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Probability You will want to be familiar with the probability formulas that are provided on the exam. When determining the probability of A or B, remember to subtract the intersection of A and B P ( A B ) P ( A) P ( B ) P ( A B ) When determining the probability of A given B, divide the intersection by the probability of P( A B) B P( A | B) P( B) The mean of a random variable is the sum of each value multiplied by the probability of each outcome. E ( X ) x xi pi The variance of a random variable is the sum each deviation squared multiplied by the probability of each outcome Var ( X ) x2 ( xi x ) 2 pi Remember standard deviation is the square root of variance! Remember for Binomial random variables, there should be two outcomes for each trial, each trial should be independent of the rest, there should be a fixed number of trials, and the probability of “success” should remain the same throughout. o If X has a binomial distribution with number of trials, n, and probability of success, p, then: n P ( X k ) p k (1 p ) n k k X np X np (1 p ) Additional rules (not included in test booklet): If two events are disjoint (mutually exclusive), then P( A B) 0 Example: Let A represent drawing a black card and B represent drawing a heart. Since hearts are red, P (Black card Heart) P ( A B ) 0 Two events are independent if and only if P ( A | B ) P ( A) or P ( A B ) P ( A) P ( B ) Example: Let A represent drawing a black card and B represent rolling a 5. Since cards in a standard deck and six-sided dice are independent, P (Black card Rolling a Five) P ( A B ) P ( A) P ( B ) 11 26 1 12 Other useful tools: Tree diagrams provide easy‐to‐interpret models for situations in which a scenario may be decomposed into multiple stages, one following another. It is important to understand that the probability on each branch in a tree diagram is a conditional probability. Two‐way (contingency) tables are particularly useful in interpreting conditional probabilities and examining independence. In such tables the entry in each cell is the joint probability for the row and column events that define the cell. The total probability for all cells must be 1, and the sum for each row (column) gives the “marginal” probability for that row (column). If you have a single row or column add to 1 you may need to rethink the formulation of the table! Example: Taking AP Physics Not taking AP Physics Total Taking AP Stats Not taking AP Stats Total 0.45 0.35 0.80 0.15 0.05 0.20 0.60 0.40 1.00 One person’s name is drawn at random from this population. Determine each 1. The probability the person is taking AP Stats. P(AP Stat.) 0.6 2. The probability the person is taking AP Stats or taking AP Physics. P (AP Stat. AP Phys.) 0.6 0.8 0.45 0.95 3. The probability the person is taking AP Stats and taking AP Physics. P (AP Stat. AP Phys.) 0.45 4. The person is taking AP Physics given that he/she is taking AP Stats. 0.45 P (AP Phys.|AP Stat.) 0.75 0.60 5. Are taking AP Stats and taking AP Physics independent? Explain. ? P (AP Phys.) P (AP Phys.|AP Stat.) No, since 0.80 0.75 , taking AP Stats and taking AP Physics are not independent events. Venn Diagrams are sometimes helpful, especially when given the probability of separate events. Rules for Random Variables: For any random variable X, where a and b are constants: E ( aX b) a E ( X ) b Var(aX + b) = a2Var(X) Example: The expected value of a six-sided die is 3.5 and its variance is 2.91667. Suppose we want to multiply the outcome of rolling a six-sided die by 5 and then add 2. In other words let Y 5 X 2 . Then the expected value of Y is E (Y ) E (5 X 2) 5 E( X ) 2 5(3.5) 2 19.5 And the variance of Y is Var (Y ) Var ( aX b ) a 2Var ( X ) 25(2.91667) 72.91667 When combining random variables X and Y, the means follow the same operation as the combination, in other words, if you are adding the means add, if you are subtracting the means subtract: E(X ± Y ) = E(X) ± E(Y ) For independent random variables X and Y, the variances always add. Be careful if you are given standard deviation!: Var(X ± Y ) = Var(X) + Var(Y ) Example: Imagine a population of ice cream containers with a mean content of 64 ounces and standard deviation of 2 ounces. Suppose we want to take a single scoop from each container. The scoop has a mean of 4 ounces with a standard deviation of 0.5 ounces. After all the containers have one scoop out of them, the mean amount in each container should be 60 ounces E ( X Y ) E ( X ) E (Y ) 64 4 60 ounces Remember that standard deviation is the square root of variance so the standard deviation of the containers after one scoop is taken out is X Y Var ( X Y ) Var ( X ) Var (Y ) 2 2 0.5 2 2.062 ounces Example: Suppose you roll a six-sided die twice and Y is the sum of those two rolls. Here, the random variable written as Y X 1 X 2 denotes the result of doing the same underlying action twice, noting the value obtained on the first action is independent of the one obtained on the second action, and adding the two values together. The mean of Y is then E (Y ) E ( X 1 ) E ( X 2 ) 3.5 3.5 7 And the standard deviation of Y is X1 X 2 Var ( X 1 ) Var ( X 2 ) (2.916667) (2.916667) 2.415 Compare the distribution of Y with the distribution of K the result of rolling one die and doubling its value. K 2 X so the mean of K is E (K ) 2E ( X ) 2(3.5) 7 And the standard deviation of K is 2 X 2 2 Var ( X ) 4(2.916667) 3.416 Thus X 1 X 2 2 X even though X , X 1 , and X 2 have the same distribution. From the examples, you can see that X 1 X 2 and 2 X have the same expected value but a different variance and standard deviation! Vocabulary Reminder The expected value of a random variable is the same as the mean of that random variable. Do not confuse “expected value” with “most likely outcome”. Calculator Use To save time on the exam, you will want to use your calculator for probability computations. Specifically, you will want to know how to: • Enter a probability distribution in lists (values in one list and probabilities in a second list) and use 1‐Var Stats to compute the mean (expected value) and standard deviation. • Use NormCdf to calculate areas associated with normal curves and use invNorm to calculate variable values associated with a given percentile. • Use binomPdf to compute the probability of a specific binomial outcome and binomCdf to compute the probability of an interval of binomial outcomes. Note: When you use your calculator for computations on Free Response questions, it will be very important to provide proper communication and support for your work. Multiple Choice Questions: Questions 1 and 2 refer to the following situation: The class of 1968 and 1998 held a joint reunion in 2008 at the local high school. Attendees were asked to fill out a survey to determine what they did after graduation. The results are summarized in the table below College Job Military Other 1968 56 73 85 7 1998 173 62 37 20 1. What is the probability that a randomly selected attendee graduated in 1998 and went into the military? A. B. C. D. E. 0.072 0.127 0.303 0.596 0.669 2. What is the probability that a randomly selected 1968 graduate went to college after graduation? A. B. C. D. E. 0.245 0.253 0.560 0.592 0.755 3. A fair die is rolled 3 times. The first 2 rolls resulted in 2 fives. What is the probability of not rolling a five on the next roll? A. 1 5 B. 6 2 3 1 5 C. 1 6 6 2 1 5 D. 6 6 E. 0 4. Which of the following statements is true for two events, each with probability greater than 0? A. B. C. D. E. If the events are mutually exclusive, they must be independent. If the events are independent, they must be mutually exclusive. If the events are not mutually exclusive, they must be independent. If the events are not independent, they must be mutually exclusive. If the events are mutually exclusive, they cannot be independent. 5. Let X represent a random variable whose distribution is normal, with a mean of 50 and a standard deviation of 5. Which of the following is equivalent to P ( X 55) ? A. B. C. D. E. P ( X 55) P ( X 55) P ( X 45) P (45 X 55) 1 P( X 45) 6. Lynn is planning to fly from New York to Los Angeles and will take the Airtight Airlines flight that leaves at 8 A.M. The website she used to make her reservation states that the probability that the flight will arrive in Los Angeles on time is 0.70. Of the following, which is the most reasonable explanation for how that probability could have been estimated? A. By using an extended weather forecast for the date of her flight, which showed a 30% chance of bad weather B. By making assumptions about how airplanes work, and factoring all of those assumptions into an equation to arrive at the probability C. From the fact that, of all airline flights arriving in California, 70% arrive on time D. From the fact that, of all airline flight in the United States, 70% arrive on time E. From the fact that, on all previous days this particular flight had been scheduled, it had arrived on time 70% of those days 7. According to a recent study on climate change, 40% of adults say that global climate change is a threat to the well-being of their country. Suppose a random sample of 25 adults is interviewed, what is the probability that more than 14 say that global climate change is a threat? A. B. C. D. E. 0.0132 0.0344 0.0778 0.9222 0.9656 8. In a game, a spinner with five equal-sized spaces is labeled from A to E. If a player spins an A they win 15 points. If any other letter is spun the player loses 4 points. What is the expected gain or loss from playing the game 40 times? A. B. C. D. E. Gain of 360 points. Gain of 55 points. Gain of 8 points. Loss of 1 point. Loss of 8 points. For questions 9 and 10 refer to the information below: The number of sweatshirts a vendor sells daily has the following probability distribution. Number of Sweatshirts X P(X) 0 0.3 1 0.2 2 0.3 3 0.1 4 0.08 5 0.02 9. If each sweatshirt sells for $25, what is the expected daily total dollar amount taken in by the vendor from the sale of sweatshirts? A. B. C. D. E. $5.00 $7.60 $35.50 $38.00 $75.00 10. Suppose the vendor constructs a weekly sales report for the 7-day dollar amount taken in by the vendor, Y. What is the standard deviation for Y, assuming X $1.33 , each sweatshirt sells for $25, and the sales on each day are independent? A. B. C. D. E. $15.26 $17.60 $76.29 $87.99 $232.80 11. Students, teachers, and parents in the drama club have been asked to decide on the fall fundraiser. They must choose from selling popcorn, suckers, or beef jerky. Their votes are given in the table below. Students Teachers Parents Total Popcorn 35 10 45 90 Suckers 35 0 55 90 Beef Jerky 80 10 0 90 Which of the following statement(s) is true? I. II. III. A. B. C. D. E. The events teacher and suckers are independent. The events beef jerky and parents are mutually exclusive. The probability of student or popcorn is approximately 0.7593 I only II only III only I and III only II and III only Total 150 20 100 270 Free Response Questions: 12. Two medicines are available as relief for motion sickness for commercial airline pilots. Medicine A is known to effectively cure the condition 40 percent of the time. A single dose of medicine A costs $10. Medicine B is known to effectively cure the condition 80 percent of the time. A single dose of medicine B costs $40. The medicines work independently of one another. Both medicines can be safely administered while in flight. The airline company intends to recommend one of the following two plans as the standard treatment for their pilots. Plan I: Treat with medicine A first. If it is not effective, then treat with medicine B. Plan II: Treat with medicine B first. If it is not effective, then treat with medicine A. (a) If a pilot uses plan I, what is the probability that the pilot will be cured? If a pilot uses plan II, what is the probability that the pilot will be cured? (b) Compute the expected cost per pilot when plan I is used for treatment. Compute the expected cost per pilot when plan II is used for treatment. (c) Based on the results in parts (a) and (b), which plan would you recommend to the airline? Explain your recommendation. 13. After winning the national championship, a local sports team is selling t-shirts. Each customer may purchase up to 4 adult size shirts and 5 children's size shirts. Let A be the number of adult shirts purchased by a single customer. The probability distribution of the number of adult shirts purchased by a single customer is given in the table below. A P( A) 0 0.1 1 0.15 2 0.55 3 0.05 4 0.15 (a) Compute the mean and standard deviation for A. (b) Suppose the mean and the standard deviation for the number of child shirts purchased by a single customer are 2 and 1.5, respectively. Assume that the numbers of adult shirts and child shirts purchased are independent random variables. Compute the mean and the standard deviation of the total number of adult and child shirts purchased by a single customer. (c) Suppose each child shirt costs $12 and each adult shirt costs $24. Compute the mean and standard deviation of the total amount spent per purchase. 14. Men’s shoe sizes are determined by their foot lengths. Suppose that men’s foot-lengths are approximately normally distributed with mean 27 centimeters and standard deviation 1.6 centimeters. A retailer sells men’s sandals in sizes S, M, L, XL, where the sandal sizes are defined in the table below. Sandal Size S M L Foot Length 24 foot length 26 26 foot length 28 28 foot length 30 XL 30 foot length 32 (a) Because the retailer only stocks the sizes listed above, what proportion of customers will find that the retailer does not carry any sandals in their sizes? Show your work. (b) Using a sketch of a normal curve, illustrate the proportion of men whose sandal size is L. Calculate this proportion. (c) Of 8 randomly selected customers, what is the probability that exactly 3 will request size L ? Show your work.