1 / 19

Assignment: Review probability topics, exercises in Chapters 15 and 16

Mar. 1 Statistic for the day: Number of years before the sequence of Easter dates repeats itself: 5,700,000. Source: http://webexhibits.org/calendars/. Assignment: Review probability topics, exercises in Chapters 15 and 16.

Télécharger la présentation

Assignment: Review probability topics, exercises in Chapters 15 and 16

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Mar. 1 Statistic for the day:Number of years before the sequence of Easter dates repeats itself: 5,700,000 Source: http://webexhibits.org/calendars/ Assignment: Review probability topics, exercises in Chapters 15 and 16 These slides were created by Tom Hettmansperger and in some cases modified by David Hunter

  2. ExpectationInsurance Example 14 p267 extended. Suppose my insurance company has 10,000 policy holders and they are all skateboarders. I collect a $500 premium each year. I pay off $1500 for a claim of a skate board accident. From past experience I know 10% will file a claim. How much do I expect to make per customer?

  3. Pr(claim) = .10 loss is $1500 - $500 = $1000 recorded as -$1000 Pr(no claim) = .90 gain is $500 -------------------------------------------------------------------------- Expected value = .10x(-1000) + .90x(500) = -100 + 450 = 350 dollars per customer -------------------------------------------------------------------------- Expected value for the 10,000 customers = 10,000x350 = 3,500,000 dollars per year

  4. Alternatively: Expect 10% or 1,000 claims for 1000x(-$1000) = -$1,000,000 loss Expect 90% or 9,000 earning $500 each or $4,500,000 gain So we expect $4,500,000 - $1,000,000 = $3,500,000 net profit

  5. Some old ideas from chapter 7 (histograms) and chapter 8 (bell shaped curves). The standard deviation is roughly (3,600,000 – 3,400,000)/4 = 50,000 Actual value is 45,000 computed from a formula.

  6. So 95% of the time the net profit will be between $3,410,00 and $3,590,000 with expected value $3,500,000

  7. Craps • Win on first roll if 7 or 11 • Lose on first roll if 2, 3, or 12 • Anything else on first roll becomes “point”: • Win on subsequent roll if point • Lose on subsequent roll if 7 • Probability of winning: 244/495, or 49.3%

  8. Betting ten dollars on a game of craps: You win $10 with probability .493 You lose $10 with probability .507 Expected winnings: .493($10) + .507(-$10) = -$.14 How is a fourteen-cent-loss relevant to your situation if you only play once? How is a fourteen-cent-gain relevant to your situation if you are the casino and you play ten thousand times a day?

  9. What would a 95% interval be? About -$500 to +$3500 Expected: $1400 Std Dev: About $1000

  10. What if the casino plays 100,000 times a day? Note: Now all values are positive

  11. Exercise 25 p276 Suppose 72% of children live with both parents, 22% live with mother only, 3% live with father only, 3% live with neither mother nor father. Pr(live with 2 parents) = .72 Pr(live with 1 parent) = .22 + .03 = .25 Pr(live with 0 parents) = .03 Expected number of parents lived with: 2x.72 + 1x.25 + 0x.03 = 1.44 + .25 + 0 = 1.69 parents If we sampled 1000 children and counted the total number of parents, we would expect around 1690 parents.

  12. Calibrating personal probabilities of experts: p. 290

  13. Anchoring • Do you think the number of cancer deaths in 1993 • was above or below 200,000? • What do you think the number of cancer deaths was • in 1993? Answer: 538,000 • Was the number of deaths due to aids • in 1993 lower or higher than 30,000? • What do you think the number of deaths due • to aids in 1993 was? Answer: 16,885

  14. Recall earlier quiz we didn’t have: • Mary likes earrings and spends time at festivals shopping • for jewelry. Her boy friend and several of her close girl • friends have tattoos. They have encouraged her to also • get a tattoo. • Unknown to you, Mary will be sitting next to you in the • next stat100.2 class. • Which of the following do you think is more likely and why? • Mary is a physics major. • Mary is a physics major with pierced ears.

  15. An answer of B (Mary is a physics major with pierced ears) is impossible and illustrates the Conjunction fallacy: assigning higher probability to a detailed scenario involving the conjunction of events than to one of the simple events that make up the conjunction. A possible cause of this fallacy is the Representative heuristic: leads people to assign higher probabilities than are warranted to scenarios that are representative of how we imagine things would happen.

  16. Forgotten base rates • Kahneman and Tversky example, p. 286 • Another possible example: Suppose that a dreaded disease affects 1% of those who get tested for it. Also suppose that the test is 99% accurate. What would you advise a patient who tests positive if the test result were the only piece of information? True probability of disease: about 50%

  17. Birthdays How many people must be in the same room to guarantee that at least 2 people will have the same birthday? Answer: 366 (not allowing for leap years) Different question: How many people must be in the same room to so that at least 2 people will have the same birthday with a high probability?

  18. Suppose there are 15 people in a room. What is the probability that at least 2 people have the same birthday? Pr(at least one match) = 1 - .75 = .25

More Related