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Sushi Roulette

Sushi Roulette.

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Sushi Roulette

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  1. Sushi Roulette • In the Japanese game show Sushi Roulette, the contestant spins a large wheel that’s divided into 12 equal sections. Nine of the sections have a sushi roll, and three have a “wasabi bomb.” When the wheel stops, the contestant must eat whatever food is on that section. To win the game, the contestant must eat one wasabi bomb. Find the probability that it takes 3 or more spins for the contestant to get a wasabi bomb. Show your method clearly. 3.1

  2. Knee Surgery • Patients receiving artificial knees often experience pain after surgery. The pain is measured on a subjective scale with possible values of 1(low) to 5 (high). Let X be the pain score for a randomly selected patient. The following table gives part of the probability distribution for X. 0.0958

  3. Find P(X = 5) • If two patients who received artificial knees are chosen at random, what’s the probability that both of them report pain scores of 1 or 2? • Compute the mean and standard deviation of X.

  4. Keno • In a game of 4-spot Keno, the player picks 4 numbers from 1 to 80. The casino randomly selects 20 winning numbers from 1 to 80. The table below shows the possible outcomes of the game and their probabilities, along with the amount of money (Payout) that the player wins for a $1 bet. If X = the payout for a single $1 bet, you can check that = $0.70 and = $6.58. 0.7787

  5. Interpret the values of and . • Jerry places a single $5 bet on 4-spot Keno. Find the expected value and the standard deviation of his winnings. • Marla plays five games of 4-spot Keno, betting $1 each time. Find the expected value and the standard deviation of her total winnings. • Based on your answers to (b) and (c), which player would the casino prefer? Justify your answer.

  6. Winner, Winner, Chicken Dinner • As a special promotion for its 20 –ounce bottles of soda, a soft drink company printed a message on the inside of each cap. Some of the caps said, “Please try again,” while others said, “You’re a winner!” The company advertised the promotion with the slogan “1 in 6 wins a prize.” Suppose the company is telling the truth and that every 20-ounce bottle of soda it fills has a 1-in-6 chance of being a winner. Seven friends each buy one 20-ounce bottle of the soda at a local convenience store. Let X = the number who win a prize. P(B|A) = P(B) = 0.5; Yes

  7. Explain why X is a binomial random variable. • Find the mean and standard deviation of X. Interpret each value in context. • The store clerk is surprised when three of the friends win a prize. Is this group of friends just lucky, or is the company’s 1-in-6 claim inaccurate? Compute P(X > 3) and use the result to justify your answer.

  8. Cranky Mower • To start her old mower, Rita has to pull a cord and hope for some luck. On any particular pull, the mower has a 20% chance of starting. • Find the probability that it takes her exactly 3 pulls to start the mower. • Find the probability that it takes her more than 10 pulls to start the mower. 0.5625

  9. 12,000 Miles & Running • A study of 12,000 able-bodied male students at the University of Illinois found that their times for the mile run were approximately Normal with mean 7.11 minutes and standard deviation 0.74 minutes. Choose a student at random from this group and call his time for the mile Y. Find P(Y < 6) and interpret the result. 47.3

  10. Too much juice? • A company has developed a drug test to detect steroid use by athletes. The test is accurate 95% of the time when an athlete has taken steroids. It is 97% accurate when an athlete hasn’t taken steroids. Suppose that the drug test will be used in a population of athletes in which 10% have actually taken steroids. Let’s choose an athlete at random and administer the drug test. 0.0668

  11. Make a tree diagram showing the sample space of this chance process. • What’s the probability that the randomly selected athlete tests positive? • Suppose that the chosen athlete tests positive. What’s the probability that he or she actually used steroids?

  12. Johnny’s Pizza • You work at Johnny’s pizza shop. You have the following information about the 7 pizzas in the oven: 3 of the 7 have thick crust, and of these 1 has only sausage and 2 have only mushrooms. The remaining 4 pizzas have regular crust, and of these 2 have only sausage and 2 have only mushrooms. Choose a pizza at random from the oven. 14.71

  13. Are the events {getting a thick-crust pizza) and {getting a pizza with mushrooms} independent? Explain. • You add an eighth pizza to the oven. This pizza has thick crust with only cheese. Now are the events {getting a thick-crust pizza) and {getting a pizza with mushrooms} independent? Explain.

  14. Too cool for the cabin? • During the winter months, the temperatures at the Young’s Colorado cabin can stay well below freezing (32 or 0) for weeks at a time. To prevent the pipes from freezing, Mrs. Young sets the thermostat at 50. She also buys a digital thermometer that records the indoor temperature each night at midnight. Unfortunately, the thermometer is programmed to measure the temperature in degrees Celsius. Based on several years’ worth of data, the temperature T in the cabin at midnight on a randomly selected night follows a Normal distribution with mean 8.5 and standard deviation 2.25. 0.1074

  15. Let Y = the temperature in the cabin at midnight on a randomly selected night in degrees Fahrenheit (recall that F = 9/5 C +32). Find the mean and standard deviation of Y. • Find the probability that the midnight temperature in the cabin is below 40 .

  16. Solutions • Sushi Roulette • Success = landing on wasabi bomb • 2 outcomes, independent trials, probability of success is the same from spin to spin, considering the number and types of sections remain the same between spins (p = .25) • We do not know the number of spins the contestant will take, making this variable a geometric random variable • Let X = # of spins until contestant lands on wasabi bomb • P(X > 3) = P(1st 2 spins are sushi rolls) = (1 - .25)2 = .5625

  17. Knee Surgery • P(X = 5) = 1 – P(X < 4) = 1 – 0.9 = 0.1 • P(X = 1 or X = 2) = P(one patient reports 1 or 2) = .1 +.2 = .3 P(two patients report 1 or 2) = P(patient 1 AND patient 2 report 1 or 2) = (.3)2 = 0.09 c)

  18. Keno • In the long run, a typical player can expect a payout of $0.70, which may vary as much as $6.58. • Jerry’s mean payout is 5 times the original mean…$3.50. The standard deviation of his winnings is 5 times the original standard deviation…$32.90 • Marla’s mean payout is .7 +.7 +.7+.7+.7 = $3.50. The standard deviations of her winnings is • The casino would prefer Marla. The standard deviations of her winnings is lower than Jerry’s. The casino would payout lower winnings more consistently.

  19. Winner, Winner, Chicken Dinner • There are two outcomes…winner or not. Whether or not one friend is a winner does not change the probability of another friends winning. The probability of success is the same from bottle to bottle (p = 1/6). There are seven friends who will purchase soda (known number of trials). • ; =0.986; If seven friends purchased bottles of soda over and over again, we would expect 1.17 of them to win, on average. This numbers of wins would vary by as much as 0.986 on average. • P(X > 3) = 1 – P(X < 2) = 1 – =0.0958 This probability suggests that we could expect 3 or more friends in a group of 7 to win about 77% of the time. The company’s claim appears to be inaccurate.

  20. Cranky Mower • This is a geometric setting… • Success = the mower cranks • Assume attempts are independent. (In practice, this may not be true.) • Probability of success does not change from attempt to attempt (p = 0 .2) • The number of attempts to be made is unknown • Let X = # of attempts until mower cranks. Then P(X = 3) = (1 – 0.2)2(0.2) = 0.128 • P(X > 10) = P(mower did not crank the 1st ten times) = (1 – 0.2)10 = 0.1074

  21. 12,000 Miles & Running • P(Y < 6) = P(z < • The probability that a randomly selected male student at the University of Illinois will run the mile in less than 6 minutes is 0.0668.

  22. Too much juice? b) P(tests positive) = P(positive and has used OR positive and has not used) = 0.95(0.1) + (0.03)(0.9) = 0.122 c) P(has used|tests positive) =

  23. Johnny’s Pizza • Let A = getting a thick-crust pizza • Let B = getting a pizza with mushrooms • P(B|A)=P(B) iff A & B are independent P(B|A) = ; P(B)=. Because the probability of getting a pizza with mushrooms changes when we know that the pizza has thick crust, these events are dependent. b) P(B|A) = ; P(B)=. Because the probability of getting a pizza with mushrooms does not change when we know that the pizza has thick crust, these events are now independent.

  24. Too cool for the cabin? b) P(Y < 40) = P(

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