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Chapter 3. Probability

Chapter 3. Probability

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64 <strong>Chapter</strong> 3: <strong>Probability</strong>8. Being Struck by LighteningNumber struck per year= 389, People in US during that year= 28,142,906P(struck by lightening this year) =389281,421,906= 0.00000138number of those struck by lighteningnumber in population=Using <strong>Probability</strong> to Identify Unusual Events. In exercises 9-12 consider an event to be “unusual” if itsprobability is less than or equal to 0.05. (This is equivalent to the same criterion commonly used in inferentialstatistics, but the value of 0.05 is not absolutely rigid, and other values such as 0.01 are sometimes used instead.)9. <strong>Probability</strong> of a Wrong Eventnumber of wrong events 5a. P ( wrong) == = 0. 0588number of trials or observations 85b. Since P= 0.0588 is greater than 0.05, it is not “unusual” for this test conclusion to bewrong for women who are pregnant.10. <strong>Probability</strong> of Wrong Resultnumber of wrong results 3a. P ( wrong result) == = 0. 214number of trials or observations 14b. Since P= 0.214 is greater than 0.05, it is not “unusual” for this test conclusion to bewrong for women who are not pregnant.11. Cholesterol-Reducing Drugnumber of flu symptom events 19 19a. P ( flu symptoms) == = = 0. 0220number of trials or observations 19 + 844 863b. Since P= 0.0220 is less than 0.05, it is considered “unusual” for patients taking this drug to experience flusymptoms.12. Guessing Birthdaysnumber of ways correct birthday can occur 1a. P (guessing birthday) == = 0. 00274number of different days 365b. Since P= 0.00274 is less than 0.05, it is considered “unusual” for Mike to guess Kelly’s birthday correctly.c. This would be such a good guess or unusual outcome that it would be reasonable to believe that Mike mayhave already known Kelly’s birthday, but it is possible that this was a good guess.d. Guessing another person’s age and being 15 years over the estimate is not flattering to anyone. In this case,Mike is not likely to get another date, but it’s not impossible.1<strong>3.</strong> <strong>Probability</strong> of a Birthdaynumber of ways correct birthday can occur 1a. P ( correct) == = 0. 00274number of different days 365

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