the probability that a Health Index Score is between 80.5 and 82. 72. If we know a Health Index Score is 81.5 or more, what is the probability that it is 82.7? 73. What is the probability that a Health Index Score is 80.7 or 82.7? 74. What is the probability that a Health Index Score is less than 80.2 given that it is already less than 81? 75. What occupation has the highest Health Index Score? 76. What occupation has the lowest emotional index score? 77. What is the range of the data? 78. Compute the average Health Index Score. 79. If all occupations are equally likely for a certain individual, what is the probability that he or she will have an occupation with lower than average Health Index Score? 234 Chapter 3 | Probability Topics 3.3 Two Basic Rules of Probability 80. On February 28, 2013, a Field Poll Survey reported that 61 percent of California registered voters approved of a law that was about to be passed. Among 18- to 39-year olds (California registered voters), the approval rating was 78 percent. Six in 10 California registered voters said that the upcoming Supreme Court’s ruling about the constitutionality of the law was either very or somewhat important to them. Out of those registered voters who supported the law, 75 percent say the ruling is important to them. In this problem, let • C = California registered voters who supported the law, • B = California registered voters who say the Supreme Court’s ruling about the law is very or somewhat important to them, and • A = California registered voters who are 18 to 39 years old. a. Find P(C). |
b. Find P(B). c. Find P(C|A). d. Find P(B|C). e. f. g. Find P(C AND B). h. i. Find P(C OR B). j. Are C and B mutually exclusive events? Show why or why not. In words, what is C|A? In words, what is B|C? In words, what is C AND B? 81. After a mayor of a major Canadian city announced his plans to cut budget costs in late 2011, researchers polled 1,046 people to measure the mayor’s popularity. Everyone polled expressed either approval or disapproval. These are the results their poll produced: • • • In early 2011, 60 percent of the population approved of the mayor's actions in office. In mid-2011, 57 percent of the population approved of his actions. In late 2011, the percentage of popular approval was measured at 42 percent. a. What is the sample size for this study? b. What proportion in the poll disapproved of the mayor, according to the results from late 2011? c. How many people polled responded that they approved of the mayor in late 2011? d. What is the probability that a person supported the mayor, based on the data collected in mid-2011? e. What is the probability that a person supported the mayor, based on the data collected in early 2011? Use the following information to answer the next three exercises. A local restaurant sells pork chops and chicken breasts. The given values below are the weights (in ounces) of pork chops and chicken breasts listed on the menu. Your server will randomly select one piece of meat (pork chop or chicken breast) that you will be served. 17 20 21 18 20 20 20 18 19 19 20 19 21 20 18 20 20 19 18 19 17 19 17 21 17 21 18 21 19 21 20 17 20 18 19 20 20 17 21 20 Pork Chops Chicken Breasts Table 3.18 82. a. List the sample space of the possible items that are on the menu. b. Find P(you will get a 17-oz. piece of meat). c. Find P(you will get a pork chop). d. Find P(you will get a 17-oz. pork chop). e. f. Find two mutually exclusive events. g. Are the events getting 17 oz. of meat and getting a pork chop independent? Is getting a pork chop the complement of getting a chicken breast? Why? |
This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 235 83. Compute the probabilities. a. P(you will get a chicken breast) b. P(you will get a 17-oz. chicken breast) c. P(you will get a chicken breast or you will not get a 17-oz. pork chop) d. P(you will not get a chicken breast and you will get an 18-oz. pork chop) e. P(you will get a piece of meat that is not 21 oz.) f. P(you will get a piece of chicken that is not 21 oz.) g. P(you will not get a chicken breast and you will not get a pork chop) 84. Compute the probabilities: a. P(you will not get a pork chop) b. P(you will get a 20-oz. pork chop) c. P(you will not get a chicken breast or you will not get an 18-oz. pork chop) d. P(you will not get a chicken breast and you will not get an 18-oz. pork chop) e. P(you will get a pork chop that is not 21 oz.) f. P(you will not get a chicken breast or you will not get a pork chop) 85. Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card. • G = card drawn is green • E = card drawn is even-numbered a. List the sample space. b. P(G) = ________ c. P(G|E) = ________ d. P(G AND E) = ________ e. P(G OR E) = ________ f. Are G and E mutually exclusive? Justify your answer numerically. 86. Roll two fair dice separately. Each die has six faces. a. List the sample space. b. Let A be the event that either a three or four is rolled first, followed by an even number. Find P(A). c. Let B be the event that the sum of the two rolls is at most seven. Find P(B). d. e. Are A and B mutually exclusive events? Explain |
your answer in one to three complete sentences, including In words, explain what P(A|B) represents. Find P(A|B). numerical justification. f. Are A and B independent events? Explain your answer in one to three complete sentences, including numerical justification. 87. A special deck of cards has 10 cards. Four are green, three are blue, and three are red. When a card is picked, its color is recorded. An experiment consists of first picking a card and then tossing a coin. a. List the sample space. b. Let A be the event that a blue card is picked first, followed by landing a head on the coin toss. Find P(A). c. Let B be the event that a red or green is picked, followed by landing a head on the coin toss. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification. d. Let C be the event that a red or blue is picked, followed by landing a head on the coin toss. Are the events A and C mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification. 88. An experiment consists of first rolling a die and then tossing a coin. a. List the sample space. b. Let A be the event that either a three or a four is rolled first, followed by landing a head on the coin toss. Find P(A). c. Let B be the event that the first and second tosses land on heads. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification. 89. An experiment consists of tossing a nickel, a dime, and a quarter. Of interest is the side the coin lands on. a. List the sample space. b. Let A be the event that there are at least two tails. Find P(A). c. Let B be the event that the first and second tosses land on heads. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including justification. 236 Chapter 3 | Probability Topics 90. Consider the following scenario: Let P(C) =.4. Let P(D) =.5. Let P(C|D) =.6. a. Find P(C AND D). b. Are C and D mutually exclusive? Why or why not? c. Are C and D independent events? Why or why not? d. |
Find P(C OR D). e. Find P(D|C). 91. Y and Z are independent events. a. Rewrite the basic Addition Rule P(Y OR Z) = P(Y) + P(Z) - P(Y AND Z) using the information that Y and Z are independent events. b. Use the rewritten rule to find P(Z) if P(Y OR Z) =.71 and P(Y) =.42. 92. G and H are mutually exclusive events. P(G) =.5 P(H) =.3 a. Explain why the following statement MUST be false: P(H|G) =.4. b. Find P(H OR G). c. Are G and H independent or dependent events? Explain in a complete sentence. 93. Approximately 281,000,000 people over age five live in the United States. Of these people, 55,000,000 speak a language other than English at home. Of those who speak another language at home, 62.3 percent speak Spanish. Let E = speaks English at home; E′ = speaks another language at home; and S = speaks Spanish. Finish each probability statement by matching the correct answer. Probability Statements Answers a. P(E′) = b. P(E) = c. P(S and E′) = d. P(S|E′) = Table 3.19 i..8043 ii..623 iii..1957 iv..1219 94. In 1994, the U.S. government held a lottery to issue 55,000 licenses of a certain type. Renate Deutsch, from Germany, was one of approximately 6.5 million people who entered this lottery. Let G = won license. a. What was Renate’s chance of winning one of the licenses? Write your answer as a probability statement. b. In the summer of 1994, Renate received a letter stating she was one of 110,000 finalists chosen. Once the finalists were chosen, assuming that each finalist had an equal chance to win, what was Renate’s chance of winning one of the licenses? Write your answer as a conditional probability statement. Let F = was a finalist. c. Are G and F independent or dependent events? Justify your answer numerically and also explain why. d. Are G and F mutually exclusive events? Justify your answer numerically and explain why. |
95. Three professors at George Washington University did an experiment to determine if economists are more likely to return found money than other people. They dropped 64 stamped, addressed envelopes with $10 cash in different classrooms on the George Washington campus. Forty-four percent were returned overall. From the economics classes 56 percent of the envelopes were returned. From the business, psychology, and history classes 31 percent were returned. Let R = money returned; E = economics classes; and O = other classes. a. Write a probability statement for the overall percentage of money returned. b. Write a probability statement for the percentage of money returned out of the economics classes. c. Write a probability statement for the percentage of money returned out of the other classes. d. e. Based upon this study, do you think that economists are more selfish than other people? Explain why or why not. Is money being returned independent of the class? Justify your answer numerically and explain it. Include numbers to justify your answer. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 237 96. The following table of data obtained from www.baseball-almanac.com shows hit information for four players. Suppose that one hit from the table is randomly selected. Name Single Double Triple Home Run Total Hits Babe Ruth 1,517 Jackie Robinson 1,054 Ty Cobb Hank Aaron Total Table 3.20 3,603 2,294 8,471 506 273 174 624 1,577 136 54 295 98 583 714 137 114 755 2,873 1,518 4,189 3,771 1,720 12,351 Are the hit being made by Hank Aaron and the hit being a double independent events? a. Yes, because P(hit by Hank Aaron|hit is a double) = P(hit by Hank Aaron) b. No, because P(hit by Hank Aaron|hit is a double) ≠ P(hit is a double) c. No, because P(hit is by Hank Aaron|hit is a double) ≠ P(hit by Hank Aaron) d. Yes, because P(hit is by Hank Aaron|hit is a double) = P(hit is a double) 97. United Blood Services is a blood bank that serves more than 500 hospitals in 18 states. According to their website, a person with type O blood and a negative Rh factor (Rh–) can donate blood |
to any person with any bloodtype. Their data show that 43 percent of people have type O blood and 15 percent of people have Rh– factor; 52 percent of people have type O or Rh– factor. a. Find the probability that a person has both type O blood and the Rh– factor. b. Find the probability that a person does not have both type O blood and the Rh– factor. 98. At a college, 72 percent of courses have final exams and 46 percent of courses require research papers. Suppose that 32 percent of courses have a research paper and a final exam. Let F be the event that a course has a final exam. Let R be the event that a course requires a research paper. a. Find the probability that a course has a final exam or a research project. b. Find the probability that a course has neither of these two requirements. 99. In a box of assorted cookies, 36 percent contain chocolate and 12 percent contain nuts. Of those, 8 percent contain both chocolate and nuts. Sean is allergic to both chocolate and nuts. a. Find the probability that a cookie contains chocolate or nuts (he can't eat it). b. Find the probability that a cookie does not contain chocolate or nuts (he can eat it). 100. A college finds that 10 percent of students have taken a distance learning class and that 40 percent of students are part-time students. Of the part-time students, 20 percent have taken a distance learning class. Let D = event that a student takes a distance learning class and E = event that a student is a part-time student. a. Find P(D AND E). b. Find P(E|D). c. Find P(D OR E). d. Using an appropriate test, show whether D and E are independent. e. Using an appropriate test, show whether D and E are mutually exclusive. 3.4 Contingency Tables Use the information in the Table 3.21 to answer the next eight exercises. The table shows the political party affiliation of each of 67 members of the U.S. Senate in June 2012, and when they would next be up for reelection. Up for Reelection: Democratic Party Republican Party Other Total November 2014 November 2016 20 10 Table 3.21 13 24 0 0 238 Chapter 3 | Probability Topics Up for Reelection: Democratic Party Republican Party Other Total Total Table 3.21 101. What is the probability that a randomly selected senator had an Other affiliation? 102. What is the probability that a |
randomly selected senator would be up for reelection in November 2016? 103. What is the probability that a randomly selected senator was a Democrat and was up for reelection in November 2016? 104. What is the probability that a randomly selected senator was a Republican or was up for reelection in November 2014? 105. Suppose that a member of the U.S. Senate is randomly selected. Given that the randomly selected senator was up for reelection in November 2016, what is the probability that this senator was a Democrat? 106. Suppose that a member of the U.S. Senate is randomly selected. What is the probability that the senator was up for reelection in November 2014, knowing that this senator was a Republican? 107. The events Republican and Up for reelection in 2016 are ________. independent a. mutually exclusive b. c. both mutually exclusive and independent d. neither mutually exclusive nor independent 108. The events Other and Up for reelection in November 2016 are ________. independent a. mutually exclusive b. c. both mutually exclusive and independent d. neither mutually exclusive nor independent Use the following information to answer the next two exercises. The table of data obtained from www.baseball-almanac.com shows hit information for four well-known baseball players. Suppose that one hit from the table is randomly selected. Name Single Double Triple Home Run Total Hits Babe Ruth 1,517 Jackie Robinson 1,054 Ty Cobb Hank Aaron TOTAL Table 3.22 3,603 2,294 8,471 506 273 174 624 1,577 136 54 295 98 583 714 137 114 755 2,873 1,518 4,189 3,771 1,720 12,351 109. Find P(Hit was made by Babe Ruth). a. b. c. d. 1,518 2,873 2,873 12,351 583 12,351 4,189 12,351 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 239 110. Find P(Hit was made by Ty Cobb|The hit was a Home Run). a. b. c. d. 4,189 12,351 114 1,720 1,720 4,189 114 12,351 111. Table 3.23 identifies a group of children by one of four hair colors, and by type of hair. Hair Type Brown Blond Black Red Totals 20 80 Wavy Straight Totals Table 3.23 15 3 12 43 |
215 15 20 a. Complete the table. b. What is the probability that a randomly selected child will have wavy hair? c. What is the probability that a randomly selected child will have either brown or blond hair? d. What is the probability that a randomly selected child will have wavy brown hair? e. What is the probability that a randomly selected child will have red hair, given that he or she has straight hair? f. g. If B is the event of a child having brown hair, find the probability of the complement of B. In words, what does the complement of B represent? 112. In a previous year, the weights of the members of a California football team and a Texas football team were published in a newspaper. The factual data were compiled into the following table. The weights in the column headings are in pounds. Shirt # ≤ 210 211–250 251–290 > 290 1–33 34–66 66–99 21 6 6 Table 3.24 5 18 12 0 7 22 0 4 5 For the following, suppose that you randomly select one player from the California team or the Texas team. a. Find the probability that his shirt number is from 1 to 33. b. Find the probability that he weighs at most 210 pounds. c. Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds. d. Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds. e. Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210 pounds. 3.5 Tree and Venn Diagrams Use the following information to answer the next two exercises. This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing three red (R), four yellow (Y), and five blue (B) beads. For the coin, P(H) = 2 3 where H is heads and T is tails. and P(T) = 1 3 240 Chapter 3 | Probability Topics Figure 3.22 113. Find P(tossing a head on the coin AND a red bead). a. b. c. d. 2 3 5 15 6 36 5 36 a. b. 114. Find P(blue bead). 15 36 10 36 10 12 6 36 d. c. 115. A box of cookies contains three chocolate and seven butter cookies. Miguel randomly selects a cookie and eats it. Then he randomly selects another |
cookie and eats it. How many cookies did he take? a. Draw the tree that represents the possibilities for the cookie selections. Write the probabilities along each branch of the tree. b. Are the probabilities for the flavor of the second cookie that Miguel selects independent of his first selection? Explain. c. For each complete path through the tree, write the event it represents and find the probabilities. d. Let S be the event that both cookies selected were the same flavor. Find P(S). e. Let T be the event that the cookies selected were different flavors. Find P(T) by two different methods by using the complement rule and by using the branches of the tree. Your answers should be the same with both methods. f. Let U be the event that the second cookie selected is a butter cookie. Find P(U). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 241 BRINGING IT TOGETHER: HOMEWORK 116. A previous year, the weights of the members of a California football team and a Texas football team were published in a newspaper The factual data are compiled into Table 3.25. Shirt# ≤ 210 211–250 251–290 290≤ 1–33 21 34–66 66–99 6 6 Table 3.25 5 18 12 0 7 22 0 4 5 For the following, suppose that you randomly select one player from the California team or the Texas team. If having a shirt number from one to 33 and weighing at most 210 pounds were independent events, then what should be true about P(Shirt# 1–33|≤ 210 pounds)? 117. The probability that a male develops some form of cancer in his lifetime is.4567. The probability that a male has at least one false-positive test result, meaning the test comes back for cancer when the man does not have it, is.51. Some of the following questions do not have enough information for you to answer them. Write not enough information for those answers. Let C = a man develops cancer in his lifetime and P = a man has at least one false-positive. a. P(C) = ______ b. P(P|C) = ______ c. P(P|C') = ______ d. If a test comes up positive, based upon numerical values, can you assume that man has cancer? Justify numerically and explain why or why not. |
118. Given events G and H: P(G) =.43; P(H) =.26; P(H AND G) =.14 a. Find P(H OR G). b. Find the probability of the complement of event (H AND G). c. Find the probability of the complement of event (H OR G). 119. Given events J and K: P(J) =.18; P(K) =.37; P(J OR K) =.45 a. Find P(J AND K). b. Find the probability of the complement of event (J AND K). c. Find the probability of the complement of event (J OR K). Use the following information to answer the next two exercises. Suppose that you have eight cards. Five are green and three are yellow. The cards are well shuffled. 120. Suppose that you randomly draw two cards, one at a time, with replacement. Let G1 = first card is green Let G2 = second card is green a. Draw a tree diagram of the situation. b. Find P(G1 AND G2). c. Find P(at least one green). d. Find P(G2|G1). e. Are G2 and G1 independent events? Explain why or why not. 121. Suppose that you randomly draw two cards, one at a time, without replacement. G1 = first card is green G2 = second card is green a. Draw a tree diagram of the situation. b. Find P(G1 AND G2). c. Find P(at least one green). d. Find P(G2|G1). e. Are G2 and G1 independent events? Explain why or why not. Use the following information to answer the next two exercises. The percent of licensed U.S. drivers (from a recent year) 242 Chapter 3 | Probability Topics who are female is 48.60. Of the females, 5.03 percent are age 19 and under; 81.36 percent are age 20–64; 13.61 percent are age 65 or over. Of the licensed U.S. male drivers, 5.04 percent are age 19 and under; 81.43 percent are age 20–64; 13.53 percent are age 65 or over. 122. Complete the following: a. Construct a table or a tree diagram of the situation. b. Find P(driver is female). c. Find P(driver is |
age 65 or over|driver is female). d. Find P(driver is age 65 or over AND female). e. f. Find P(driver is age 65 or over). g. Are being age 65 or over and being female mutually exclusive events? How do you know? In words, explain the difference between the probabilities in Part c and Part d. 123. Suppose that 10,000 U.S. licensed drivers are randomly selected. a. How many would you expect to be male? b. Using the table or tree diagram, construct a contingency table of gender versus age group. c. Using the contingency table, find the probability that out of the age 20–64 group, a randomly selected driver is female. 124. Approximately 86.5 percent of Americans commute to work by car, truck, or van. Out of that group, 84.6 percent drive alone and 15.4 percent drive in a carpool. Approximately 3.9 percent walk to work and approximately 5.3 percent take public transportation. a. Construct a table or a tree diagram of the situation. Include a branch for all other modes of transportation to work. b. Assuming that the walkers walk alone, what percent of all commuters travel alone to work? c. Suppose that 1,000 workers are randomly selected. How many would you expect to travel alone to work? d. Suppose that 1,000 workers are randomly selected. How many would you expect to drive in a carpool? 125. When the euro coin was introduced in 2002, two math professors had their statistics students test whether the Belgian one euro-coin was a fair coin. They spun the coin rather than tossing it and found that out of 250 spins, 140 showed a head (event H) while 110 showed a tail (event T). On that basis, they claimed that it is not a fair coin. a. Based on the given data, find P(H) and P(T). b. Use a tree to find the probabilities of each possible outcome for the experiment of spinning the coin twice. c. Use the tree to find the probability of obtaining exactly one head in two spins of the coin. d. Use the tree to find the probability of obtaining at least one head. 126. Use the following information to answer the next two exercises. The following are real data from Santa Clara County, California. As of a certain time, there had been a total of 3,059 documented cases of a disease in the county. They were grouped into the following categories, |
with risk factors of becoming ill with the disease labeled as Methods A, B, and C and Other: Method A Method B Method C Other Totals Female 0 Male 2,146 Totals ____ Table 3.26 70 463 ____ 136 60 ____ 49 135 ____ ____ ____ ____ Suppose a person with a disease in Santa Clara County is randomly selected. a. Find P(Person is female). b. Find P(Person has a risk factor of method C). c. Find P(Person is female OR has a risk factor of method B). d. Find P(Person is female AND has a risk factor of method A). e. Find P(Person is male AND has a risk factor of method B). f. Find P(Person is female GIVEN person got the disease from method C). g. Construct a Venn diagram. Make one group females and the other group method C. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 243 127. Answer these questions using probability rules. Do NOT use the contingency table. Three thousand fifty-nine cases of a disease had been reported in Santa Clara County, California, through a certain date. Those cases will be our population. Of those cases, 6.4 percent obtained the disease through method C and 7.4 percent are female. Out of the females with the disease, 53.3 percent got the disease from method C. a. Find P(Person is female). b. Find P(Person obtained the disease through method C). c. Find P(Person is female GIVEN person got the disease from method C) d. Construct a Venn diagram representing this situation. Make one group females and the other group method C. Fill in all values as probabilities. REFERENCES 3.1 Terminology Worldatlas. (2013). Countries list by continent. Retrieved from http://www.worldatlas.com/cntycont.htm 3.2 Independent and Mutually Exclusive Events Gallup. (n.d.). Retrieved from www.gallup.com/ Lopez, S., and Sidhu, P. (2013, March 28). U.S. teachers love their lives, but struggle in the workplace. Gallup Wellbeing. http://www.gallup.com/poll/161516/teachers-love-lives-struggle-workplace.aspx 3. |
3 Two Basic Rules of Probability Baseball Almanac. (2013). Retrieved from www.baseball-almanac.com DiCamillo, Mark, and Field, M. The file poll. Field Research Corporation. Retrieved from http://www.field.com/ fieldpollonline/subscribers/Rls2443.pdf Field Research Corporation. (n.d.). Retrieved from www.field.com/fieldpollonline Forum Research. (n.d.). Mayor’s approval down. Retrieved from http://www.forumresearch.com/forms/News Archives/ News Releases/74209_TO_Issues_-_Mayoral_Approval_%28Forum_Research%29%2820130320%29.pdf Rider, D. (2011, Sept. 14). Ford support plummeting, poll suggests. The Star. Retrieved from http://www.thestar.com/news/ gta/2011/09/14/ford_support_plummeting_poll_suggests.html Shin, H. B., and Kominski, R. A. (2010 April). Language use in the United States: 2007 (American Community Survey Reports, ACS-12). Washington, DC: United States Census Bureau. Retrieved from http://www.census.gov/hhes/socdemo/ language/data/acs/ACS-12.pdf The Roper Center for Public Opinion Research. (n.d.). Archives. Retrieved from http://www.ropercenter.uconn.edu/ The Wall Street Journal. (n.d.). Retrieved from https://www.wsj.com/ U.S. Census Bureau. (n.d.). Retrieved from https://www.census.gov/ Wikipedia. (n.d.). Roulette. Retrieved from http://en.wikipedia.org/wiki/Roulette 3.4 Contingency Tables American Red Cross. (2013). Blood Types. Retrieved from http://www.redcrossblood.org/learn-about-blood/bloodtypes Centers for Disease Control and Prevention/National Center for Health Statistics, United States Department of Health and Human Services. (n.d.). Retrieved from https://www.cdc.gov/nchs/ Haiman, C. A., et al. (2006, Jan. 26). Ethnic and racial differences in the smoking-related risk of lung cancer. The New England Journal of Medicine. Retrieved from http:// |
www.nejm.org/doi/full/10.1056/NEJMoa033250 Samuels, T. M. (2013). Strange facts about RH negative blood. eHow Health. Retrieved from http://www.ehow.com/ facts_5552003_strange-rh-negative-blood.html The Disaster Center Crime Pages. (n.d.). United States: Uniform crime report – state statistics from 1960–2011. Retrieved 244 Chapter 3 | Probability Topics from http://www.disastercenter.com/crime/ United Blood Services. (2011). Human blood types. Retrieved from http://www.unitedbloodservices.org/learnMore.aspx United States Senate. (n.d.). Retrieved from www.senate.gov 3.5 Tree and Venn Diagrams American Cancer Society. (n.d.). Retrieved from https://www.cancer.org/ Clara County Public Health Department. (n.d.). Retrieved from https://www.sccgov.org/sites/sccphd/en-us/pages/phd.aspx Federal Highway Administration, U.S. Department of Transportation. (n.d.). Retrieved from https://www.fhwa.dot.gov/ The Data and Story Library. (1996). Retrieved from http://lib.stat.cmu.edu/DASL/ The Roper Center http://www.ropercenter.uconn.edu/data_access/data/search_for_datasets.html Public Opinion Research. (2013). for Search for datasets. Retrieved from USA Today. (n.d.). Retrieved from https://www.usatoday.com/ U.S. Census Bureau. (n.d.). Retrieved from https://www.census.gov/ World Bank Group. (2013). Environment. Available online at http://data.worldbank.org/topic/environment SOLUTIONS 1 a. P(L′) = P(S) b. P(M OR S) c. P(F AND L) d. P(M|L) e. P(L|M) f. P(S|F) g. P(F|L) h. P(F OR L) i. P(M AND S) j. P(F) 3 P(N) = 15 42 = 5 14 =.36 5 P(C) = 5 42 =.12 7 P |
(G) = 20 150 = 2 15 =.13 9 P(R) = 22 150 = 11 75 =.15 11 P(O) = 150 − 22 − 38 − 20 − 28 − 26 150 = 16 150 = 8 75 =.11 13 P(E) = 47 194 =.24 15 P(N) = 23 194 =.12 17 P(S) = 12 194 = 6 97 =.06 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 245 19 13 52 = 1 4 =.25 21 3 6 = 1 2 =.5 23 P(R) = 4 8 =.5 25 P(O OR H) 27 P(H|I) 29 P(N|O) 31 P(I OR N) 33 P(I) 35 The likelihood that an event will occur given that another event has already occurred. 37 1 39 the probability of landing on an even number or a multiple of three 41 P(J) =.3 43 P(Q AND R) = P(Q)P(R).1 = (.4)P(R) P(R) =.25 45 0.376 47 C|L means, given the person chosen is a Latino Californian, the person is a registered voter who prefers life in prison without parole for a person convicted of first degree murder. 49 L AND C is the event that the person chosen is a voter of the ethnicity in question who prefers life without parole over the death penalty for a person convicted of first degree murder. 51.6492 53 No, because P(L AND C) does not equal 0. 55 P(musician is a male AND had private instruction) = 15 130 = 3 26 =.12 57 P(being a female musician AND learning music in school) = 38 130 = 19 65 =.29 P(being a female musician)P(learning music in school) = ⎛ ⎝ 72 130 ⎞ ⎛ ⎝ ⎠ ⎞ ⎠ 62 130 = 4, 464 16, 900 = 1, 116 4, 225 =.26 No, they are not independent because P(being a female musician AND learning music in school) is not equal to P(being a female musician)P(learning music in school). Chapter 3 | Probability Topics 246 58 Figure 3.23 60 35 |
,065 100,450 62 To pick one person from the study who is Japanese American AND uses the product 21 to 30 times a day means that the person has to meet both criteria: both Japanese American and uses the product 21 to 30 times a day. The sample space 4,715 100,450 should include everyone in the study. The probability is. 64 To pick one person from the study who is Japanese American given that person uses the product 21 to 30 times a day, means that the person must fulfill both criteria and the sample space is reduced to those who uses the product 21 to 30 times a day. The probability is 4715 15,273. 67 a. You can't calculate the joint probability knowing the probability of both events occurring, which is not in the information given; the probabilities should be multiplied, not added; and probability is never greater than 100 percent b. A home run by definition is a successful hit, so he has to have at least as many successful hits as home runs. 69 0 71.3571 73.2142 75 Physician (83.7) 77 83.7 − 79.6 = 4.1 79 P(Occupation < 81.3) =.5 81 a. The Forum Research surveyed 1,046 Torontonians. b. 58 percent c. 42 percent of 1,046 = 439 (rounding to the nearest integer) d. e..57.60. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 247 82 a. yes; P(getting a pork chop) = P(not getting a chicken breast) b. getting a pork chop and getting a chicken breast c. no 83 a. 20/40 = 1/2 b. 5/40 = 1/8 c. 39/40 d. 4/40 = 1/10 e. 33/40 f. 15/40 = 3/8 g. 0/40 = 0 84 Compute the probabilities. a. 20/40 = 1/2 b. 8/40 = 1/5 c. 40/40 = 1 d. 16/40 = 2/5 e. 18/40 = 9/20 f. 40/40 = 1 85 a. {G1, G2, G3, G4, G5, Y1, Y2, Y3} b. c. d. e. No, because |
P(G AND E) does not equal 0. 87 NOTE The coin toss is independent of the card picked first. a. {(G,H) (G,T) (B,H) (B,T) (R,H) (R,T)} b. P(A) = P(blue)P(head) = ⎛ ⎝ ⎞ ⎠ 3 10 ⎞ ⎠ ⎛ ⎝ 1 2 = 3 20 c. Yes, A and B are mutually exclusive because they cannot happen at the same time; you cannot pick a card that is both blue and also (red or green). P(A AND B) = 0. d. No, A and C are not mutually exclusive because they can occur at the same time. In fact, C includes all of the outcomes of A; if the card chosen is blue it is also (red or blue). P(A AND C) = P(A) = 3 20. 248 89 Chapter 3 | Probability Topics a. S = {(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)} b. 4 8 c. Yes, because if A has occurred, it is impossible to obtain two tails. In other words, P(A AND B) = 0. 91 a. If Y and Z are independent, then P(Y AND Z) = P(Y)P(Z), so P(Y OR Z) = P(Y) + P(Z) – P(Y)P(Z). b..5 93 iii; i; iv; ii 95 a. P(R) =.44 b. P(R|E) =.56 c. P(R|O) =.31 d. No, whether the money is returned is not independent of which class the money was placed in. There are several ways to justify this mathematically, but one is that the money placed in economics classes is not returned at the same overall rate; P(R|E) ≠ P(R). e. No, this study definitely does not support that notion; in fact, it suggests the opposite. The money placed in the economics classrooms was returned at a higher rate than the money place in all classes collectively; P(R|E) > P(R). 97 a. P(type O OR Rh– |
) = P(type O) + P(Rh–) – P(type O AND Rh–) 0.52 = 0.43 + 0.15 – P(type O AND Rh–); solve to find P(type O AND Rh–) =.06 6 percent of people have type O, Rh– blood b. P(NOT(type O AND Rh–)) = 1 – P(type O AND Rh–) = 1 –.06 =.94 94 percent of people do not have type O, Rh– blood 99 a. Let C = be the event that the cookie contains chocolate. Let N = the event that the cookie contains nuts. b. P(C OR N) = P(C) + P(N) – P(C AND N) =.36 +.12 –.08 =.40 c. P(NEITHER chocolate NOR nuts) = 1 – P(C OR N) = 1 –.40 =.60 101 0 103 10 67 105 10 34 107 d 110 b 112 a. b. c. 26 106 33 106 21 106 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 249 d. e. ⎛ ⎝ 26 106 ⎞ ⎠ + ⎛ ⎝ ⎞ ⎠ 33 106 – ⎛ ⎝ ⎞ ⎠ 21 106 = ⎛ ⎝ ⎞ ⎠ 38 106 21 33 114 a 117 a. P(C) =.4567 b. not enough information c. not enough information d. no, because over half (0.51) of men have at least one false-positive text 119 a. P(J OR K) = P(J) + P(K) − P(J AND K);.45 =.18 +.37 – P(J AND K); solve to find P(J AND K) =.10 b. P(NOT (J AND K)) = 1 – P(J AND K) = 1 – 010 =.90 c. P(NOT (J OR K)) = 1 – P(J OR K) = 1 –.45 =.55 120 Figure 3.24 a. b. P(GG = 25 64 c. P(at least one green) = P(GG) + P(GY) |
+ P(YG) = 25 64 + 15 64 + 15 64 = 55 64 d. P(G|G) = 5 8 e. Yes, they are independent because the first card is placed back in the bag before the second card is drawn. The composition of cards in the bag remains the same from draw one to draw two. 122a. 250 Chapter 3 | Probability Topics <20 20–64 >64 Totals Female.0244.3954.0661.486 Male.0259.4186.0695.514 Totals.0503.8140.1356 1 Table 3.27 b. P(F) =.486 c. P(>64|F) =.1361 d. P(>64 and F) = P(F) P(>64|F) = (.486)(.1361) =.0661 e. P(>64|F) is the percentage of female drivers who are 65 or older and P(>64 and F) is the percentage of drivers who are female and 65 or older. f. P(>64) = P(>64 and F) + P(>64 and M) =.1356 g. No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 and F) =.0661. 124 a. Car, Truck or Van Walk Public Transportation Other Totals Alone.7318 Not Alone.1332 Totals.8650 Table 3.28.0390.0530.0430 1 b. If we assume that all walkers are alone and that none from the other two groups travel alone (which is a big assumption) we have: P(Alone) =.7318 +.0390 =.7708. c. Make the same assumptions as in (b) we have: (.7708)(1,000) = 771 d. (.1332)(1,000) = 133 126 The completed contingency table is as follows: Method A Method B Method C Other Totals Female 0 Male 2,146 Totals 2,146 Table 3.29 70 463 533 136 60 196 49 135 184 255 2,804 3,059 a. b. c. 255 3059 196 3059 718 3059 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | |
Probability Topics 251 d. 0 e. f. 463 3059 136 196 g. Figure 3.25 252 Chapter 3 | Probability Topics This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 253 4 | DISCRETE RANDOM VARIABLES Figure 4.1 You can use probability and discrete random variables to calculate the likelihood of lightning striking the ground five times during a half-hour thunderstorm. (credit: Leszek Leszczynski) Introduction Chapter Objectives By the end of this chapter, the student should be able to do the following: • Recognize and understand discrete probability distribution functions, in general. • Calculate and interpret expected values. • Recognize the binomial probability distribution and apply it appropriately. • Recognize the poisson probability distribution and apply it appropriately. • Recognize the geometric probability distribution and apply it appropriately. • Recognize the hypergeometric probability distribution and apply it appropriately. • Classify discrete word problems by their distributions. A student takes a 10-question, true-false quiz. Because the student had such a busy schedule, he or she could not study and guesses randomly at each answer. What is the probability of the student passing the test with at least a 70 percent? Small companies might be interested in the number of long-distance phone calls their employees make during the peak time of the day. Suppose the average is 20 calls. What is the probability that the employees make more than 20 long-distance phone calls during the peak time? 254 Chapter 4 | Discrete Random Variables These two examples illustrate two different types of probability problems involving discrete random variables. Recall that discrete data are data that you can count. A random variable is a variable whose values are numerical outcome of a probability experiment. We always describe a random variable in words and its values in numbers. The values of a random variable can vary with each repetition of an experiment. Random Variable Notation Uppercase letters such as X or Y denote a random variable. Lowercase letters like x or y denote the value of a random variable. If X is a random variable, then X is written in words, and x is given as a number. The following are examples of random variables: Example 1: Suppose a jar contains three marbles, one blue, one red, and one white. Randomly draw one marble from the jar. Let X = the possible number of red marbles to |
be drawn. The sample space for the drawing is red, white, and blue. Then, x = 0,1. If the marble we draw is red, then x = 1; otherwise, x = 0. Example 2: Let X = the number of female children in a randomly selected family with only two kids. Here we are only interested in families with two kids, not families with one kid or more than two kids. The sample space for the genders of two-kid families is MM, MF, FM, FF. Here the first letter represents the gender of the older child and the second letter represents the gender of the younger child. F represents a female child and M represents a male child. For example, FM represents that the older child is a girl and the younger child is a boy, while MF represents that the older child is a boy and the younger child is a girl. Then, x = 0,1,2. A family has 0 female children if it has two boys (MM), a family has one female child if it has one boy and one girl (MF or FM), and a family has two female children if both kids are girls (FF). Example 3: Let X = the number of heads you get when you toss three fair coins. The sample space for the toss of three fair coins is TTT, THH, HTH, HHT, HTT, THT, TTH, HHH. Here the first letter represents the result of the first toss, the second letter represents the result of the second toss, and the third letter represents the result of the third toss. T represents a tail and H represents a head. For example, THH means we get a tail in the first toss but a head in the second and third toss, while HHT means we get a head in the first and second toss but a tail in the third toss. Then, x = 0, 1, 2, 3. There are 0 heads if the result is TTT, one head if the result is THT, TTH, or HTT, two heads if the result is THH, HTH, or HHT, and three heads if the result is HHH. Toss a coin 10 times and record the number of heads. After all members of the class have completed the experiment (tossed a coin 10 times and counted the number of heads), fill in Table 4.1. Let X = the number of heads in 10 tosses of the coin. x Frequency of |
x Relative Frequency of x Table 4.1 a. Which value(s) of x occurred most frequently? b. If you tossed the coin 1,000 times, what values could x take on? Which value(s) of x do you think would occur most frequently? c. What does the relative frequency column sum to? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 255 4.1 | Probability Distribution Function (PDF) for a Discrete Random Variable There are two types of random variables, discrete random variables and continuous random variables. The values of a discrete random variable are countable, which means the values are obtained by counting. All random variables we discussed in previous examples are discrete random variables. We counted the number of red balls, the number of heads, or the number of female children to get the corresponding random variable values. The values of a continuous random variable are uncountable, which means the values are not obtained by counting. Instead, they are obtained by measuring. For example, let X = temperature of a randomly selected day in June in a city. The value of X can be 68°, 71.5°, 80.6°, or 90.32°. These values are obtained by measuring by a thermometer. Another example of a continuous random variable is the height of a randomly selected high school student. The value of this random variable can be 5'2", 6'1", or 5'8". Those values are obtained by measuring by a ruler. A discrete probability distribution function has two characteristics: 1. Each probability is between zero and one, inclusive. 2. The sum of the probabilities is one. Example 4.1 A child psychologist is interested in the number of times a newborn baby's crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let X = the number of times per week a newborn baby's crying wakes its mother after midnight. For this example, x = 0, 1, 2, 3, 4, 5. P(x) = probability that X takes on a value x. x P(x) 0 1 2 3 4 5 P(x = 0) = 2 50 P(x = 1) = 11 50 P(x = 2) = 23 50 P(x = 3) = 9 50 P(x = 4) = 4 50 P(x = 5 |
) = 1 50 Table 4.2 X takes on the values 0, 1, 2, 3, 4, 5. This is a discrete PDF because we can count the number of values of x and also because of the following two reasons: a. Each P(x) is between zero and one, therefore inclusive b. The sum of the probabilities is one, that is, 2 50 + 11 50 + 23 50 + 9 50 + 4 50 + 1 50 = 1 256 Chapter 4 | Discrete Random Variables 4.1 A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. Let X = the number of times a patient rings the nurse during a 12-hour shift. For this exercise, x = 0, 1, 2, 3, 4, 5. P(x) = the probability that X takes on value x. Why is this a discrete probability distribution function (two reasons)? X P(x) 0 P(x = 0) = 4 50 1 P(x = 1) = 8 50 2 P(x = 2) = 16 50 3 P(x = 3) = 14 50 4 P(x = 4) = 6 50 5 P(x = 5) = 2 50 Table 4.3 Example 4.2 Suppose Nancy has classes three days a week. She attends classes three days a week 80 percent of the time, two days 15 percent of the time, one day 4 percent of the time, and no days 1 percent of the time. Suppose one week is randomly selected. Describe the random variable in words. Let X = the number of days Nancy ________. Solution 4.2 a. Let X = the number of days Nancy attends class per week. b. In this example, what are possible values of X? Solution 4.2 b. 0, 1, 2, and 3 c. Suppose one week is randomly chosen. Construct a probability distribution table (called a PDF table) like the one in Example 4.1. The table should have two columns labeled x and P(x). Solution 4.2 c. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 257 x P(x) 0 1 2 3.01.04.15.80 Table 4.4 The sum of |
the P(x) column is 0.01+0.04+0.15+0.80 = 1.00. 4.2 Jeremiah has basketball practice two days a week. 90 percent of the time, he attends both practices. Eight percent of the time, he attends one practice. Two percent of the time, he does not attend either practice. What is X and what values does it take on? 4.2 | Mean or Expected Value and Standard Deviation The expected value of a discrete random variable X, symbolized as E(X), is often referred to as the long-term average or mean (symbolized as μ). This means that over the long term of doing an experiment over and over, you would expect this average. For example, let X = the number of heads you get when you toss three fair coins. If you repeat this experiment (toss three fair coins) a large number of times, the expected value of X is the number of heads you expect to get for each three tosses on average. NOTE To find the expected value, E(X), or mean μ of a discrete random variable X, simply multiply each value of the random variable by its probability and add the products. The formula is given as E(X) = μ = ∑ xP(x). Here x represents values of the random variable X, P(x) represents the corresponding probability, and symbol ∑ represents the sum of all products xP(x). Here we use symbol μ for the mean because it is a parameter. It represents the mean of a population. Example 4.3 A men's soccer team plays soccer zero, one, or two days a week. The probability that they play zero days is.2, the probability that they play one day is.5, and the probability that they play two days is.3. Find the long-term average or expected value, μ, of the number of days per week the men's soccer team plays soccer. To do the problem, first let the random variable X = the number of days the men's soccer team plays soccer per week. X takes on the values 0, 1, 2. Construct a PDF table adding a column x*P(x), the product of the value x with the corresponding probability P(x). In this column, you will multiply each x value by its probability. 258 Chapter 4 | Discrete Random Variables x P(x) x*P(x) 0 1 |
2.2.5.3 (0)(.2) = 0 (1)(.5) =.5 (2)(.3) =.6 Table 4.5 Expected Value Table This table is called an expected value table. The table helps you calculate the expected value or longterm average. Add the last column x * P(x) to get the expected value/mean of the random variable X. E(X) = μ = ∑ xP(x) = 0 +.5 +.6 = 1.1 The expected value/mean is 1.1. The men's soccer team would, on the average, expect to play soccer 1.1 days per week. The number 1.1 is the long-term average or expected value if the men's soccer team plays soccer week after week after week. As you learned in Chapter 3, if you toss a fair coin, the probability that the result is heads is 0.5. This probability is a theoretical probability, which is what we expect to happen. This probability does not describe the short-term results of an experiment. If you flip a coin two times, the probability does not tell you that these flips will result in one head and one tail. Even if you flip a coin 10 times or 100 times, the probability does not tell you that you will get half tails and half heads. The probability gives information about what can be expected in the long term. To demonstrate this, Karl Pearson once tossed a fair coin 24,000 times! He recorded the results of each toss, obtaining heads 12,012 times. The relative frequency of heads is 12,012/24,000 =.5005, which is very close to the theoretical probability.5. In his experiment, Pearson illustrated the law of large numbers. The law of large numbers states that, as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency approaches zero (the theoretical probability and the relative frequency get closer and closer together). The relative frequency is also called the experimental probability, a term that means what actually happens. In the next example, we will demonstrate how to find the expected value and standard deviation of a discrete probability distribution by using relative frequency. Like data, probability distributions have variances and standard deviations. The variance of a probability distribution is symbolized as σ 2 and the standard deviation of a probability distribution is symbolized as σ. Both are parameters since they summarize information about a population. |
To find the variance σ 2 of a discrete probability distribution, find each deviation from its expected value, square it, multiply it by its probability, and add the products. To find the standard deviation σ of a probability distribution, simply take the square root of variance σ 2. The formulas are given as below. NOTE The formula of the variance σ 2 of a discrete random variable X is σ 2 = ∑ (x − μ)2 P(x). Here x represents values of the random variable X, μ is the mean of X, P(x) represents the corresponding probability, and symbol ∑ represents the sum of all products (x − μ)2 P(x). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 259 To find the standard deviation, σ, of a discrete random variable X, simply take the square root of the variance x − μ)2 P(x) Example 4.4 A researcher conducted a study to investigate how a newborn baby’s crying after midnight affects the sleep of the baby's mother. The researcher randomly selected 50 new mothers and asked how many times they were awakened by their newborn baby's crying after midnight per week. Two mothers were awakened zero times, 11 mothers were awakened one time, 23 mothers were awakened two times, nine mothers were awakened three times, four mothers were awakened four times, and one mother was awakened five times. Find the expected value of the number of times a newborn baby's crying wakes its mother after midnight per week. Calculate the standard deviation of the variable as well. To do the problem, first let the random variable X = the number of times a mother is awakened by her newborn’s crying after midnight per week. X takes on the values 0, 1, 2, 3, 4, 5. Construct a PDF table as below. The column of P(x) gives the experimental probability of each x value. We will use the relative frequency to get the probability. For example, the probability that a mother wakes up zero times is 2 50 since there are two mothers out of 50 who were awakened zero times. The third column of the table is the product of a value and its probability, xP(x). x P(x) xP(x) 0 P(x = 0) = 2 50 ⎛ (0) ⎝ 2 50 ⎞ � |
�� = 0 1 P(x = 1) = 11 50 ⎛ (1) ⎝ 11 50 ⎞ ⎠ = 11 50 2 P(x = 2) = 23 50 ⎛ (2) ⎝ 23 50 ⎞ ⎠ = 46 50 3 P(x = 3) = 9 50 ⎛ (3) ⎝ 9 50 ⎞ ⎠ = 27 50 4 P(x = 4) = 4 50 ⎛ (4) ⎝ 4 50 ⎞ ⎠ = 16 50 5 P(x = 5) = 1 50 ⎛ (5) ⎝ 1 50 ⎞ ⎠ = 5 50 Table 4.6 We then add all the products in the third column to get the mean/expected value of X. E(X) = μ = ∑ xP(x) = 0 + 11 50 + 46 50 + 27 50 + 16 50 + 5 50 = 105 50 = 2.1 Therefore, we expect a newborn to wake its mother after midnight 2.1 times per week, on the average. To calculate the standard deviation σ, we add the fourth column (x-μ)2 and the fifth column (x - μ)2 ∙ P(x) to get the following table: 260 Chapter 4 | Discrete Random Variables x P(x) xP(x) (x-µ)2 (x-µ)2•P(x) 0 P(x = 0) = 2 50 ⎛ (0) ⎝ 2 50 ⎞ ⎠ = 0 (0 − 2.1)2 = 4.41 4.41 • 2 50 =.1764 1 P(x = 1) = 11 50 ⎛ (1) ⎝ 11 50 ⎞ ⎠ = 11 50 (1 − 2.1)2 = 1.21 1.21 • 11 50 =.2662 2 P(x = 2) = 23 50 ⎛ (2) ⎝ 23 50 ⎞ ⎠ = 46 50 (2 − 2.1)2 =.01 3 P(x = 3) = 9 50 ⎛ (3) ⎝ 9 50 ⎞ ⎠ = 27 50 (3 − 2.1)2 =.81.01 • 23 50.81 • |
9 50 =.0046 =.1458 4 P(x = 4) = 4 50 ⎛ (4) ⎝ 4 50 ⎞ ⎠ = 16 50 (4 − 2.1)2 = 3.61 3.61 • 4 50 =.2888 5 P(x = 5) = 1 50 ⎛ (5) ⎝ 1 50 ⎞ ⎠ = 5 50 (5 − 2.1)2 = 8.41 8.41 • 1 50 =.1682 Table 4.7 We then add all the products in the 5th column to get the variance of X. σ 2 =.1764 +.2662 +.0046 +.1458 +.2888 +.1682 = 1.05 To get the standard deviation σ, we simply take the square root of variance σ2. σ = σ 2 = 1.05 ≈ 1.0247 4.4 A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. What is the expected value? x P(x) 0 1 2 3 4 5 P(x = 0) = 4 50 P(x = 1) = 8 50 P(x = 2) = 16 50 P(x = 3) = 14 50 P(x = 4) = 6 50 P(x = 5) = 2 50 Table 4.8 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 261 Example 4.5 Suppose you play a game of chance in which five numbers are chosen from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. A computer randomly selects five numbers from zero to nine with replacement. You pay $2 to play and could profit $100,000 if you match all five numbers in order (you get your $2 back plus $100,000). Over the long term, what is your expected profit of playing the game? To do this problem, set up a PDF table for the amount of money you can profit. Let X = the amount of money you profit. If your five numbers match in order, you will win the game and will get your $2 back plus $100,000 |
. That means your profit is $100,000. If your five numbers do not match in order, you will lose the game and lose your $2. That means your profit is -$2. Therefore, X takes on the values $100,000 and –$2. That is the second column x in the PDF table below. To win, you must get all five numbers correct, in order. The probability of choosing the correct first number is 1 10 because there are 10 numbers (from zero to nine) and only one of them is correct. The probability of choosing the correct second number is also 1 10 because the selection is done with replacement and there are still 10 numbers (from zero to nine) for you to choose. Due to the same reason, the probability of choosing the correct third number, the correct fourth number, and the correct fifth number are also 1 10. The selection of one number does not affect the selection of another number. That means the five selections are independent. The probability of choosing all five correct numbers and in order is equal to the product of the probabilities of choosing each number correctly. ⎝choosing all five numbers correctly ⎞ P⎛ ⎝choosing 2nd number correctly⎞ P⎛ ⎠ • P⎛ ⎠ • P⎛ = ( 1 10 ) • ( 1 10 ) • ( 1 10 ⎝choosing 1st number correctly⎞ ⎝choosing 5th number correctly⎞ ) • ( 1 10 ) • ( 1 10 ) ⎠ ⎠ • =.00001 Therefore, the probability of winning is.00001 and the probability of losing is 1 −.00001 =.99999. That is how we get the third column P(x) in the PDF table below. To get the fourth column xP(x) in the table, we simply multiply the value x with the corresponding probability P(x). The PDF table is as follows: x P(x) x*P(x) Loss –2.99999 (–2)(.99999) = –1.99998 Profit 100,000.00001 (100000)(.00001) = 1 Table 4.9 We then add all the products in the last column to get the mean/expected value of X. E(X) = μ = ∑ xP(x) = − 1.99998 + 1 = −.99 |
98. Since –.99998 is about –1, you would, on average, expect to lose approximately $1 for each game you play. However, each time you play, you either lose $2 or profit $100,000. The $1 is the average or expected loss per game after playing this game over and over. 262 Chapter 4 | Discrete Random Variables 4.5 You are playing a game of chance in which four cards are drawn from a standard deck of 52 cards. You guess the suit of each card before it is drawn. The cards are replaced in the deck on each draw. You pay $1 to play. If you guess the right suit every time, you get your money back and $256. What is your expected profit of playing the game over the long term? Example 4.6 Suppose you play a game with a biased coin. You play each game by tossing the coin once. P(heads) = 2 3 and. If you toss a head, you pay $6. If you toss a tail, you win $10. If you play this game many times, P(tails) = 1 3 will you come out ahead? a. Define a random variable X. Solution 4.6 a. X = amount of profit b. Complete the following expected value table. Solution 4.6 b. x ____ ____ WIN 10 1 3 LOSE ____ ____ ____ –12 3 Table 4.10 x P(x) xP(x) 1 3 2 3 WIN 10 LOSE –6 Table 4.11 10 3 –12 3 c. What is the expected value, μ? Do you come out ahead? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 263 Solution 4.6 c. Add the last column of the table. The expected value E(X) = μ = 10 3 + ⎛ ⎝− 12 67. You lose, on average, about 67 cents each time you play the game, so you do not come out ahead. 4.6 Suppose you play a game with a spinner. You play each game by spinning the spinner once. P(red) = 2 5, P(blue) = 2 5, and P(green) = 1 5. If you land on red, you pay $10. If you land on blue, you don't pay or |
win anything. If you land on green, you win $10. Complete the following expected value table. –20 5 x P(x) Red Blue Green 10 Table 4.12 2 5 Generally for probability distributions, we use a calculator or a computer to calculate μ and σ to reduce rounding errors. For some probability distributions, there are shortcut formulas for calculating μ and σ. Example 4.7 Toss a fair, six-sided die twice. Let X = the number of faces that show an even number. Construct a table like Table 4.12 and calculate the mean μ and standard deviation σ of X. Solution 4.7 Tossing one fair six-sided die twice has the same sample space as tossing two fair six-sided dice. The sample space has 36 outcomes. (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) Table 4.13 Use the sample space to complete the following table. 264 Chapter 4 | Discrete Random Variables x P(x) xP(x) (x – μ)2 ⋅ P(x) 0 1 2 9 36 18 36 9 36 0 18 36 18 36 (0 – 1)2 ⋅ 9 36 = 9 36 (1 – 1)2 ⋅ 18 36 = 0 (2 – 1)2 ⋅ 9 36 = 9 36 Table 4.14 Calculating μ and σ. Add the values in the third column to find the expected value: μ = 36 36 = 1. Use this value to complete the fourth column. Add the values in the fourth column and take the square root of the sum: σ = 18 36 ≈.7071. Some |
of the more common discrete probability functions are binomial, geometric, hypergeometric, and Poisson. Most elementary courses do not cover the geometric, hypergeometric, and Poisson. Your instructor will let you know if he or she wishes to cover these distributions. A probability distribution function is a pattern. You try to fit a probability problem into a pattern or distribution in order to perform the necessary calculations. These distributions are tools to make solving probability problems easier. Each distribution has its own special characteristics. Learning the characteristics enables you to distinguish among the different distributions. 4.3 | Binomial Distribution (Optional) There are three characteristics of a binomial experiment: 1. There are a fixed number of trials. Think of trials as repetitions of an experiment. The letter n denotes the number of trials. 2. There are only two possible outcomes, called success and failure, for each trial. The outcome that we are measuring is defined as a success, while the other outcome is defined as a failure. The letter p denotes the probability of a success on one trial, and q denotes the probability of a failure on one trial. p + q = 1. 3. The n trials are independent and are repeated using identical conditions. Because the n trials are independent, the outcome of one trial does not help in predicting the outcome of another trial. Another way of saying this is that for each individual trial, the probability, p, of a success and probability, q, of a failure remain the same. Let us look at several examples of a binomial experiment. Example 1: Toss a fair coin once and record the result. This is a binomial experiment since it meets all three characteristics. The number of trials n = 1. There are only two outcomes, a head or a tail, of each trial. We can define a head as a success if we are measuring number of heads. For a fair coin, the probabilities of getting head or tail are both.5. So, p = q −.5. Both p and q remain the same from trial to trial. This experiment is also called a Bernoulli trial, named after Jacob Bernoulli who, in the late 1600s, studied such trials extensively. Any experiment that has characteristics two and three and where n = 1 is called a Bernoulli trial. A binomial experiment takes place when the number of successes is counted in one or more Bernoulli trials. Example 2: Randomly guess a multiple choice question has A, B, C and D four |
options. This is a binomial experiment since it meets all three characteristics. The number of trials n = 1. There are only two outcomes, guess correctly or guess wrong, of each trial. We can define guess correctly as a success. For a random This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 265 guess (you have no clue at all), the probability of guessing correct should be 1 4 because there are four options and only one option is correct. So, and p = 1 4 and. Both p and q remain the same from trial to trial. This experiment is also a Bernoulli trial. It meets the characteristics two and three and n = 1. Example 3: Toss a fair coin five times and record the result. This is a binomial experiment since it meets all three characteristics. The number of trials n = 5. There are only two outcomes, head or tail, of each trial. If we define head as a success, then p = q = 0.5. Both p and q remain the same for each trial. Since n = 5, this experiment is not a Bernoulli trial although it meets the characteristics two and three. Example 4: Randomly guess 10 multiple choice questions in an exam. Each question has A, B, C and D four options. This is a binomial experiment since it meets all three characteristics. The number of trials n = 10. There are only two outcomes, guess correctly or guess wrong, of each trial. We can define guess correctly as a success. As we explained in example 2, p = 1 4. Both p and q remain the same for each guess. Since n = 10, this and experiment is not a Bernoulli trial. The next two experiments are not binomial experiments. Example 5: Randomly select two balls from a jar with five red balls and five blue balls without replacement. This means we select the first ball, and then without returning the selected ball into the jar, we will select the second ball. This is not a binomial experiment since the third characteristic is not met. The number of trials n = 2. There are only two outcomes, a red ball or a blue ball, of each trial. If we define selecting a red ball as a success, then selecting a blue ball is a failure. The probability of getting the first ball red is 5 10 since there are five red balls out of 10 |
balls. So, p = 5 10 and q = 1 − p = 1 − 5 10 = 5 10. However, p and q do not remain the same for the second trial. If the first ball selected is red, then the probability of getting the second ball red is 4 9 since there are only four red balls out of nine balls. But if the first ball selected is blue, then the probability of getting the second ball red is 5 9 since there are still five red balls out of nine balls. Example 6: Toss a fair coin until a head appears. This is not a binomial experiment since the first characteristic is not met. The number of trials n is not fixed. n could be 1 if a head appears from the first toss. n could be 2 if the first toss is a tail and the second toss is a head. So on and so forth. More examples of binomial and non-binomial experiments will be discussed in this section later. The outcomes of a binomial experiment fit a binomial probability distribution. The random variable X = the number of successes obtained in the n independent trials. There are shortcut formulas for calculating mean μ, variance σ2, and standard deviation σ of a binomial probability distribution. The formulas are given as below. The deriving of these formulas will not be discussed in this book. Here n is the number of trials, p is the probability of a success, and q is the probability of a failure. μ = np, σ 2 = npq, σ = npq. Example 4.8 At ABC High School, the withdrawal rate from an elementary physics course is 30 percent for any given term. 266 Chapter 4 | Discrete Random Variables This implies that, for any given term, 70 percent of the students stay in the class for the entire term. The random variable X = the number of students who withdraw from the randomly selected elementary physics class. Since we are measuring the number of students who withdrew, a success is defined as an individual who withdrew. 4.8 The state health board is concerned about the amount of fruit available in school lunches. Forty-eight percent of schools in the state offer fruit in their lunches every day. This implies that 52 percent do not. What would a success be in this case? Example 4.9 Suppose you play a game that you can only either win or lose. The probability that you win any game is 55 percent, and the probability that you lose is 45 percent. Each game you play |
is independent. If you play the game 20 times, write the function that describes the probability that you win 15 of the 20 times. Here, if you define X as the number of wins, then X takes on the values 0, 1, 2, 3,..., 20. The probability of a success is p = 0.55. The probability of a failure is q =.45. The number of trials is n = 20. The probability question can be stated mathematically as P(x = 15). If you define X as the number of losses, then a success is defined as a loss and a failure is defined as a win. A success does not necessarily represent a good outcome. It is simply the outcome that you are measuring. X still takes on the values of 0, 1, 2, 3,..., 20. The probability of a success is p =.45. The probability of a failure is q =.55. 4.9 A trainer is teaching a dolphin to do tricks. The probability that the dolphin successfully performs the trick is 35 percent, and the probability that the dolphin does not successfully perform the trick is 65 percent. Out of 20 attempts, you want to find the probability that the dolphin succeeds 12 times. State the probability question mathematically. Example 4.10 A fair coin is flipped 15 times. Each flip is independent. What is the probability of getting more than 10 heads? Let X = the number of heads in 15 flips of the fair coin. X takes on the values 0, 1, 2, 3,..., 15. Since the coin is fair, p =.5 and q =.5. The number of trials n = 15. State the probability question mathematically. Solution 4.10 P(x > 10) 4.10 A fair, six-sided die is rolled 10 times. Each roll is independent. You want to find the probability of rolling a one more than three times. State the probability question mathematically. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 267 Example 4.11 Approximately 70 percent of statistics students do their homework in time for it to be collected and graded. Each student does homework independently. In a statistics class of 50 students, what is the probability that at least 40 will do their homework on time? Students are selected randomly. a. This is a binomial problem because |
there is only a success or a ________, there are a fixed number of trials, and the probability of a success is.70 for each trial. Solution 4.11 a. failure b. If we are interested in the number of students who do their homework on time, then how do we define X? Solution 4.11 b. X = the number of statistics students who do their homework on time c. What values does x take on? Solution 4.11 c. 0, 1, 2,..., 50 d. What is a failure, in words? Solution 4.11 d. Failure is defined as a student who does not complete his or her homework on time. The probability of a success is p =.70. The number of trials is n = 50. e. If p + q = 1, then what is q? Solution 4.11 e. q =.30 f. The words at least translate as what kind of inequality for the probability question P(x ____ 40)? Solution 4.11 f. greater than or equal to (≥) The probability question is P(x ≥ 40). 4.11 Sixty-five percent of people pass the state driver’s exam on the first try. A group of 50 individuals who have taken the driver’s exam is randomly selected. Give two reasons why this is a binomial problem. Notation for the Binomial: B = Binomial Probability Distribution Function X ~ B(n, p) Read this as X is a random variable with a binomial distribution. The parameters are n and p: n = number of trials, p = probability of a success on each trial. 268 Chapter 4 | Discrete Random Variables Example 4.12 It has been stated that about 41 percent of adult workers have a high school diploma but do not pursue any further education. If 20 adult workers are randomly selected, find the probability that at most 12 of them have a high school diploma but do not pursue any further education. How many adult workers do you expect to have a high school diploma but do not pursue any further education? Let X = the number of workers who have a high school diploma but do not pursue any further education. X takes on the values 0, 1, 2,..., 20 where n = 20, p =.41, and q = 1 –.41 =.59. X ~ B(20,.41) Find P(x ≤ 12). There is a formula |
to define the probability of a binomial distribution P(x). We can use the formula to find P(x ≤ 12). But the calculation is tedious and time consuming, and people usually use a graphing calculator, software, or binomial table to get the answer. Use a graphing calculator, you can get P(x ≤ 12) =.9738. The instruction of TI-83, 83+, 84, 84+ is given below. Go into 2nd DISTR. The syntax for the instructions are as follows: To calculate the probability of a value P(x = value) : use binompdf(n, p, number). Here binompdf represents binomial probability density function. It is used to find the probability that a binomial random variable is equal to an exact value. n is the number of trials, p is the probability of a success, and number is the value. If number is left out, which means use binompdf(n, p), then all the probabilities P(x = 0), P(x = 1), …, P(x = n) will be calculated. To calculate the cumulative probability P(x ≤ value) : use binomcdf(n, p, number). Here binomcdf represents binomial cumulative distribution function. It is used to determine the probability of at most type of problem, the probability that a binomial random variable is less than or equal to a value. n is the number of trials, p is the probability of a success, and number is the value. If number is left out, all the cumulative probabilities P(x ≤ 0), P(x ≤ 1), …, P(x ≤ n) will be calculated. To calculate the cumulative probability P(x ≥ value) : use 1 - binomcdf(n, p, number). n is the number of trials, p is the probability of a success, and number is the value. TI calculators do not have a built-in function to find the probability that a binomial random variable is greater than a value. However, we can use the fact that P(x > value) = 1 − P(x ≤ value) to find the answer. For this problem: After you are in 2nd DISTR, arrow down to binomcdf. Press ENTER. Enter 20,.41,12). The result is P(x ≤ 12) =.9738. NOTE If you want to find P(x = 12), use |
the pdf (binompdf). If you want to find P(x > 12), use 1 − binomcdf(20,.41,12). The probability that at most 12 workers have a high school diploma but do not pursue any further education is.9738. The graph of X ~ B(20,.41) is as follows. The previous graph is called a probability distribution histogram. It is made of a series of vertical bars. The x-axis of each bar is the value of X = the number of workers who have only a high school diploma, and the height of that bar is the probability of that value occurring. The number of adult workers that you expect to have a high school diploma but not pursue any further education This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 269 is the mean, μ = np = (20)(.41) = 8.2. The formula for the variance is σ2 = npq. The standard deviation is σ = npq. σ = (20)(.41)(.59) = 2.20. The following is the interpretation of the mean μ = 8.2 and standard deviation σ = 2.20 : If you randomly select 20 adult workers, and do that over and over, you expect around eight adult workers out of 20 to have a high school diploma but do not pursue any further education on average. And you expect that to vary by about two workers on average. 4.12 About 32 percent of students participate in a community volunteer program outside of school. If 30 students are selected at random, find the probability that at most 14 of them participate in a community volunteer program outside of school. Use the TI-83+ or TI-84 calculator to find the answer. Example 4.13 A store releases a 560-page art supply catalog. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let X = the number of pages that feature signature artists. a. What values does x take on? b. What is the probability distribution? Find the following probabilities: i. ii. the probability that two pages feature signature artists the probability that at most six pages feature signature artists iii. the probability that more than three pages feature signature artists c. Using the formulas, calculate the (i) mean and (ii) standard deviation. Solution 4.13 a. x = 0, 1 |
, 2, 3, 4, 5, 6, 7, 8 b. This is a binomial experiment since all three characteristics are met. Each page is a trial. Since we sample 100 pages, the number of trials is n = 100. For each page, there are two possible outcomes, features signature artists or does not feature signature artists. Since we are measuring the number of pages that feature signature artists, a page that features signature artists is defined as a success and a page that does not feature signature artists is defined as a failure. There are 8 out of 560 pages that feature signature artists. Therefore the probability of a success p = 8 560 and the probability of a failure q = 1 − p = 1 − 8 560 = 552 560. Both p and q remain the same for each page. Therefore, X is a binomial random variable, and it can be ⎛ ⎝100, 8 written as X ~ B 560 ⎞ ⎠. We can use a graphing calculator to answer Parts i to iii. i. P(x = 2) = binompdf ⎛ ⎝100, 8 560, 2 ⎞ ⎠ =.2466 ii. P(x ≤ 6) = binomcdf ⎛ ⎝100, 8 560, 6 ⎞ ⎠ =.9994 270 Chapter 4 | Discrete Random Variables iii. P(x > 3) = 1 – P(x ≤ 3) = 1 – binomcdf ⎛ ⎝100, 8 560, 3 ⎞ ⎠ = 1 –.9443 =.0557 c. i. mean = np = (100) ⎛ ⎝ ⎞ ⎠ 8 560 = 800 560 ≈ 1.4286 ii. ⎛ standard deviation = npq = (100) ⎝ 8 560 ⎞ ⎛ ⎝ ⎠ ⎞ ⎠ 552 560 ≈ 1.1867 4.13 According to a poll, 60 percent of American adults prefer saving over spending. Let X = the number of American adults out of a random sample of 50 who prefer saving to spending. a. What is the probability distribution for X? b. Use your calculator to find the following probabilities: i. The probability that 25 adults in the sample prefer saving over spending ii. The probability that at most 20 adults prefer saving iii. The |
probability that more than 30 adults prefer saving c. Using the formulas, calculate the (i) mean and (ii) standard deviation of X. Example 4.14 The lifetime risk of developing a specific disease is about 1 in 78 (1.28 percent). Suppose we randomly sample 200 people. Let X = the number of people who will develop the disease. a. What is the probability distribution for X? b. Using the formulas, calculate the (i) mean and (ii) standard deviation of X. c. Use your calculator to find the probability that at most eight people develop the disease. d. Is it more likely that five or six people will develop the disease? Justify your answer numerically. Solution 4.14 a. This is a binomial experiment since all three characteristics are met. Each person is a trial. Since we sample 200 people, the number of trials is n = 200. For each person, there are two possible outcomes: will develop the disease or not. Since we are measuring the number of people who will develop the disease, a person who will develop the disease is defined as a success and a person who will not develop the disease is defined as a failure. The risk of developing the disease is 1.28 percent. Therefore the probability of a success, p = 1.28 percent,.0128, and the probability of a failure0128 =.9872. Both p and q remain the same for each person. Therefore, X is a binomial random variable and it can be written as X ~ B(200,.0128). We can use a graphing calculator to answer Questions c and d. b. i. Mean = np = 200(.0128) = 2.56 ii. Standard Deviation = npq = (200)(0.128)(.9872) ≈ 1.5897 c. Using the TI-83, 83+, 84 calculator with instructions as provided in Example 4.12: P(x ≤ 8) = binomcdf(200,.0128, 8) =.9988 d. P(x = 5) = binompdf(200,.0128, 5) =.0707 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 271 P(x = 6) = binompdf(200,.0128, 6) =.0298 So P |
(x = 5) > P(x = 6); it is more likely that five people will develop the disease than six. 4.14 During the 2013 regular basketball season, a player had the highest field goal completion rate in the league. This player scored with 61.3 percent of his shots. Suppose you choose a random sample of 80 shots made by this player during the 2013 season. Let X = the number of shots that scored points. a. What is the probability distribution for X? b. Using the formulas, calculate the (i) mean and (ii) standard deviation of X. c. Use your calculator to find the probability that this player scored with 60 of these shots. d. Find the probability that this player scored with more than 50 of these shots. Example 4.15 The following example illustrates a problem that is not binomial. It violates the condition of independence. ABC High School has a student advisory committee made up of 10 staff members and six students. The committee wishes to choose a chairperson and a recorder. What is the probability that the chairperson and recorder are both students? The names of all committee members are put into a box, and two names are drawn without replacement. The first name drawn determines the chairperson and the second name the recorder. There are two trials. However, the trials are not independent because the outcome of the first trial affects the outcome of the second trial. The probability of a student on the first draw is 6 16 because there are six students out of 16 members (10 staff members + six students). If the first draw selects a student, then the probability of a student on the second draw is 5 16 because there are only five students out of 15 members. If the first draw selects a staff member, then the probability of a student on the second draw is 6 15 because there are still six students out of 15 members. The probability of drawing a student's name changes for each of the trials and, therefore, violates the condition of independence. 4.15 A lacrosse team is selecting a captain. The names of all the seniors are put into a hat, and the first three that are drawn will be the captains. The names are not replaced once they are drawn (one person cannot be two captains). You want to see if the captains all play the same position. State whether this problem is binomial or not and state why. 4.4 | Geometric Distribution (Optional) There are three main characteristics of a geometric experiment: 1. Repeating independent Bernoulli |
trials until a success is obtained. Recall that a Bernoulli trial is a binomial experiment with number of trials n = 1. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bull's-eye until you hit the bull's-eye. The first time you hit the bull's-eye is a success so you stop throwing the dart. It might take six tries until you hit the bull's-eye. You can think of the trials as failure, failure, failure, failure, failure, success, stop. 272 Chapter 4 | Discrete Random Variables 2. In theory, the number of trials could go on forever. There must be at least one trial. 3. The probability, p, of a success and the probability, q, of a failure do not change from trial to trial. p + q = 1 and q = 1 − p. For example, the probability of rolling a three when you throw one fair die is 1 6. This is true no matter how many times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls one through four, you do not get a face with a three. The probability for each of the rolls is q = 5 6, the probability of a failure. The probability of getting a three on the fifth roll is ⎛ ⎝ 0804. X = the number of independent trials until the first success. p = the probability of a success, q = 1 – p = the probability of a failure. There are shortcut formulas for calculating mean μ, variance σ2, and standard deviation σ of a geometric probability distribution. The formulas are given as below. The deriving of these formulas will not be discussed in this book. μ = 1 p, σ 2 = ( 1 p)( 1 p − 1), σ = ( 1 p)( 1 p − 1) Example 4.16 Suppose a game has two outcomes, win or lose. You repeatedly play that game until you lose. The probability of losing is p = 0.57. If we let X = the number of games you play until you lose (includes the losing game), then X is a geometric random variable. All three characteristics are met. Each game you play is a Bernoulli trial, either win or lose. You would need to play at least one game before you stop. X takes on the values 1, |
2, 3,... (could go on indefinitely). Since we are measuring the number of games you play until you lose, we define a success as losing a game and a failure as winning a game. The probability of a success p =.57 and the probability of a failure q = 1 – p = 1 – 0.57 = 0.43. Both p and q remain the same from game to game. If we want to find the probability that it takes five games until you lose, then the probability could be written as P(x = 5). We will explain how to find a geometric probability later in this section. 4.16 You throw darts at a board until you hit the center area. Your probability of hitting the center area is p = 0.17. You want to find the probability that it takes eight throws until you hit the center. What values does X take on? Example 4.17 A safety engineer feels that 35 percent of all industrial accidents in her plant are caused by failure of employees to follow instructions. She decides to look at the accident reports (selected randomly and replaced in the pile after reading) until she finds one that shows an accident caused by failure of employees to follow instructions. If we let X = the number of accidents the safety engineer must examine until she finds a report showing an accident caused by employee failure to follow instructions, then X is a geometric random variable. All three characteristics are met. Each accident report she reads is a Bernoulli trial: the accident was either caused by failure of employees to follow instructions or not. She would need to read at least one accident report before she stops. X takes on the values 1, 2, 3,... (could go on indefinitely). Since we are measuring the number of reports she needs to read until one that shows an accident caused by failure of employees to follow instructions, we define a success as an accident caused by failure of employees to follow instructions. If an accident was caused by another reason, the report is defined as a failure. The probability of a success p =.35 and the probability of a failure q = 1 − p = 1 −.35 =.65. Both p and q remain the same from report to report. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 273 If we want to find the probability that the safety engineer will have to examine at least three reports until she finds a |
report showing an accident caused by employee failure to follow instructions, then the probability could be written as p =.35. If we want to find how many reports, on average, the safety engineer would expect to look at until she finds a report showing an accident caused by employee failure to follow instructions, we need to find the expected value E(x). We will explain how to solve these questions later in this section. 4.17 An instructor feels that 15 percent of students get below a C on their final exam. She decides to look at final exams (selected randomly and replaced in the pile after reading) until she finds one that shows a grade below a C. We want to know the probability that the instructor will have to examine at least 10 exams until she finds one with a grade below a C. What is the probability question stated mathematically? Example 4.18 Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55 percent of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he or she lives within five miles of you. What is the probability that you need to contact four people? This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if he or she lives within five miles of you. There is no definite number of trials (number of times you ask a student). a. Let X = the number of ________ you must ask ________ one says yes. Solution 4.18 a. Let X = the number of students you must ask until one says yes. b. What values does X take on? Solution 4.18 b. 1, 2, 3,..., (total number of students) c. What are p and q? Solution 4.18 c. p =.55; q =.45 d. The probability question is P(_______). Solution 4.18 d. P(x = 4) 4.18 You need to find a store that carries a special printer ink. You know that of the stores that carry printer ink, 10 274 Chapter 4 | Discrete Random Variables percent of them carry the special ink. You randomly call each store until one has the ink you need. What are p and q? Notation for the Geometric: G = Geometric Probability Distribution Function X ~ G(p) |
Read this as X is a random variable with a geometric distribution. The parameter is p; p = the probability of a success for each trial. Example 4.19 Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective? Let X = the number of computer components tested until the first defect is found. X takes on the values 1, 2, 3,... where p =.02. X ~ G(.02) Find P(x = 7). There is a formula to define the probability of a geometric distribution P(x). We can use the formula to find P(x = 7). But since the calculation is tedious and time consuming, people usually use a graphing calculator or software to get the answer. Using a graphing calculator, you can get P(x = 7) =.0177. The instruction of TI83, 83+, 84, 84+ is given below. Go into 2nd DISTR. The syntax for the instructions are as follows: To calculate the probability of a value P(x = value), use geometpdf(p, number). Here geometpdf represents geometric probability density function. It is used to find the probability that a geometric random variable is equal to an exact value. p is the probability of a success and number is the value. To calculate the cumulative probability P(x ≤ value), use geometcdf(p, number). Here geometcdf represents geometric cumulative distribution function. It is used to determine the probability of “at most” type of problem, the probability that a geometric random variable is less than or equal to a value. p is the probability of a success and number is the value. To find P(x = 7), enter 2nd DISTR, arrow down to geometpdf(. Press ENTER. Enter.02,7). The result is P(x = 7) =.0177. If we need to find P(x ≤ 7) enter 2nd DISTR, arrow down to geometcdf(. Press ENTER. Enter.02,7). The result is (x ≤ = 7) =.1319. The graph of X ~ G(.02) is This OpenStax book is available for free at http://cnx.org/content/col30309/1 |
.8 Chapter 4 | Discrete Random Variables 275 Figure 4.2 The previous probability distribution histogram gives all the probabilities of X. The x-axis of each bar is the value of X = the number of computer components tested until the first defect is found, and the height of that bar is the probability of that value occurring. For example, the x value of the first bar is 1 and the height of the first bar is 0.02. That means the probability that the first computer components tested is defective is.02. p = 1.02 The expected value or mean of X is E(X) = μ = 1 = 50. The variance of X is σ 2 = ( 1 p)( 1 p − 1) = ( 1.02 )( 1.02 − 1) = (50)(49) = 2,450 The standard deviation of X is σ = σ 2 = 2,450 = 49.5 Here is how we interpret the mean and standard deviation. The number of components that you would expect to test until you find the first defective one is 50 (which is the mean). And you expect that to vary by about 50 computer components (which is the standard deviation) on average. 4.19 The probability of a defective steel rod is.01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer. Example 4.20 The lifetime risk of developing pancreatic cancer is about one in 78 (1.28 percent). Let X = the number of people you ask until one says he or she has pancreatic cancer. Then X is a discrete random variable with a geometric distribution: X ~ G ⎛ ⎝ or X ~ G(.0128). ⎞ ⎠ 1 78 a. What is the probability that you ask 10 people before one says he or she has pancreatic cancer? b. What is the probability that you must ask 20 people? c. Find the (i) mean and (ii) standard deviation of X. 276 Chapter 4 | Discrete Random Variables Solution 4.20 a. P(x = 10) = geometpdf(.0128, 10) =.0114 b. P(x = 20) = geometpdf(.0128, 20) =.01 c. i. Mean = μ = 1 p = 1.0128 = 78 ii01 |
28 ⎞ ⎛ ⎝ ⎠ 1.0128 − 1 ⎞ ⎠ = (78)(78 − 1) = 6,006 = 77.4984 ≈ 77 The number of people whom you would expect to ask until one says he or she has pancreatic cancer is 78. And you expect that to vary by about 77 people on average. 4.20 The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in Afghanistan is 12 percent. Let X = the number of Afghani women you ask until one says that she is literate. a. What is the probability distribution of X? b. What is the probability that you ask five women before one says she is literate? c. What is the probability that you must ask 10 women? d. Find the (i) mean and (ii) standard deviation of X. 4.5 | Hypergeometric Distribution (Optional) There are five characteristics of a hypergeometric experiment: 1. You take samples from two groups. 2. You are concerned with a group of interest, called the first group. 3. You sample without replacement from the combined groups. For example, you want to choose a softball team from a combined group of 11 men and 13 women. The team consists of 10 players. 4. Each pick is not independent, since sampling is without replacement. In the softball example, the probability of picking a woman first is 13 24. The probability of picking a man second is 11 23 if a woman was picked first. It is 10 23 if a man was picked first. The probability of the second pick depends on what happened in the first pick. 5. You are not dealing with Bernoulli trials. The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution. The random variable X = the number of items from the group of interest. Example 4.21 A candy dish contains 100 jelly beans and 80 gumdrops. Fifty candies are picked at random. What is the probability that 35 of the 50 are gumdrops? The two groups are jelly beans and gumdrops. Since the probability question asks for the probability of picking gumdrops, the group of interest (first group) is gumdrops. The size of the group of interest (first group) is 80. The size of the second group is 100. The size of the sample is 50 (jelly beans or gumdrops). Let X |
= the number of gumdrops in the sample of 50. X takes on the values x = 0, 1, 2,..., 50. What is the probability statement written mathematically? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 277 Solution 4.21 P(x = 35) 4.21 A bag contains letter tiles. 44 of the tiles are vowels, and 56 are consonants. Seven tiles are picked at random. You want to know the probability that four of the seven tiles are vowels. What is the group of interest, the size of the group of interest, and the size of the sample? Example 4.22 Suppose a shipment of 100 DVD players is known to have 10 defective players. An inspector randomly chooses 12 for inspection. He is interested in determining the probability that, among the 12 players, at most two are defective. The two groups are the 90 non-defective DVD players and the 10 defective DVD players. The group of interest (first group) is the defective group because the probability question asks for the probability of at most two defective DVD players. The size of the sample is 12 DVD players. They may be non-defective or defective. Let X = the number of defective DVD players in the sample of 12. X takes on the values 0, 1, 2,..., 10. X may not take on the values 11 or 12. The sample size is 12, but there are only 10 defective DVD players. Write the probability statement mathematically. Solution 4.22 P(x ≤ 2) 4.22 A gross of eggs contains 144 eggs. A particular gross is known to have 12 cracked eggs. An inspector randomly chooses 15 for inspection. She wants to know the probability that, among the 15, at most three are cracked. What is X, and what values does it take on? Example 4.23 You are president of an on-campus special events organization. You need a committee of seven students to plan a special birthday party for the president of the college. Your organization consists of 18 women and 15 men. You are interested in the number of men on your committee. If the members of the committee are randomly selected, what is the probability that your committee has more than four men? This is a hypergeometric problem because you are choosing your committee from two groups (men and women). a. Are you choosing with or |
without replacement? Solution 4.23 a. without b. What is the group of interest? Solution 4.23 b. the men 278 Chapter 4 | Discrete Random Variables c. How many are in the group of interest? Solution 4.23 c. 15 men d. How many are in the other group? Solution 4.23 d. 18 women e. Let X = ________ on the committee. What values does X take on? Solution 4.23 e. Let X = the number of men on the committee. x = 0, 1, 2,..., 7. f. The probability question is P(_______). Solution 4.23 f. P(x > 4) 4.23 A palette has 200 milk cartons. Of the 200 cartons, it is known that 10 of them have leaked and cannot be sold. A stock clerk randomly chooses 18 for inspection. He wants to know the probability that among the 18, no more than two are leaking. Give five reasons why this is a hypergeometric problem. Notation for the Hypergeometric: H = Hypergeometric Probability Distribution Function X ~ H(r, b, n) Read this as X is a random variable with a hypergeometric distribution. The parameters are r, b, and n: r = the size of the group of interest (first group), b = the size of the second group, n = the size of the chosen sample. Example 4.24 A school site committee is to be chosen randomly from six men and five women. If the committee consists of four members chosen randomly, what is the probability that two of them are men? How many men do you expect to be on the committee? Let X = the number of men on the committee of four. The men are the group of interest (first group). X takes on the values 0, 1, 2, 3, 4, where r = 6, b = 5, and n = 4. X ~ H(6, 5, 4) Find P(x = 2). P(x = 2) =.4545 (calculator or computer) NOTE Currently, the TI-83+ and TI-84 do not have hypergeometric probability functions. There are a number This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 279 of computer packages, including Microsoft Excel, that do. The probability that |
there are two men on the committee is about.45. The graph of X ~ H(6, 5, 4) is Figure 4.3 The y-axis contains the probability of X, where X = the number of men on the committee. You would expect m = 2.18 (about two) men on the committee. The formula for the mean is μ = nr r + b = (4)(6) 6 + 5 = 2.18. 4.24 An intramural basketball team is to be chosen randomly from 15 boys and 12 girls. The team has 10 slots. You want to know the probability that eight of the players will be boys. What is the group of interest and the sample? 4.6 | Poisson Distribution (Optional) There are two main characteristics of a Poisson experiment. 1. The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. For example, a book editor might be interested in the number of words spelled incorrectly in a particular book. It might be that, on the average, there are five words spelled incorrectly in 100 pages. The interval is the 100 pages. 2. The Poisson distribution may be used to approximate the binomial if the probability of success is small (such as.01) and the number of trials is large (such as 1,000). You will verify the relationship in the homework exercises. n is the number of trials, and p is the probability of a success. The random variable X = the number of occurrences in the interval of interest. Example 4.25 The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12. Of interest is the number of loaves of bread put on the shelf in five minutes. The time interval of interest is five minutes. What is the probability that the number of loaves, selected randomly, put on the shelf in five minutes is three? Let X = the number of loaves of bread put on the shelf in five minutes. If the average number of loaves put on the 280 Chapter 4 | Discrete Random Variables shelf in 30 minutes (half-hour) is 12, then the average number of loaves put on the shelf in five minutes is ⎛ ⎝ ⎞ ⎠ 5 30 (12) = 2 loaves of bread. The probability question asks |
you to find P(x = 3). 4.25 The average number of fish caught in an hour is eight. Of interest is the number of fish caught in 15 minutes. The time interval of interest is 15 minutes. What is the average number of fish caught in 15 minutes? Example 4.26 A bank expects to receive six bad checks per day, on average. What is the probability of the bank getting fewer than five bad checks on any given day? Of interest is the number of checks the bank receives in one day, so the time interval of interest is one day. Let X = the number of bad checks the bank receives in one day. If the bank expects to receive six bad checks per day then the average is six checks per day. Write a mathematical statement for the probability question. Solution 4.26 P(x < 5) 4.26 An electronics store expects to have 10 returns per day on average. The manager wants to know the probability of the store getting fewer than eight returns on any given day. State the probability question mathematically. Example 4.27 You notice that a news reporter says "uh," on average, two times per broadcast. What is the probability that the news reporter says "uh" more than two times per broadcast? This is a Poisson problem because you are interested in knowing the number of times the news reporter says "uh" during a broadcast. a. What is the interval of interest? Solution 4.27 a. one broadcast b. What is the average number of times the news reporter says "uh" during one broadcast? Solution 4.27 b. 2 c. Let X = ________. What values does X take on? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 281 Solution 4.27 c. Let X = the number of times the news reporter says "uh" during one broadcast. x = 0, 1, 2, 3,... d. The probability question is P(______). Solution 4.27 d. P(x > 2) 4.27 An emergency room at a particular hospital gets an average of five patients per hour. A doctor wants to know the probability that the ER gets more than five patients per hour. Give the reason why this would be a Poisson distribution. Notation for the Poisson: P = Poisson Probability Distribution Function X ~ P(μ) Read this as X is a |
random variable with a Poisson distribution. The parameter is μ (or λ); μ (or λ) = the mean for the interval of interest. Example 4.28 Leah's answering machine receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes? Let X = the number of calls Leah receives in 15 minutes. The interval of interest is 15 minutes or 1 4 hour. x = 0, 1, 2, 3,... If Leah receives, on the average, six telephone calls in two hours, and there are eight 15-minute intervals in two hours, then Leah receives (6) =.75 calls in 15 minutes, on average. So, μ =.75 for this problem(.75) Find P(x > 1). P(x > 1) =.1734 (calculator or computer) NOTE The TI calculators use λ (lambda) for the mean. • Press 1 – and then press 2nd DISTR. • Arrow down to poissoncdf. Press ENTER. • Enter (.75,1). • The result is P(x > 1) =.1734. 282 Chapter 4 | Discrete Random Variables The probability that Leah receives more than one telephone call in the next 15 minutes is about.1734 or P(x > 1) = 1 − poissoncdf(.75, 1). The graph of X ~ P(.75) is Figure 4.4 The y-axis contains the probability of x where X = the number of calls in 15 minutes. 4.28 A customer service center receives about 10 emails every half-hour. What is the probability that the customer service center receives more than four emails in the next six minutes? Use the TI-83+ or TI-84 calculator to find the answer. Example 4.29 According to Baydin, an email management company, an email user gets, on average, 147 emails per day. Let X = the number of emails an email user receives per day. The discrete random variable X takes on the values x = 0, 1, 2.... The random variable X has a Poisson distribution: X ~ P(147). The mean is 147 emails. a. What is the probability that an email user receives exactly 160 emails per day? b. What is the probability that an email user receives at most 160 emails per day? c. What is |
the standard deviation? Solution 4.29 a. P(x = 160) = poissonpdf(147, 160) ≈.0180 b. P(x ≤ 160) = poissoncdf(147, 160) ≈.8666 c. Standard Deviation = σ = μ = 147 ≈ 12.1244 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 283 4.29 According to a recent poll girls between the ages of 14 and 17 send an average of 187 text messages each day. Let X = the number of texts that a girl aged 14 to 17 sends per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: X ~ P(187). The mean is 187 text messages. a. What is the probability that a teen girl sends exactly 175 texts per day? b. What is the probability that a teen girl sends at most 150 texts per day? c. What is the standard deviation? Example 4.30 Text message users receive or send an average of 41.5 text messages per day. a. How many text messages does a text message user receive or send per hour? b. What is the probability that a text message user receives or sends two messages per hour? c. What is the probability that a text message user receives or sends more than two messages per hour? Solution 4.30 a. Let X = the number of texts that a user sends or receives in one hour. The average number of texts received per hour is 41.5 24 ≈ 1.7292. b. X ~ P(1.7292), so P(x = 2) = poissonpdf(1.7292, 2) ≈.2653 c. P(x > 2) = 1 – P(x ≤ 2) = 1 – poissoncdf(1.7292, 2) ≈ 1 –.7495 =.2505 4.30 Scientists recently researched the busiest airport in the world. On average, there are 2,500 arrivals and departures each day. a. How many airplanes arrive and depart the airport per hour? b. What is the probability that there are exactly 100 arrivals and departures in one hour? c. What is the probability that there are at most 100 arrivals and departures in one hour? Example |
4.31 On May 13, 2013, starting at 4:30 p.m., the probability of low seismic activity for the next 48 hours in Alaska was reported as about 1.02 percent. Use this information for the next 200 days to find the probability that there will be low seismic activity in 10 of the next 200 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close? Solution 4.31 Let X = the number of days with low seismic activity. Using the binomial distribution • P(x = 10) = binompdf(200,.0102, 10) ≈.000039 Using the Poisson distribution 284 Chapter 4 | Discrete Random Variables • Calculate μ = np = 200(.0102) ≈ 2.04 • P(x = 10) = poissonpdf(2.04, 10) ≈.000045 We expect the approximation to be good because n is large (greater than 20) and p is small (less than.05). The results are close—both probabilities reported are almost 0. 4.31 On May 13, 2013, starting at 4:30 p.m., the probability of moderate seismic activity for the next 48 hours in the Kuril Islands off the coast of Japan was reported at about 1.43 percent. Use this information for the next 100 days to find the probability that there will be low seismic activity in 5 of the next 100 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close? 4.7 | Discrete Distribution (Playing Card Experiment) This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 285 4.1 Discrete Distribution (Playing Card Experiment) Student Learning Outcomes • The student will compare empirical data and a theoretical distribution to determine if an everyday experiment fits a discrete distribution. • The student will compare technology-generated simulation and a theoretical distribution. • The student will demonstrate an understanding of long-term probabilities. Supplies • One full deck of playing cards • Programmable calculator Procedure for Empirical Data The experimental procedure for empirical data is to pick one card from a deck of shuffled cards. 1. The theoretical probability of picking a diamond from a deck is ________. 2. Shuffle a deck of cards. 3. Pick one card from it. 4. Record whether it was a diamond or |
not a diamond. 5. Put the card back and reshuffle. 6. Do this a total of 10 times. 7. Record the number of diamonds picked. 8. Let X = number of diamonds. Theoretically, X ~ B(_____,_____) Procedure for Simulation Repeat the experimental procedure using a programmable calculator. 1. Use the randInt function to generate data. Consider 1 to be spades, 2 to be hearts, 3 to be diamonds, and 4 to be clubs. Generate 10 draws of cards with four suits with randInt(1,4,10). 2. Let X = number of diamonds. Theoretically, X ~ B(_____,_____). Organize the Empirical Data 1. Record the number of diamonds picked for your class with playing cards in Table 4.15. Then calculate the relative frequency. x Frequency Relative Frequency 0 1 2 3 4 5 __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ 286 Chapter 4 | Discrete Random Variables x 6 7 8 9 Frequency Relative Frequency __________ __________ __________ __________ __________ __________ __________ __________ 10 __________ __________ Table 4.15 2. Calculate the following: a. b. x¯ = ________ s = ________ 3. Construct a histogram of the empirical data. Figure 4.5 Organize the Simulation Data 1. Use Table 4.16 to record the number of diamonds picked for your class using the calculator simulation. Calculate the relative frequency. X Frequency Relative Frequency 0 1 2 3 4 5 6 __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 4.16 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 287 X Frequency Relative Frequency 7 8 9 __________ __________ __________ __________ __________ __________ 10 __________ __________ Table 4.16 2. Calculate the following: a. b. x¯ = ________ s = ________ 3. Construct a histogram of the simulation data. Figure 4.6 Theoretical Distribution a. Build the |
theoretical PDF chart based on the distribution in the Procedure section. x P(x 288 Chapter 4 | Discrete Random Variables x P(x) 9 10 Table 4.17 b. Calculate the following: a. μ = ____________ b. σ = ____________ c. Construct a histogram of the theoretical distribution. Figure 4.7 Using the Data NOTE RF = relative frequency Use the table from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places. • P(x = 3) = ________ • P(1 < x < 4) = ________ • P(x ≥ 8) = ________ Use the data from the Organize the Empirical Data section to calculate the following answers. Round your answers to four decimal places. • RF(x = 3) = ________ • RF(1 < x < 4) = ________ • RF(x ≥ 8) = ________ Use the data from the Organize the Simulation Data section to calculate the following answers. Round your answers to four decimal places. • RF(x = 3) = ________ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 289 • RF(1 < x < 4) = ________ • RF(x ≥ 8) = ________ Discussion Questions For Questions 1 and 2, think about the shapes of the two graphs, the probabilities, the relative frequencies, the means, and the standard deviations. 1. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical, empirical, and simulation distributions. Use complete sentences. 2. Describe the three most significant differences between the graphs or distributions of the theoretical, empirical, and simulation distributions. 3. Using your answers from Questions 1 and 2, does it appear that the two sets of data fit the theoretical distribution? In complete sentences, explain why or why not. 4. Suppose that the experiment had been repeated 500 times. Would you expect Table 4.15, Table 4.16, or Table 4.17 to change, and how would it change? Why? Why wouldn’t the other table(s) change? 4.8 | Discrete Distribution (Lucky Dice Experiment) 290 Chapter 4 | Discrete Random Variables 4.2 Discrete Distribution (Lucky Dice Experiment) Student Learning Outcomes • The student will compare empirical data and a theoretical distribution |
to determine if a Tet gambling game fits a discrete distribution. • The student will demonstrate an understanding of long-term probabilities. Supplies • One “Lucky Dice” game or three regular dice • One programming calculator Procedure Round answers to relative frequency and probability problems to four decimal places. 1. The experimental procedure is to bet on one object. Then, roll three Lucky Dice and count the number of matches. The number of matches will decide your profit. 2. What is the theoretical probability of one die matching the object? 3. Choose one object to place a bet on. Roll the three Lucky Dice. Count the number of matches. 4. Let X = number of matches. Theoretically, X ~ B(______,______) 5. Let Y = profit per game. Organize the Data In Table 4.18, fill in the y-value that corresponds to each x-value. Next, record the number of matches picked for your class. Then, calculate the relative frequency. 1. Complete the table. y Frequency Relative Frequency x 0 1 2 3 Table 4.18 2. Calculate the following: a. b. c. d. x¯ = _______ sx = ________ y¯ = _______ sy = _______ 3. Explain what x¯ represents. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 291 4. Explain what y¯ represents. 5. Based upon the experiment, answer the following questions: a. What was the average profit per game? b. Did this represent an average win or loss per game? c. How do you know? Answer in complete sentences. 6. Construct a histogram of the empirical data. Figure 4.8 Theoretical Distribution Build the theoretical PDF chart for x and y based on the distribution from the Procedure section. 1. y P(x) = P(y) x 0 1 2 3 Table 4.19 2. Calculate the following: a. μx = ________ b. σx = ________ c. μx = ________ 3. Explain what μx represents. 4. Explain what μy represents. 5. Based upon theory, answer the following questions: a. What was the expected profit per game? b. Did the expected profit represent an average win or loss per game? c. How do you know? Answer in complete sentences. 292 Chapter 4 | Discrete |
Random Variables 6. Construct a histogram of the theoretical distribution. Figure 4.9 Use the Data NOTE RF = relative frequency Use the data from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places. 1. P(x = 3) = ________ 2. P(0 < x < 3) = ________ 3. P(x ≥ 2) = ________ Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places. 1. RF(x = 3) = ________ 2. RF(0 < x < 3) = ________ 3. RF(x ≥ 2) = ________ Discussion Question For Questions 1 and 2, consider the graphs, the probabilities, the relative frequencies, the means, and the standard deviations. 1. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical and empirical distributions. Use complete sentences. 2. Describe the three most significant differences between the graphs or distributions of the theoretical and empirical distributions. 3. Thinking about your answers to Questions 1 and 2, does it appear that the data fit the theoretical distribution? In complete sentences, explain why or why not. 4. Suppose that the experiment had been repeated 500 times. Would you expect Table 4.18 or Table 4.19 to change, and how would it change? Why? Why wouldn’t the other table change? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 293 KEY TERMS Bernoulli trials an experiment with the following characteristics: 1. There are only two possible outcomes called success and failure for each trial 2. The probability p of a success is the same for any trial (so the probability q = 1 − p of a failure is the same for any trial) binomial experiment a statistical experiment that satisfies the following three conditions: 1. There are a fixed number of trials, n 2. There are only two possible outcomes, called success and, failure, for each trial; the letter p denotes the probability of a success on one trial, and q denotes the probability of a failure on one trial 3. The n trials are independent and are repeated using identical conditions a discrete random variable (RV) that arises from Bernoulli trials; there are a fixed binomial probability distribution number, n, of independent trials Independent means that the result of any |
trial (for example, trial one) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV X is defined as the number of successes in n trials. The notation is: X ~ B(n, p). The mean is μ = np and the standard deviation is σ = npq. The probability of the following exactly x successes in n trials is n P(X = x) = ⎛ ⎝ x ⎞ ⎠ pxqn − x expected value expected arithmetic average when an experiment is repeated many times; also called the mean; notations μ; for a discrete random variable (RV) with probability distribution function P(x),the definition can also be written in the form μ = ∑ xP(x) geometric distribution until the first success. The geometric variable X is defined as the number of trials until the first success. Notation X ~ G(p). The mean is μ a discrete random variable (RV) that arises from the Bernoulli trials; the trials are repeated = 1 p and the standard deviation is ⎞ ⎠. The probability of exactly x failures before the first success is given by the formula. P(X = x) = p(1 – p) x – 1 geometric experiment a statistical experiment with the following properties: 1. There are one or more Bernoulli trials with all failures except the last one, which is a success 2. In theory, the number of trials could go on foreve; there must be at least one trial 3. The probability, p, of a success and the probability, q, of a failure do not change from trial to trial hypergeometric experiment a statistical experiment with the following properties: 1. You take samples from two groups 2. You are concerned with a group of interest, called the first group 3. You sample without replacement from the combined groups 4. Each pick is not independent, since sampling is without replacement 5. You are not dealing with Bernoulli trials hypergeometric probability a discrete random variable (RV) that is characterized by the following: 1. The experiment uses a fixed number of trials. 2. The probability of success is not the same from trial to trial We sample from two groups of items when we are interested in only one group. X is defined as the number of 294 Chapter 4 | Discrete Random Variables successes out of the total number of items chosen |
. Notation X ~ H(r, b, n), where r = the number of items in the group of interest, b = the number of items in the group not of interest, and n = the number of items chosen. mean a number that measures the central tendency; a common name for mean is average The term mean is a shortened form of arithmetic mean. By definition, the mean for a sample (denoted by x¯ ) is x¯ = is μ = and the mean for a population (denoted by μ) Sum of all values in the sample Number of values in the sample Sum of all values in the population Number of values in the population. mean of a probability distribution the long-term average of many trials of a statistical experiment Poisson probability distribution a discrete random variable (RV) that counts the number of times a certain event will occur in a specific interval; characteristics of the variable: • The probability that the event occurs in a given interval is the same for all intervals • The events occur with a known mean and independently of the time since the last event The distribution is defined by the mean μ of the event in the interval. Notation X ~ P(μ). The mean is μ = np. The μ x x! standard deviation is σ = μ. The probability of having exactly x successes in r trials is P(X = x) = (e −μ ). The Poisson distribution is often used to approximate the binomial distribution, when n is large and p is small (a general rule is that n should be greater than or equal to 20 and p should be less than or equal to.05). probability distribution function (PDF) a mathematical description of a discrete random variable (RV), given either in the form of an equation (formula) or in the form of a table listing all the possible outcomes of an experiment and the probability associated with each outcome random variable (RV) a characteristic of interest in a population being studied; common notation for variables are uppercase Latin letters X, Y, Z,... ; common notation for a specific value from the domain (set of all possible values of a variable) are lowercase Latin letters x, y, and z For example, if X is the number of children in a family, then x represents a specific integer 0, 1, 2, 3,... ; variables in statistics differ from variables in intermediate algebra in the two following ways: • The domain of the random variable (RV |
) is not necessarily a numerical set; the domain may be expressed in words; for example, if X = hair color then the domain is {black, blond, gray, green, orange} • We can tell what specific value x the random variable X takes only after performing the experiment standard deviation of a probability distribution experiment are from the mean of the distribution a number that measures how far the outcomes of a statistical the law of large numbers as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency probability approaches zero CHAPTER REVIEW 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable The characteristics of a probability distribution function (PDF) for a discrete random variable are as follows: 1. Each probability is between zero and one, inclusive (inclusive means to include zero and one) 2. The sum of the probabilities is one 4.2 Mean or Expected Value and Standard Deviation The expected value, or mean, of a discrete random variable predicts the long-term results of a statistical experiment that has been repeated many times. The standard deviation of a probability distribution is used to measure the variability of possible outcomes. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 295 4.3 Binomial Distribution (Optional) A statistical experiment can be classified as a binomial experiment if the following conditions are met: 1. There are a fixed number of trials, n 2. There are only two possible outcomes, called success and failure, for each trial; the letter p denotes the probability of a success on one trial and q denotes the probability of a failure on one trial 3. The n trials are independent and are repeated using identical conditions The outcomes of a binomial experiment fit a binomial probability distribution. The random variable X = the number of successes obtained in the n independent trials. The mean of X can be calculated using the formula μ = np, and the standard deviation is given by the formula σ = npq. 4.4 Geometric Distribution (Optional) There are three characteristics of a geometric experiment: 1. There are one or more Bernoulli trials with all failures except the last one, which is a success 2. In theory, the number of trials could go on forever; there must be at least one trial 3. The probability, p, of a success and the probability, q, of a failure are the same |
for each trial In a geometric experiment, define the discrete random variable X as the number of independent trials until the first success. We say that X has a geometric distribution and write X ~ G(p) where p is the probability of success in a single trial. The mean of the geometric distribution X ~ G(p) is μ = 1 − p p2 =.5 Hypergeometric Distribution (Optional) A hypergeometric experiment is a statistical experiment with the following properties: 1. You take samples from two groups 2. You are concerned with a group of interest, called the first group 3. You sample without replacement from the combined groups 4. Each pick is not independent, since sampling is without replacement 5. You are not dealing with Bernoulli trials The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution. The random variable X = the number of items from the group of interest. The distribution of X is denoted X ~ H(r, b, n), where r = the size of the group of interest (first group), b = the size of the second group, and n = the size of the chosen sample. It follows that n ≤ r + b. The mean of X is μ = nr r + b and the standard deviation is σ = rbn(r + b − n) (r + b)2 (r + b − 1). 4.6 Poisson Distribution (Optional) A Poisson probability distribution of a discrete random variable gives the probability of a number of events occurring in a fixed interval of time or space, if these events happen at a known average rate and independently of the time since the last event. The Poisson distribution may be used to approximate the binomial, if the probability of success is small (less than or equal to.05) and the number of trials is large (greater than or equal to 20). FORMULA REVIEW 296 Chapter 4 | Discrete Random Variables 4.2 Mean or Expected Value and Standard Deviation Mean or Expected Value: μ = ∑ x ∈ X xP(x) Standard Deviation: σ = ∑ x ∈ X (x − μ)2 P(x) 4.3 Binomial Distribution (Optional) X ~ B(n, p) means that the discrete random variable X has a binomial probability distribution with n trials and probability of success p. X = the number of successes in n independent trials n = the number of independent |
and n = the size of the chosen sample. X = the number of items from the group of interest that are in the chosen sample, and X may take on the values x = 0, 1,..., up to the size of the group of interest. The minimum value for X may be larger than zero in some instances. n ≤ r + b The mean of X is given by the formula μ = nr r + b and the standard deviation is = rbn(r + b − n) (r + b)2(r + b − 1). 4.6 Poisson Distribution (Optional) X ~ P(μ) means that X has a Poisson probability distribution where X = the number of occurrences in the interval of interest. X takes on the values x = 0, 1, 2, 3,... The mean μ is typically given. The variance is σ2 = μ, and the standard deviation is σ = μ. When P(μ) is used to approximate a binomial distribution, μ = np where n represents the number of independent trials and p represents the probability of success in a single trial. 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable Use the following information to answer the next five exercises: A company wants to evaluate its attrition rate, or in other words, how long new hires stay with the company. Over the years, the company has established the following probability distribution: Let X = the number of years a new hire will stay with the company. Let P(x) = the probability that a new hire will stay with the company x years. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 297 1. Complete Table 4.20 using the data provided. x P(x) 0 1 2 3 4 5 6.12.18.30.15.10.05 Table 4.20 2. P(x = 4) = ________ 3. P(x ≥ 5) = ________ 4. On average, how long would you expect a new hire to stay with the company? 5. What does the column “P(x)” sum to? Use the following information to answer the next four exercises: A baker is deciding how many batches of muffins to make to sell in his bakery. He wants to make enough to sell every one and no fewer. |
Through observation, the baker has established a probability distribution. x P(x) 1 2 3 4.15.35.40.10 Table 4.21 6. Define the random variable X. 7. What is the probability the baker will sell more than one batch? P(x > 1) = ________ 8. What is the probability the baker will sell exactly one batch? P(x = 1) = ________ 9. On average, how many batches should the baker make? Use the following information to answer the next two exercises: Ellen has music practice three days a week. She practices for all of the three days 85 percent of the time, two days 8 percent of the time, one day 4 percent of the time, and no days 3 percent of the time. One week is selected at random. 10. Define the random variable X. 11. Construct a probability distribution table for the data. 12. We know that for a probability distribution function to be discrete, it must have two characteristics. One is that the sum of the probabilities is one. What is the other characteristic? Use the following information to answer the next five exercises: Javier volunteers in community events each month. He does not do more than five events in a month. He attends exactly five events 35 percent of the time, four events 25 percent 298 Chapter 4 | Discrete Random Variables of the time, three events 20 percent of the time, two events 10 percent of the time, one event 5 percent of the time, and no events 5 percent of the time. 13. Define the random variable X. 14. What values does x take on? 15. Construct a PDF table. 16. Find the probability that Javier volunteers for fewer than three events each month. P(x < 3) = ________ 17. Find the probability that Javier volunteers for at least one event each month. P(x > 0) = ________ 4.2 Mean or Expected Value and Standard Deviation 18. Complete the expected value table. x P(x) x*P(x) 0 1 2 3.2.2.4.2 Table 4.22 19. Find the expected value from the expected value table. 20. Find the standard deviation. x P(x) x*P(x) 2 4 6 8.1.3.4.2 2(.1) =.2 4(.3) = 1.2 6(.4) = 2.4 8(.2) |
= 1.6 Table 4.23 x P(x) x*P(x) (x – μ)2P(x) 2 4 6 8 0.1 0.3 0.4 0.2 2(.1) =.2 (2–5.4)2(.1) = 1.156 4(.3) = 1.2 (4–5.4)2(.3) =.588 6(.4) = 2.4 (6–5.4)2(.4) =.144 8(.2) = 1.6 (8–5.4)2(.2) = 1.352 Table 4.24 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 299 21. Identify the mistake in the probability distribution table. x P(x) x*P(x) 1 2 3 4 5.15.25.30.20.15.15.50.90.80.75 Table 4.25 22. Identify the mistake in the probability distribution table. x P(x) x*P(x) 1 2 3 4 5.15.25.25.20.15.15.40.65.85 1 Table 4.26 Use the following information to answer the next five exercises: A physics professor wants to know what percent of physics majors will spend the next several years doing postgraduate research. He has the following probability distribution: x P(x) x*P(x) 1 2 3 4 5 6.35.20.15.10.05 Table 4.27 23. Define the random variable X. 24. Define P(x), or the probability of x. 25. Find the probability that a physics major will do postgraduate research for four years. P(x = 4) = ________ 26. Find the probability that a physics major will do postgraduate research for at most three years. P(x ≤ 3) = ________ 27. On average, how many years would you expect a physics major to spend doing postgraduate research? Use the following information to answer the next seven exercises: A ballet instructor is interested in knowing what percent of each year's class will continue on to the next so that she can plan what classes to offer. Over the years, she has established the following probability distribution: • Let X |
= the number of years a student will study ballet with the teacher. 300 Chapter 4 | Discrete Random Variables • Let P(x) = the probability that a student will study ballet x years. 28. Complete Table 4.28 using the data provided. x P(x) x*P(x) 1 2 3 4 5 6 7.10.05.10.30.20.10 Table 4.28 29. In words, define the random variable X. 30. P(x = 4) = ________ 31. P(x < 4) = ________ 32. On average, how many years would you expect a child to study ballet with this teacher? 33. What does the column P(x) sum to and why? 34. What does the column x*P(x) sum to and why? 35. You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win $30. If it is not a face card, you pay $2. There are 12 face cards in a deck of 52 cards. What is the expected value of playing the game? 36. You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win $30. If it is not a face card, you pay $2. There are 12 face cards in a deck of 52 cards. Should you play the game? 4.3 Binomial Distribution (Optional) Use the following information to answer the next eight exercises: Researchers collected data from 203,967 incoming firsttime, full-time freshmen from 270 four-year colleges and universities in the United States. Of those students, 71.3 percent replied that, yes, they agreed with a recent federal law that was passed. Suppose that you randomly pick eight first-time, full-time freshmen from the survey. You are interested in the number who agreed with that law. 37. In words, define the random variable X. 38. X ~ _____(_____,_____) 39. What values does the random variable X take on? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 301 40. Construct the probability distribution function (PDF). x P(x) Table 4.29 41. On average (μ), how many would you expect to answer yes? 42. |
What is the standard deviation (σ)? 43. What is the probability that at most five of the freshmen reply yes? 44. What is the probability that at least two of the freshmen reply yes? 4.4 Geometric Distribution (Optional) Use the following information to answer the next six exercises: Researchers collected data from 203,967 incoming firsttime, full-time freshmen from 270 four-year colleges and universities in the United States. Of those students, 71.3 percent replied that, yes, they agree with a recent law that was passed. Suppose that you randomly select freshman from the study until you find one who replies yes. You are interested in the number of freshmen you must ask. 45. In words, define the random variable X. 46. X ~ _____(_____,_____) 47. What values does the random variable X take on? 48. Construct the probability distribution function (PDF). Stop at x = 6. x P(x) 1 2 3 4 5 6 Table 4.30 49. On average (μ), how many freshmen would you expect to have to ask until you found one who replies yes? 50. What is the probability that you will need to ask fewer than three freshmen? 4.5 Hypergeometric Distribution (Optional) Use the following information to answer the next five exercises: Suppose that a group of statistics students is divided into 302 Chapter 4 | Discrete Random Variables two groups: business majors and non-business majors. There are 16 business majors in the group and seven non-business majors in the group. A random sample of nine students is taken. We are interested in the number of business majors in the sample. 51. In words, define the random variable X. 52. X ~ _____(_____,_____) 53. What values does X take on? 54. Find the standard deviation. 55. On average (μ), how many would you expect to be business majors? 4.6 Poisson Distribution (Optional) Use the following information to answer the next six exercises: On average, a clothing store gets 120 customers per day. 56. Assume the event occurs independently in any given day. Define the random variable X. 57. What values does X take on? 58. What is the probability of getting 150 customers in one day? 59. What is the probability of getting 35 customers in the first four hours? Assume the store is open 12 hours each day. 60. What is the probability that the store will have more than 12 customers |
in the first hour? 61. What is the probability that the store will have fewer than 12 customers in the first two hours? 62. Which type of distribution can the Poisson model be used to approximate? When would you do this? Use the following information to answer the next six exercises: On average, eight teens in the United States die from motor vehicle injuries per day. As a result, states across the country are debating raising the driving age. 63. Assume the event occurs independently in any given day. In words, define the random variable X. 64. X ~ _____(_____,_____) 65. What values does X take on? 66. For the given values of the random variable X, fill in the corresponding probabilities. 67. Is it likely that there will be no teens killed from motor vehicle injuries on any given day in the United States? Justify your answer numerically. 68. Is it likely that there will be more than 20 teens killed from motor vehicle injuries on any given day in the United States? Justify your answer numerically. HOMEWORK This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 303 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable 69. Suppose that the PDF for the number of years it takes to earn a bachelor of science (B.S.) degree is given in Table 4.31. x P(x) 3 4 5 6 7.05.40.30.15.10 Table 4.31 In words, define the random variable X. a. b. What does it mean that the values 0, 1, and 2 are not included for x in the PDF? 4.2 Mean or Expected Value and Standard Deviation 70. A theater group holds a fund-raiser. It sells 100 raffle tickets for $5 apiece. Suppose you purchase four tickets. The prize is two passes to a Broadway show, worth a total of $150. In words, define the random variable X. a. What are you interested in here? b. c. List the values that X may take on. d. Construct a PDF. e. If this fund-raiser is repeated often and you always purchase four tickets, what would be your expected average winnings per raffle? 71. A game involves selecting a card from a regular 52-card deck and tossing a coin. The coin |
is a fair coin and is equally likely to land on heads or tails. • • • If the card is a face card, and the coin lands on heads, you win $6. If the card is a face card, and the coin lands on tails, you win $2. If the card is not a face card, you lose $2, no matter what the coin shows. a. Find the expected value for this game (expected net gain or loss). b. Explain what your calculations indicate about your long-term average profits and losses on this game. c. Should you play this game to win money? 72. You buy a ticket to a raffle that costs $10 per ticket. There are only 100 tickets available to be sold in this raffle. In this raffle there are one $500 prize, two $100 prizes, and four $25 prizes. Find your expected gain or loss. 73. Complete the PDF and answer the questions. x P(x) xP(x) 0 1 2 3.3.2.4 Table 4.32 a. Find the probability that x = 2. b. Find the expected value. 304 Chapter 4 | Discrete Random Variables 74. Suppose that you are offered the following deal: You roll a die. If you roll a six, you win $10. If you roll a four or five, you win $5. If you roll a one, two, or three, you pay $6. In words, define the random variable X. a. What are you ultimately interested in here (the value of the roll or the money you win)? b. c. List the values that X may take on. d. Construct a PDF. e. Over the long run of playing this game, what are your expected average winnings per game? f. Based on numerical values, should you take the deal? Explain your decision in complete sentences. 75. A venture capitalist, willing to invest $1,000,000, has three investments to choose from: The first investment, a software company, has a 10 percent chance of returning $5,000,000 profit, a 30 percent chance of returning $1,000,000 profit, and a 60 percent chance of losing the million dollars. The second company, a hardware company, has a 20 percent chance of returning $3,000,000 profit, a 40 percent chance of returning $1,000,000 profit, and a 40 percent chance of losing the million dollars. |
The third company, a biotech firm, has a 10 percent chance of returning $6,000,000 profit, a 70 percent of no profit or loss, and a 20 percent chance of losing the million dollars. a. Construct a PDF for each investment. b. Find the expected value for each investment. c. Which is the safest investment? Why do you think so? d. Which is the riskiest investment? Why do you think so? e. Which investment has the highest expected return, on average? 76. Suppose that 20,000 married adults in the United States were randomly surveyed as to the number of children they have. The results are compiled and are used as theoretical probabilities. Let X = the number of children married people have. x 0 1 2 3 4 5 P(x) xP(x).10.20.30.10.05 6 (or more).05 Table 4.33 In words, what does the expected value in this example represent? a. Find the probability that a married adult has three children. b. c. Find the expected value. d. Is it more likely that a married adult will have two to three children or four to six children? How do you know? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 305 77. Suppose that the PDF for the number of years it takes to earn a bachelor of science (B.S.) degree is given as in Table 4.34. x P(x) 3 4 5 6 7.05.40.30.15.10 Table 4.34 On average, how many years do you expect it to take for an individual to earn a B.S.? 306 Chapter 4 | Discrete Random Variables 78. People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video to Go is given in the following table. There is a five-video limit per customer at this store, so nobody ever rents more than five DVDs. x P(x) 0 1 2 3 4 5.03.50.24.70.04 Table 4.35 a. Describe the random variable X in words. b. Find the probability that a customer rents three DVDs. c. Find the probability that a customer rents at least four DVDs. d. Find the probability that a customer rents at most two DVDs. Another shop, |
Entertainment Headquarters, rents DVDs and video games. The probability distribution for DVD rentals per customer at this shop is given as follows. They also have a five-DVD limit per customer. x P(x) 0 1 2 3 4 5.35.25.20.10.05.05 Table 4.36 e. At which store is the expected number of DVDs rented per customer higher? f. If Video to Go estimates that they will have 300 customers next week, how many DVDs do they expect to rent next week? Answer in sentence form. If Video to Go expects 300 customers next week, and Entertainment Headquarters projects that they will have 420 customers, for which store is the expected number of DVD rentals for next week higher? Explain. g. h. Which of the two video stores experiences more variation in the number of DVD rentals per customer? How do you know that? 79. A “friend” offers you the following deal: For a $10 fee, you may pick an envelope from a box containing 100 seemingly identical envelopes. However, each envelope contains a coupon for a free gift. • Ten of the coupons are for a free gift worth $6. • Eighty of the coupons are for a free gift worth $8. • Six of the coupons are for a free gift worth $12. • Four of the coupons are for a free gift worth $40. Based upon the financial gain or loss over the long run, should you play the game? a. Yes, I expect to come out ahead in money. b. No, I expect to come out behind in money. It doesn’t matter. I expect to break even. c. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 307 80. A university has 14 statistics classes scheduled for its Summer 2013 term. One class has space available for 30 students, eight classes have space for 60 students, one class has space for 70 students, and four classes have space for 100 students. a. What is the average class size assuming each class is filled to capacity? b. Space is available for 980 students. Suppose that each class is filled to capacity and select a statistics student at random. Let the random variable X equal the size of the student’s class. Define the PDF for X. c. Find the mean of X. d. Find the standard deviation of X. 81. In a raffle |
, there are 250 prizes of $5, 50 prizes of $25, and 10 prizes of $100. Assuming that 10,000 tickets are to be issued and sold, what is a fair price to charge to break even? 4.3 Binomial Distribution (Optional) 82. According to a recent article the average number of babies born with significant hearing loss (deafness) is approximately two per 1,000 babies in a healthy baby nursery. The number climbs to an average of 30 per 1,000 babies in an intensive care nursery. Suppose that 1,000 babies from healthy baby nurseries were randomly surveyed. Find the probability that exactly two babies were born deaf. Use the following information to answer the next four exercises: Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about 4 percent. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu. 83. Define the random variable and list its possible values. 84. State the distribution of X. 85. Find the probability that at least four of the 25 patients actually have the flu. 86. On average, for every 25 patients calling in, how many do you expect to have the flu? 87. People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video to Go is given Table 4.37. There is a five-video limit per customer at this store, so nobody ever rents more than five DVDs. x P(x) 0 1 2 3 4 5.03.50.24.07.04 Table 4.37 a. Describe the random variable X in words. b. Find the probability that a customer rents three DVDs. c. Find the probability that a customer rents at least four DVDs. d. Find the probability that a customer rents at most two DVDs. 308 Chapter 4 | Discrete Random Variables 88. A school newspaper reporter decides to randomly survey 12 students to see if they will attend Tet (Vietnamese New Year) festivities this year. Based on past years, she knows that 18 percent of students attend Tet festivities. We are interested in the number of students who will attend the festivities. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X |
~ _____(_____,_____) d. How many of the 12 students do we expect to attend the festivities? e. Find the probability that at most four students will attend. f. Find the probability that more than two students will attend. Use the following information to answer the next three exercises: The probability that a local hockey team will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain date). An upcoming monthly schedule contains 12 games. 89. What is the expected number of wins for that upcoming month? a. 1.67 b. 12 c. 382 1043 d. 4.43 Let X = the number of games won in that upcoming month. 90. What is the probability that the team wins six games in that upcoming month? a. b. c. d..1476.2336.7664.8903 91. What is the probability that the team wins at least five games in that upcoming month a. b. c. d..3694.5266.4734.2305 92. A student takes a 10-question true-false quiz, but did not study and randomly guesses each answer. Find the probability that the student passes the quiz with a grade of at least 70 percent of the questions correct. 93. A student takes a 32-question multiple choice exam, but did not study and randomly guesses each answer. Each question has three possible choices for the answer. Find the probability that the student guesses more than 75 percent of the questions correctly. 94. Six different colored dice are rolled. Of interest is the number of dice that show a one. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. On average, how many dice would you expect to show a one? e. Find the probability that all six dice show a one. f. Is it more likely that three or that four dice will show a one? Use numbers to justify your answer numerically. In words, define the random variable X. 95. More than 96 percent of the very largest colleges and universities (more than 15,000 total enrollments) have some online offerings. Suppose you randomly pick 13 such institutions. We are interested in the number that offer distance learning courses. a. b. List the values that X may take |
on. c. Give the distribution of X. X ~ _____(_____,_____) d. On average, how many schools would you expect to offer such courses? e. Find the probability that at most 10 offer such courses. f. Is it more likely that 12 or that 13 will offer such courses? Use numbers to justify your answer numerically and answer in a complete sentence. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 309 96. Suppose that about 85 percent of graduating students attend their graduation. A group of 22 graduating students is randomly chosen. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many are expected to attend their graduation? e. Find the probability that 17 or 18 attend. f. Based on numerical values, would you be surprised if all 22 attended graduation? Justify your answer numerically. 97. At the Fencing Center, 60 percent of the fencers use the foil as their main weapon. We randomly survey 25 fencers at the Fencing Center. We are interested in the number of fencers who do not use the foil as their main weapon. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many are expected to not to use the foil as their main weapon? e. Find the probability that six do not use the foil as their main weapon. f. Based on numerical values, would you be surprised if all 25 did not use foil as their main weapon? Justify your answer numerically. 98. Approximately 8 percent of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number who participated in after-school sports all four years of high school. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many seniors are expected to have participated in after-school sports all four years of high school? e. Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all In words, |
define the random variable X. four years of high school? Justify your answer numerically. f. Based upon numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. In words, define the random variable X. 99. The chance of an IRS audit for a tax return reporting more than $25,000 in income is about 2 percent per year. We are interested in the expected number of audits a person with that income has in a 20-year period. Assume each year is independent. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many audits are expected in a 20-year period? e. Find the probability that a person is not audited at all. f. Find the probability that a person is audited more than twice. 100. It has been estimated that only about 30 percent of California residents have adequate earthquake supplies. Suppose you randomly survey 11 California residents. We are interested in the number who have adequate earthquake supplies. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. What is the probability that at least eight have adequate earthquake supplies? e. f. How many residents do you expect will have adequate earthquake supplies? Is it more likely that none or that all of the residents surveyed will have adequate earthquake supplies? Why? 310 Chapter 4 | Discrete Random Variables 101. There are two similar games played for Chinese New Year and Vietnamese New Year. In the Chinese version, fair dice with numbers 1, 2, 3, 4, 5, and 6 are used, along with a board with those numbers. In the Vietnamese version, fair dice with pictures of a gourd, fish, rooster, crab, crayfish, and deer are used. The board has those six objects on it, also. We will play with bets being $1. The player places a bet on a number or object. The house rolls three dice. If none of the dice show the number or object that was bet, the house keeps the $1 bet. If one of the dice shows the number or object bet (and the other two do not show it), the player gets back his or her $1 bet, plus $1 |
profit. If two of the dice show the number or object bet (and the third die does not show it), the player gets back his or her $1 bet, plus $2 profit. If all three dice show the number or object bet, the player gets back his or her $1 bet, plus $3 profit. Let X = number of matches and Y = profit per game. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. List the values that Y may take on. Then, construct one PDF table that includes both X and Y and their probabilities. e. Calculate the average expected matches over the long run of playing this game for the player. f. Calculate the average expected earnings over the long run of playing this game for the player. g. Determine who has the advantage, the player or the house. 102. According to the World Bank, only 9 percent of the population of Uganda had access to electricity as of 2009. Suppose we randomly sample 150 people in Uganda. Let X = the number of people who have access to electricity. a. What is the probability distribution for X? b. Using the formulas, calculate the mean and standard deviation of X. c. Use your calculator to find the probability that 15 people in the sample have access to electricity. d. Find the probability that at most 10 people in the sample have access to electricity. e. Find the probability that more than 25 people in the sample have access to electricity. 103. The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate in Afghanistan is 28.1 percent. Suppose you choose 15 people in Afghanistan at random. Let X = the number of people who are literate. a. Sketch a graph of the probability distribution of X. b. Using the formulas, calculate the (i) mean and (ii) standard deviation of X. c. Find the probability that more than five people in the sample are literate. Is it more likely that three people or four people are literate? 4.4 Geometric Distribution (Optional) 104. A consumer looking to buy a used red sports car will call dealerships until she finds a dealership that carries the car. She estimates the probability that any independent dealership will have the car will be 28 percent. We are interested in the number of dealerships she must call |
. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. On average, how many dealerships would we expect her to have to call until she finds one that has the car? e. Find the probability that she must call at most four dealerships. f. Find the probability that she must call three or four dealerships. 105. Suppose that the probability that an adult in America will watch the Super Bowl is 40 percent. Each person is considered independent. We are interested in the number of adults in America we must survey until we find one who will watch the Super Bowl. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many adults in America do you expect to survey until you find one who will watch the Super Bowl? e. Find the probability that you must ask seven people. f. Find the probability that you must ask three or four people. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 311 106. It has been estimated that only about 30 percent of California residents have adequate earthquake supplies. Suppose we are interested in the number of California residents we must survey until we find a resident who does not have adequate earthquake supplies. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. What is the probability that we must survey just one or two residents until we find a California resident who does not have adequate earthquake supplies? e. What is the probability that we must survey at least three California residents until we find a California resident who does not have adequate earthquake supplies? f. How many California residents do you expect to need to survey until you find a California resident who does not have adequate earthquake supplies? g. How many California residents do you expect to need to survey until you find a California resident who does have adequate earthquake supplies? 107. In one of its spring catalogs, a retailer advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked |
more than once. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many pages do you expect to advertise footwear on them? e. Is it probable that all 20 will advertise footwear on them? Why or why not? f. What is the probability that fewer than 10 will advertise footwear on them? g. Reminder: A page may be picked more than once. We are interested in the number of pages that we must randomly survey until we find one that has footwear advertised on it. Define the random variable X and give its distribution. h. What is the probability that you only need to survey at most three pages in order to find one that advertises footwear on it? i. How many pages do you expect to need to survey in order to find one that advertises footwear? 108. Suppose that you are performing the probability experiment of rolling one fair six-sided die. Let F be the event of rolling a four or a five. You are interested in how many times you need to roll the die to obtain the first four or five as the outcome. • p = probability of success (event F occurs) • q = probability of failure (event F does not occur) a. Write the description of the random variable X. b. What are the values that X can take on? c. Find the values of p and q. d. Find the probability that the first occurrence of event F (rolling a four or five) is on the second trial. 109. Ellen has music practice three days a week. She practices for all of the three days 85 percent of the time, two days 8 percent of the time, one day 4 percent of the time, and no days 3 percent of the time. One week is selected at random. What values does X take on? 110. Researchers investigate the prevalence of a particular infectious disease in countries around the world. According to their data, “Prevalence of this disease refers to the percentage of people ages 15 to 49 who are infected with it.” In South Africa, the prevalence of this disease is 17.3 percent. Let X = the number of people you test until you find a person infected with this disease. a. Sketch a graph of the distribution of the discrete random variable X. b. What is the probability that you must test 30 people to find one with this disease? c. |
What is the probability that you must ask 10 people? d. Find the (i) mean and (ii) standard deviation of the distribution of X. 312 Chapter 4 | Discrete Random Variables 111. According to a recent poll, 75 percent of millennials (people born between 1981 and 1995) have a profile on a social networking site. Let X = the number of millennials you ask until you find a person without a profile on a social networking site. a. Describe the distribution of X. b. Find the (i) mean and (ii) standard deviation of X. c. What is the probability that you must ask 10 people to find one person without a social networking site? d. What is the probability that you must ask 20 people to find one person without a social networking site? e. What is the probability that you must ask at most five people? 4.5 Hypergeometric Distribution (Optional) 112. A group of martial arts students is planning on participating in an upcoming demonstration. Six are students of tae kwon do, and seven are students of shotokan karate. Suppose that eight students are randomly picked to be in the first demonstration. We are interested in the number of shotokan karate students in that first demonstration. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many shotokan karate students do we expect to be in that first demonstration? 113. In one of its spring catalogs, a retailer advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked at most once. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many pages do you expect to advertise footwear on them? e. Calculate the standard deviation. 114. Suppose that a technology task force is being formed to study technology awareness among instructors. Assume that 10 people will be randomly chosen to be on the committee from a group of 28 volunteers, 20 who are technically proficient and eight who are not. We are interested in the number on the committee who are not technically proficient. In words, define the random variable X. a. b. List the values that X may |
take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many instructors do you expect on the committee who are not technically proficient? e. Find the probability that at least five on the committee are not technically proficient. f. Find the probability that at most three on the committee are not technically proficient. 115. Suppose that nine Massachusetts athletes are scheduled to appear at a charity benefit. The nine are randomly chosen from eight volunteers from the local basketball team and four volunteers from the local football team. We are interested in the number of football players picked. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. Are you choosing the nine athletes with or without replacement? 116. A bridge hand is defined as 13 cards selected at random and without replacement from a deck of 52 cards. In a standard deck of cards, there are 13 cards from each suit: hearts, spades, clubs, and diamonds. What is the probability of being dealt a hand that does not contain a heart? a. What is the group of interest? b. How many are in the group of interest? c. How many are in the other group? d. Let X = _________. What values does X take on? e. The probability question is P(_______). f. Find the probability in question. g. Find the (i) mean and (ii) standard deviation of X. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 313 4.6 Poisson Distribution (Optional) 117. The switchboard in a Minneapolis law office gets an average of 5.5 incoming phone calls during the noon hour on Mondays. Experience shows that the existing staff can handle up to six calls in an hour. Let X = the number of calls received at noon. a. Find the mean and standard deviation of X. b. What is the probability that the office receives at most six calls at noon on Monday? c. Find the probability that the law office receives six calls at noon. What does this mean to the law office staff who get, on average, 5.5 incoming phone calls at noon? d. What is the probability that the office receives more than eight calls at noon? 118. The maternity ward at a hospital in the Philippines |
is one of the busiest in the world with an average of 60 births per day. Let X = the number of births in an hour. a. Find the mean and standard deviation of X. b. Sketch a graph of the probability distribution of X. c. What is the probability that the maternity ward will deliver three babies in one hour? d. What is the probability that the maternity ward will deliver at most three babies in one hour? e. What is the probability that the maternity ward will deliver more than five babies in one hour? 119. A manufacturer of decorative string lights knows that 3 percent of its bulbs are defective. Using both the binomial and Poisson distributions, find the probability that a string of 100 lights contains at most four defective bulbs. 120. The average number of children a Japanese woman has in her lifetime is 1.37. Suppose that one Japanese woman is randomly chosen. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. Find the probability that she has no children. e. Find the probability that she has fewer children than the Japanese average. f. Find the probability that she has more children than the Japanese average. 121. The average number of children a Spanish woman has in her lifetime is 1.47. Suppose that one Spanish woman is randomly chosen. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. Find the probability that she has no children. e. Find the probability that she has fewer children than the Spanish average. f. Find the probability that she has more children than the Spanish average. 122. Fertile, female cats produce an average of three litters per year. Suppose that one fertile, female cat is randomly chosen. Answer the questions about the cat's probability of litters in one year. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _______ d. Find the probability that she has no litters in one year. e. Find the probability that she has at least two litters in one year. f. Find the probability that she has exactly three litters in one year. 123. The chance of having an extra fortune in a fortune cookie is about |
3 percent. Given a bag of 144 fortune cookies, we are interested in the number of cookies with an extra fortune. Two distributions may be used to solve this problem, but only use one distribution to solve the problem. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many cookies do we expect to have an extra fortune? e. Find the probability that none of the cookies have an extra fortune. f. Find the probability that more than three have an extra fortune. g. As n increases, what happens involving the probabilities using the two distributions? Explain in complete sentences. 314 Chapter 4 | Discrete Random Variables 124. According to the South Carolina Department of Mental Health website, for every 200 U.S. women, the average number who suffer from a particular disease is one. Out of a randomly chosen group of 600 U.S. women. Determine the following: In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many are expected to suffer from this disease? e. Find the probability that no one suffers from this disease. f. Find the probability that more than four suffer from this disease. 125. The chance of an IRS audit for a tax return reporting more than $25,000 in income is about 2 percent per year. Suppose that 100 people with tax returns over $25,000 are randomly picked. We are interested in the number of people audited in one year. Use a Poisson distribution to anwer the following questions. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many are expected to be audited? e. Find the probability that no one was audited. f. Find the probability that at least three were audited. 126. Approximately 8 percent of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number who participated in after-school sports all four years of high school. In words, define the random variable X. a. b. List the values that X may take on. c. Give the |
distribution of X. X ~ _____(_____,_____) d. How many seniors are expected to have participated in after-school sports all four years of high school? e. Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. f. Based on numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. 127. On average, Pierre, an amateur chef, drops three pieces of eggshell into every two cake batters he makes. Suppose that you buy one of his cakes. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. On average, how many pieces of eggshell do you expect to be in the cake? e. What is the probability that there will not be any pieces of eggshell in the cake? f. Let’s say that you buy one of Pierre’s cakes each week for six weeks. What is the probability that there will not be any eggshell in any of the cakes? g. Based upon the average given for Pierre, is it possible for there to be seven pieces of shell in the cake? Why? Use the following information to answer the next two exercises: The average number of times per week that Mrs. Plum’s cats wake her up at night because they want to play is 10. We are interested in the number of times her cats wake her up each week. 128. In words, what is the random variable X? a. b. c. d. the number of times Mrs. Plum’s cats wake her up each week the number of times Mrs. Plum’s cats wake her up each hour the number of times Mrs. Plum’s cats wake her up each night the number of times Mrs. Plum’s cats wake her up 129. Find the probability that her cats will wake her up no more than five times next week. a. b. c. d..5000.9329.0378.0671 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 315 4.7 Discrete Distribution (Playing Card |
Experiment) 130. Use a programmable calculator to simulate a binomial distribution. a. How would you use the randInt function to simulate the number of successes in five trials of an experiment with two outcomes, each of which has a.5 probability of occurring? b. Use the randInt function to simulate 10 observations of the random variable in Part A. c. Find the sample mean and sample standard deviation. d. Compare the sample mean and sample standard deviation to the theoretical mean and the theoretical standard deviation. REFERENCES 4.2 Mean or Expected Value and Standard Deviation Florida State University. (n.d.). Class catalogue at the Florida State University. Retrieved from https://apps.oti.fsu.edu/ RegistrarCourseLookup/SearchFormLegacy World Earthquakes. http://www.worldearthquakes.com/index.php?option=ethq_prediction earthquakes: Live (2012). World earthquake news and highlights. Retrieved from 4.3 Binomial Distribution (Optional) American Cancer Society. http://www.cancer.org/cancer/pancreaticcancer/detailedguide/pancreatic-cancer-key-statistics (2013). What are the key statistics about pancreatic cancer? Retrieved from Central Intelligence Agency. (n.d.). The world factbook. Retrieved from https://www.cia.gov/library/publications/theworldfactbook/geos/af.html ESPN NBA. (2013). NBA statistics – 2013. Retrieved from http://espn.go.com/nba/statistics/_/seasontype/2 Newport, F. (2013). Americans still enjoy saving rather than spending: Few demographic differences seen in these views other than by income. GALLUP Economy. Retrieved from http://www.gallup.com/poll/162368/americansenjoy-savingrather-spending.aspx Pryor, J. H., et al. (2011). The American freshman: National norms fall 2011. Los Angeles, CA: Cooperative Institutional Research Program, Higher Education Research Institute. Retrieved from http://heri.ucla.edu/PDFs/pubs/TFS/Norms/ Monographs/TheAmericanFreshman2011.pdf Wikipedia. (n.d.). Distance education. Retrieved from http://en.wikipedia.org/wiki/Distance_education World Bank Group. (2013). Access to electricity (% of population). Retrieved from http://data.worldbank |
.org/indicator/ EG.ELC.ACCS.ZS?order=wbapi_data_value_2009%20wbapi_data_value%20wbapi_data_value-first&sort=asc 4.4 Geometric Distribution (Optional) Central Intelligence Agency. (n.d.). The world factbook. Retrieved from https://www.cia.gov/library/publications/theworldfactbook/geos/af.html Pew Research Center. (n.d.). Millennials: A portrait of generation next. Retrieved from http://www.pewsocialtrends.org/ files/2010/10/millennials-confident-connected-open-to-change.pdf Pew Research. (2013). Millennials: confident. Executive Summary: Pew Research Social & Demographic Trends. Retrieved from http://www.pewsocialtrends.org/2010/02/24/millennials-confident-connected-open-tochange/ Pryor, J. H., et al. (2011). The American freshman: National norms fall 2011. Los Angeles: Cooperative Institutional Research Program, Higher Education Research Institute. Retrieved from http://heri.ucla.edu/PDFs/pubs/TFS/Norms/ Monographs/ TheAmericanFreshman2011.pdf The European Union and ICON-Institute. (2007/8). Summary of the national risk and vulnerability assessment 2007/ http://ec.europa.eu/europeaid/where/asia/documents/ profile 8: afgh_brochure_summary_en.pdf Afghanistan. Retrieved from of A The World Bank. (2013). Prevalence of HIV, total (% of populations ages 15-49). Retrieved from http://data.worldbank.org/ 316 Chapter 4 | Discrete Random Variables indicator/SH.DYN.AIDS.ZS?order=wbapi_data_value_2011+wbapi_data_value+wbapi_data_value-last&sort=desc UNICEF Television. (n.d.). UNICEF reports on female literacy centers in Afghanistan established to teach women and girls basic reading and writing skills. (Video). Retrieved from http://www.unicefusa.org/assets/video/afghan-femaleliteracycenters.html 4.6 Poisson Distribution (Optional) Centers for Disease Control and Prevention. (2012, |
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