Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
251y0322 10/27/03 ECO251 QBA1 SECOND HOUR EXAM March 21, 2003 Name: ____KEY_____________ Social Security Number: _____________________ Part I: (48 points) Do all the following: All questions are 2 points each except as marked. Exam is normed on 50 points including take-home. Please re-read, ‘Things that You should never do on an Exam or Anywhere Else, ‘ and especially recall that a probability cannot be above 1! The following joint probability table shows the relation between two sets of events. Let event A be that the individual is below 22 ( A is over 21), and the event B 0 be that the individual had no traffic violations in the last 18 months, the event B1 be that the individual has one traffic violation in the last 18 months and the event B 2 be that the individual has 2 traffic violations over the last 18 months. No individuals in this pool of drivers has more than 2 violations. B0 B1 B2 A .22 .12 .06 A .41 .18 .01 B0 A Note, to do the problems below, at least total the rows and columns - A 1. B1 B2 .22 .12 .06 .40 .41 .18 .01 .60 .63 .30 .07 1.00 The probability that someone who is over 21 has no traffic violations is (To 2 decimal places): a) .63 b) .60. c) .41 P B0 A .41 d) *.68 You have been asked for P B0 A .683 .60 PA e) None of the above. 2. The probability that someone picked at random is over 21 and has no traffic violations is (To 2 decimal places): a) .63 b) .60. c) *.41 Joint probabilities are what the table shows! d) .68 e) None of the above. 3. The probability that someone chosen at random is either under 22 or has 2 violations is: a) .40 b) .06 c) .46 d) .47 e) *.41 P A B2 P A PB2 P A B2 .40 .07 .06 .41 f) None of the above 1 251y0322 10/27/03 B0 4. B1 B2 A .22 .12 .06 .40 A .41 .18 .01 .60 .63 .30 .07 1.00 Which two events are independent? a) A and A Note that ‘mutually exclusive’ and ‘independent are almost opposites. b) A and B 2 c) * A and B1 . P A .60 and P A B1 The definition of independence is P A B1 P A . In this case P A B1 .18 .60 . But a better way to do this is to note PB1 .30 that P A B1 P A PB1 .60.30 .18 . d) A and B 0 e) f) 5. A and B 2 None of these. Which two events are mutually exclusive? a) * A and A Complements are always mutually exclusive. None of the other pairs have a joint probability of zero. b) A and B 2 A and B1 . d) A and B 0 e) A and B 2 f) None of these. c) In questions 6 and 7 you need to know what PB0 , PB1 and PB2 are to do the problems. Show your work. Event Px xPx x 2 P x x 0 .63 0 0 Solution: We can use the following table. x 1 .30 0.30 0.30 x 2 .07 0.14 0.28 Total 1.00 0.44 0.58 6. What is the probability that a person picked at random has at least one violation? Solution: Px 1 P1 P2 .30 .07 .37 ‘For the 200th time, ‘at least one’ and ‘exactly one’ are rarely the same thing. 7. What is the mean and the standard deviation of the number of violations our drivers have? (6) 18 Solution: From the work above .3864 0.6216 xPx 0.44 . 2 E x 2 2 .58 .44 2 .3864 This can’t be x and s . These are a sample mean and variance, but there is no sample. 2 251y0322 10/27/03 8. Which of the following statements must be true if P A .6 , PB .4 and P A B 0 (i) A and B are mutually exclusive. Explanation: P A B 0 (ii) A and B are independent. Explanation: P A B P A PB (iii) A and B are collectively exhaustive. Explanation: P A B 1 (iv) P A B 0 a) All are true but (i) b) *All are true but (ii) c) All are true but (iii) d) All are true but (iv) e) All are true f) None are true. Explanation: P A B P A B P B 9. According to a survey of American households, the probability that the residents own 2 cars if annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over $25,000 and 70% had 2 cars. The probability that the residents of a household own 2 cars and have an income less than or equal to $25,000 a year is (3): a) 0.12. b) 0.18. c) *0.22. d) 0.48. Solution: The easiest way I know is to start with the given facts. Let ‘ two ’ be the event that a household owns two cars, and ‘ 25 ’ be the event that a family has an income over $25000. The problem says P two 25 .80 , P25 .60 , Ptwo .70 and asks for P two 25 Using the second two facts we get two two 25 25 __ __ __ .60 .70 __ __ . But 80% of the 60% of families 1.00 that own two cars have incomes over 25000, that is Ptwo 25 P two 25 P25 .80 .60 .48 . So now we have two two 25 25 .48 .22 .12 .60 __ __ .70 1.00 25 25 .48 __ two two and, finally, __ .60 two two __ __ .70 1.00 25 25 .48 .22 .12 .60 If we fill in more, we get .18 .40 .70 .30 1.00 . From the table we read P two 25 .22 3 251y0322 10/27/03 10. According to a survey of American households, the probability that the residents own 2 cars if annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over $25,000 and 70% had 2 cars. The probability that annual household income is over $25,000 if the residents of a household own 2 cars is: (4) a) 0.42. b) 0.48. c) 0.50. P25 two .48 d) 0.69. If we look at the table above P 25 two .6857 . Or we Ptwo .70 Ptwo 25 P25 .80 .60 .6857 could use Bayes’ rule to say P25 two Ptwo .70 11. If you toss a coin 6 times, what is the chance that you get at least one head? (Show your work) (4) Solution: The probability of at least one head is 1 PTTTTTT 1 .56 1 .015625 .984375 . 12. If you have five pennies and four dimes in your pocket and you pick three coins, what is the chance that you get exactly 2 dimes? (Show your work) (4) n! Solution: Remember that C rn The number of ways you can get one penny from 5 is n r ! r! C15 5 . The number of ways you can get two dimes from four is C 24 of ways that you can get 3 coins from 9 is C 39 4! 4 3 6 . The number 2!2! 2 1 9! 9 8 7 84 . our solution, which is an example 6!3! 3 2 1 56 0.3571 . We could also say that your 84 chance of getting a dime on the first and second try is 4 3 5 60 10 , your chance of getting a 9 8 7 504 84 of the Hypergeometric distribution is P2 C15 C 24 C 39 4 5 3 60 10 and your chance of getting a dime on the second 9 8 7 504 84 5 4 3 60 10 10 .3571 and third try is . Thus, your chance of getting exactly $0.20 is 3 9 8 7 504 84 84 dime on the first and third try is 13. Show, by using the appropriate formula, that the number of ways that you can pick 52 items from 55, if order is not important, is the same as the number of ways you can pick 3 items from 55.(4) 55! 55! 55! 55! 55 C 52 and C 355 . These are obviously the same. 55 3!3! 52!3! 55 52 !52! 3!52! 4 251y0322 10/27/03 TABLE 4-3 A survey is taken among customers of a fast-food restaurant to determine preference for hamburger or chicken. Of 200 respondents selected, 75 were children and 125 were adults. 120 preferred hamburger and 80 preferred chicken. 55 of the children preferred hamburger. Let ‘ H ' be hamburger, ‘ C ’ be chicken, ‘ A ’ be Adult and ‘ A ’ be child. The numbers given are as H C H C A __ __ 125 A 65 60 125 follows. . Fill in the table and get or, if you divide by A 55 __ 75 A 55 20 75 120 80 200 120 80 200 H C A .325 .300 .625 200, A .275 .100 .375 .600 .400 1.000 14. Referring to Table 4-3, the probability that a randomly selected individual is a child and prefers 20 .100 chicken is __________. Solution: P A C 200 15. Referring to Table 4-3, the probability that a randomly selected individual is a child or prefers 75 120 55 hamburger is __________. Solution: P A H P A PH P A H 200 200 200 140 .70 or P A H P A PH P A H .375 .600 .275 .700 200 16. Referring to Table 4-3, assume we know that a person has ordered chicken. The probability that 60 P A C 200 60 .75 or this individual is an adult is __________. Solution: P A C 80 PC 80 200 P A C .300 P AC .75 .400 PC 20. Assume that z is a standardized random variable. What is the mean and standard deviation of 6 z 3 ? Solution: Remember that Ez 0 and Varz 1. The outline says E ax b aEx b and Varax b a 2Varx , Substitute -6 for a , z for x and 3 for b and we have E 6 z 3 6Ez 3 60 3 3 and Var 6 z 3 62 Var z 36 1 36 . The standard deviation is thus 36 6. 5 251y0322 10/27/03 ECO251 QBA1 SECOND EXAM March 21, 2003 TAKE HOME SECTION Name: ____KEY ________________ Social Security Number: _________________________ Throughout this exam show your work! Please indicate clearly what sections of the problem you are answering and what formulas you are using. Part II. Do all the Following (9 Points) Show your work! 1. My Social Security Number is 265398248. Take the following set of numbers: 10 23 17 16 32 35 44 45 33 21 and use your Social Security Number to provide the first digit of the first nine numbers, so that, for me, the numbers would become 210, 623, 517, 316, 932, 835, 244, 445, 833, 21. Compute a sample standard deviation for the resulting numbers. (3- 2point penalty for not doing) Solution: Please recall the following. x x , (Computational Formula) s x 2 n So doing it both ways, we compute Row 1 2 3 4 5 6 7 8 9 10 x2 x 210 623 517 316 932 835 244 445 833 21 4976 44100 388129 267289 99856 868624 697225 59536 198025 693889 441 3317114 xx -287.6 125.4 19.4 -181.6 434.4 337.4 -253.6 -52.6 335.4 476.6 0.0 2 nx 2 n 1 x x 2 82714 15725 376 32979 188703 113839 64313 2767 112493 227148 841056 (Definitional formula) s 2 So x 4976 , x 2 3717114 x x 2 n 1 x x 0 (A check on the validity of the mean.), x x 2 841056 (with a slight rounding error) and n 10 . This means x s 2 x 4976 497 .6 . n x 10 2 nx 2 3317114 10 497 .62 9 n 1 841056 .4 93450 .71111 9 s 93450.7111 305.697 251y0322 10/27/03 6 2. At a local university 30% of women, 40% of men and 80% of monsters are business majors. 60% of students are women, 30% of students are men and 10% of students are monsters. (6) a. What percent of students are business majors? b. What percent of business majors are monsters? c. List the following events W - a person is a woman, Ma - a person is a man, Mo - a person is a monster, B a person is a business major. Find P B Mo , P Mo B , PB , PMo and show that your solution in b. satisfies Bayes’ rule. Solution: According to the problem statement P B W .30 , P B Ma .40 and P B Mo .80 . Also PW .60 , PMa .30 and PMo .10 . If 100 students walk into a room, 60 will be women, 30 W Ma Mo Total B men and 10 monsters. We can show this as Of the 60 women, 30% or B Total 60 30 10 100 18 will be business majors. Of the 30 men, 40% or 12 will be business majors. Of the 10 monsters, W Ma Mo Total B 18 12 8 38 80% or 8 will be business majors. We now have B Total 60 30 10 100 W Ma Mo Total B 18 12 8 38 We really ought to finish our table. or we might present the joint B 42 18 2 62 Total 60 30 10 100 W Ma Mo Total B .18 .12 .08 .38 probability table . This serves as a check that our probabilities B .42 .18 .02 .62 Total .60 .30 .10 1.00 add to 1. a) Thus 38 out of 100 or 38% are business majors. PB .38 b) 8 out of 38 business majors are monsters. c) From the work above: PB Mo 8 .2105 or 21.05%. P Mo B .2105 38 8 .80 , P Mo B .2105 , PB .38 and PMo .10 . 10 PB AP A P Mo B PB Since Bayes’ rule says PA B we can either write P B Mo and say P B PMo P B Mo PMo .2105 .38 .80 .10 or we can try P Mo B as .2105 . Both of these are true P B .10 .38 except for rounding error. .80 7