Chapter 16 Random Variables

50 Part IV Randomness and Probability Chapter 16 Random Variables 1. Epected value. µ = EY ( )= 10(0.3) + 0(0.5) + 30(0.) = 19 µ = EY ( )= (0.3) + 4(0.4) + 6(0.) + 8(0.1) = 4.. Epected value. µ = EY ( )= 0(0.) + 1(0.4) + (0.4) = 1. µ = EY ( )= 100(0.1) + 00(0.) + 300(0.5) + 400(0.) = 80 3. Pick a card, any card. Win $0 $5 $10 $30 P(amount won) 6 5 13 5 1 5 1 5 µ = E( amount won) = $ 0 6 + $ 5 13 + $ 10 1 + $ 30 1 $ 413. 5 5 5 5 Answers may vary. In the long run, the epected payoff of this game is $4.13 per play. Any amount less than $4.13 would be a reasonable amount to pay in order to play. Your decision should depend on how long you intend to play. If you are only going to play a few times, you should risk less. 4. You bet! Win $100 $50 $0 1 1 5 5 P(amount won) 6 6 6 = 5 5 36 6 6 = 5 36 µ = = + + E( amount won) $ 100 1 $ 50 5 $ 0 5 $ 3. 61 6 6 6 Answers may vary. In the long run, the epected payoff of this game is $3.61 per play. Any amount less than $3.61 would be a reasonable amount to pay in order to play. Your decision should depend on how long you intend to play. If you are only going to play a few times, you should risk less. 5. Kids. Kids 1 3 P(Kids) 0.5 0.5 0.5

Chapter 16 Random Variables 51 µ = E (Kids) = 1(0.5) + (0.5) + 3(0.5) = 1.75 kids Boys 0 1 3 P(boys) 0.5 0.5 0.15 0.15 6. Carnival. µ = E (Boys) = 0(0.5) + 1(0.5) + (0.15) + 3(0.15) = 0.875 boys Net winnings $95 $90 $85 $80 -$0 number of darts 1 dart darts 3 darts 4 darts (win) 4 darts (lose) P(Amount won) 1 10 = 01. 9 1 10 10 = 009. 9 1 10 10 = 0. 081 3 9 1 10 10 = 0. 079 4 9 10 = 0. 6561 µ = E (number of darts) = 1(0.1) + (0.09) + 3(0.081) + 4(0.079) + 4(0.6561) 3.44 darts µ = E (winnings) = $95(0.1) + $90(0.09) + $85(0.081) + $80(0.079) $0(0.6561) $17.0 7. Software. Since the contracts are awarded independently, the probability that the company will get both contracts is (0.3)(0.6) = 0.18. Organize the disjoint events in a Venn diagram. Profit larger only smaller only both neither $50,000 $0,000 $70,000 $0 P(profit) 0.1 0.4 0.18 0.8 µ = E (profit) = $50,000(0.1) + $0,000(0.4) + $70,000(0.18) = $7,000 8. Racehorse. Assuming that the two races are independent events, the probability that the horse wins both races is (0.)(0.3) = 0.06. Organize the disjoint events in a Venn diagram. Profit 1 st only nd only both neither $30,000 $30,000 $80,000 $10,000 P(profit) 0.14 0.4 0.06 0.56 µ = E (profit) = $30,000(0.14) + $30,000(0.4) + $80,000(0.06) $10,000(0.56) = $10,600

5 Part IV Randomness and Probability 9. Variation 1. = Var() Y = ( 10 19)(.) 0 3 + ( 0 19)(.) 0 5 + ( 30 19)(.) 0 = 49 = SD( Y) = Var( Y) = 49 = 7 = Var( Y) = ( 4.)(.) 03 + ( 4 4.)(.) 04 + ( 6 4.)(.) 0 + ( 8 4.)(.) 01 = 356. = SD( Y) = Var( Y) = 356. 189. 10. Variation. = Var( Y) = ( 0 1.)(.) 0 + ( 1 1.)(.) 04 + ( 1.)(.) 04 = 056. = SD( Y) = Var( Y) = 056. 075. = Var() Y = ( 100 80)(.) 0 1 + ( 00 80)(.) 0 + ( 300 80)(.) 0 5 + ( 400 80)(.) 0 = 7600 = SD( Y) = Var( Y) = 7600 87. 18 11. Pick another card. Answers may vary slightly (due to rounding of the mean) 6 = Var( Won) = ( 0 4. 13) + ( 5 4. 13) 5 13 5 (. ) ( 30 4. 13). + 10 4 13 1 + 5 1 9 5396 5 = SD( Won) = Var( Won) = 9. 5396 $ 5. 44 1. The die. Answers may vary slightly (due to rounding of the mean) = SD( Won) = Var( Won) 1456. 4043 $ 38. 16 13. Kids. 1 5 5 = = 100 3 61 50 3 61 0 361 1456 4043 6 + Var( Won) (. ) (. ) + (. ). 36 36 = Var( Kids) = ( 1 1. 75)(.) 0 5 + ( 1. 75)(. 0 5) + ( 3 1. 75)(. 0 5) = 0. 6875 = SD( Kids) = Var( Kids ) = 0. 6875 0. 83 kids 14. Darts. = Var( Winnings) = ( 95 17. 0) ( 0. 1) + ( 90 17. 0) ( 0. 09) + ( 85 17. 0) ( 0. 081) + (. )(. ) + (. )(. ). 80 17 0 0 079 0 17 0 0 6561 650 057 = SD( Winnings) = Var( Winnings) 650. 057 $ 51. 48

15. Repairs. µ = E( Number of Repair Calls) = 001 (.) + 103 (.) + 04 (.) + 30 (.) = 17. calls Chapter 16 Random Variables 53 = Var( Calls) = ( 0 17.)(.) 01 + ( 1 17.)(.) 03 + ( 17.)(.) 04 + ( 3 17.)(.) 0 = 081. = SD( Calls) = Var( Calls ) = 081. = 09. calls 16. Red lights. µ = E( Red lights) = 0( 0. 05) + 1( 0. 5) + ( 0. 35) + 3( 0. 15) + 4( 0. 15) + 5( 0. 05) =. 5 red lights = Var( Red lights) = ( 0. 5)(. 0 05) + ( 1. 5)(. 0 5) + (. 5)(. 0 35) + (. )(. ) + (. )(. ) + (. )(. ) =. 3 5 0 15 4 5 0 15 5 5 0 05 1 5875 = SD( Red lights) = Var( Red lights) = 1. 5875 16. red lights 17. Defects. The percentage of cars with no defects is 61%. µ = E( Defects) = 0( 0. 61) + 1( 0. 1) + ( 0. 11) + 3( 0. 07) = 0. 64 defects = Var( Defects) = ( 0 064. )(. 061) + ( 1 064. )(. 01) + ( 0. 64) ( 0. 11) + ( 3 0. 64) ( 0. 07) 0. 8704 = SD( Defects) = Var( Defects ) 0. 8704 0. 93 defects 18. Insurance. µ = E( Profit) = 100( 0. 9975) 9900( 0. 0005) 900( 0. 00) = $ 89 Profit $100 $9900 $900 P(Profit) 0.9975 0.0005 0.00 = Var( Profit) = ( 100 89) ( 0. 9975) + ( 9900 89) ( 0. 0005) + ( 900 89) ( 0. 00) = 67, 879 = SD( Profit) = Var( Profit) 67, 879 $ 60. 54 19. Contest. The two games are not independent. The probability that you win the second depends on whether or not you win the first. P( losing both games) = P(losing the first) P(losing the second first was lost) = (.)(.) 06 07 = 0.4

54 Part IV Randomness and Probability d) P( winning both games) = P(winning the first) P(winning the second first was won) = (.)(.) 04 0 = 0.08 X 0 1 PX ( = ) 0.4 0.50 0.08 e) µ = E( X) = 0( 0. 4) + 1( 0. 50) + ( 0. 08) = 0. 66 games = Var( X) = ( 0 0. 66)(. 0 4) + ( 1 0. 66)(. 0 50) + ( 0. 66)(. 0 08) = 0. 3844 = SD( X) = Var( X) = 0. 3844 = 0. 6 games 0. Contracts. The contracts are not independent. The probability that your company wins the second contract depends on whether or not your company wins the first contract. P( getting both contracts) = P(getting #1) P(getting # got #1) = (.)(.) 08 0 = 0.16 P(getting no contract) = P(not getting #1) P( not getting # didn't get #1) = (.)(.) 0 07 = 0.14 d) X 0 1 PX ( = ) 0.14 0.70 0.16 e) µ = E( X) = 0( 0. 14) + 1( 0. 70) + ( 0. 16) = 1. 0 contracts = Var( X) = ( 0 1. 0)(. 0 14) + ( 1 1. 0)(. 0 70) + ( 1. 0)(. 0 16) = 0. 996 = SD( X) = Var( X) = 0. 996 0. 55 contracts 1. Batteries. Number good 0 1 P(number good) 3 6 10 9 = 3 7 7 3 4 90 10 9 + 10 9 = 7 6 4 90 10 9 = 90

Chapter 16 Random Variables 55 µ = = + + E( number good ) 0 6 1 4 4 = 14. batteries 90 90 90 6 4 4 = = ( 0 14) 1 14 1 4 0 3733 + ( ) 90 + ( ) 90 Var( number good).... 90 = SD( number good) = Var( number good) 0. 3733 0. 61 batteries.. Kittens. Number of males 0 1 P(number of males) 3 6 7 6 = 4 3 3 4 4 7 6 + 7 = 6 4 4 4 3 1 7 6 = 4 µ = = + + E( number of males) 0 6 1 4 1 114. males 4 4 4 Answers may vary slightly (due to rounding of the mean) 6 4 1 = = ( 0 114) 1 1 14 114 0 408 + ( ) 4 + ( ) 4 Var( number of males).... 4 = SD( number of males) = Var( number of males) 0. 408 0. 64 males 3. Random variables. µ = E( 3X) = 3( E( X)) = 3( 10) = 30 = SD( 3X) = 3( SD( X)) = 3( ) = 6 d) µ = E( X + Y) = E( X) + E( Y) = 10 + 0 = 30 = SD( X + Y) = Var( X) + Var( Y) e) = + 5 5. 39 µ = EX ( 1 + X) = EX ( ) + EX ( ) = 10 + 10 = 0 = SD( X + X ) = VarX ( ) + VarX ( ) 1 = + 83. 4. Random variables. µ = E( X 0) = E( X) 0 = 80 0 = 60 = SD( X 0) = SD( X) = 1 µ = EY ( + 6) = EY ( ) + 6 = 0 + 6 = 6 = SD( Y + 6) = SD( Y) = 5 µ = E( X Y) = E( X) E( Y) = 10 0 = 10 = SD( X Y) = Var( X) + Var( Y) = + 5 5. 39 µ = E( 05. Y) = 05. ( E( Y)) = 051. ( ) = 6 = SD( 05. Y) = 05. ( SD( Y)) = 053. ( ) = 15.

56 Part IV Randomness and Probability d) µ = E( X + Y) = E( X) + E( Y) = 80 + 1 = 9 = SD( X + Y) = Var( X) + Var( Y) e) = 1 + 3 1. 37 µ = EY ( 1 + Y) = EY ( ) + EY ( ) = 1 + 1 = 4 = SD( Y + Y ) = VarY ( ) + VarY ( ) 1 = 3 + 3 44. 5. Random variables. µ = E( 08. Y) = 08. ( E( Y)) = 08300. ( ) = 40 = SD(. 08Y) = 08.( SD()) Y = 0816.( ) = 18. d) µ = E( X + Y) = E( X) + ( E( Y)) = 10 + ( 300) = 70 e) = SD( X + Y ) = Var( X) + Var( Y) = 1 + ( 16 ) 34. 18 µ = EY ( 1 + Y) = EY ( ) + EY ( ) = 300 + 300 = 600 = SD( Y + Y ) = Var( Y ) + Var( Y ) 1 = 16 + 16. 63 6. Random variables. µ = E( Y + 0) = ( E( Y)) + 0 = (1) + 0 = 44 = SD( Y + 0) = ( SD( Y)) = ( 3) = 6 d) µ = E( 05. X + Y) = 05. ( E( X)) + E( Y) = 0.5(80) +1 = 3 e) = SD(. 05X + Y) = 05. Var( X) + Var() Y = 0.5 ( 1 ) + 3 4. 4 µ = EX ( 1 + X + X3) = EX ( ) + EX ( ) + EX ( ) = 80 + 80 + 80 = 40 = SD( X + X + X ) = Var( X ) + Var( X ) + Var( X ) 1 3 1 3 = 1 + 1 + 1 0. 78 µ = E( X Y) = E( X) E( Y) = 80 1 = 68 = SD( X Y) = Var( X) + Var( Y) = 1 + 3 1. 37 µ = E( X 100) = ( E( X)) 100 = 140 = SD( X 100) = ( SD( X)) = ( 1) = 4 µ = E( 3X Y) = 3( E( X)) E( Y) = 3( 10) 300 = 60 = SD( 3X Y) = 3 Var( X) + Var( Y) = 3 ( 1 ) + 16 39. 40 µ = E( 3X) = 3( E( X)) = 3( 80) = 40 = SD( 3X) = 3( SD( X)) = 3( 1) = 36 µ = E( X 5Y) = E( X) 5( E( Y)) = 80 5( 1) = 0 = SD( X 5Y ) = Var( X) + 5 Var( Y) = 1 + 5 ( 3 ) 19. 1

Chapter 16 Random Variables 57 7. Eggs. µ = E( Broken eggs in 3 dozen) = 3( E( Broken eggs in 1 dozen)) = 306 (. ) = 18. eggs = SD( Broken eggs in 3 dozen) = 05. + 05. + 05. 087. eggs The cartons of eggs must be independent of each other. 8. Garden. µ = E( bad seeds in 5 packets) = 5( E( bad seeds in 1 packet)) = 5( ) = 10 bad seeds = SD( bad seeds in 5 packets) = 1. + 1. + 1. + 1. + 1. 68. bad seeds The packets of seeds must be independent of each other. If you buy an assortment of seeds, this assumption is probably OK. If you buy all of one type of seed, the store probably has seed packets from the same batch or lot. If some are bad, the others might tend to be bad as well. 9. Repair calls. µ = E( calls in 8 hours) = 8( E( calls in 1 hour) = 8( 1. 7) = 13. 6 calls = SD( calls in 8 hours) = 8( Var( calls in 1 hour)) = 809 (. ) 55. calls This is only valid if the hours are independent of one another. 30. Stop! µ = E( red lights in 5 days) = 5( E( red lights each day)) = 55 (. ) = 115. red lights = SD( red lights in 5 days) = 5( Var( red lights each day) = 5( 16. ) 8. red lights Standard deviation may vary slightly due to rounding of the standard deviation of the number of red lights each day, and may only be calculated if the days are independent of each other. This seems reasonable. 31. Fire! The standard deviation is large because the profits on insurance are highly variable. Although there will be many small gains, there will occasionally be large losses, when the insurance company has to pay a claim. µ = E( two policies) = ( E( one policy)) = ( 150) = $ 300 = SD( two policies) = ( Var( one policy)) = ( 6000 ) $ 8, 485. 8 µ = E( 10, 000 policies) = 10, 000( E( one policy)) = 10, 000( 150) = $ 1, 500, 000 = SD( 10, 000 policies) = 10, 000( Var( one policy)) = 10, 000( 6000 ) = $ 600, 000

58 Part IV Randomness and Probability d) If the company sells 10,000 policies, they are likely to be successful. A profit of $0, is.5 standard deviations below the epected profit. This is unlikely to happen. However, if the company sells fewer policies, then the likelihood of turning a profit decreases. In an etreme case, where only two policies are sold, a profit of $0 is more likely, being only a small fraction of a standard deviation below the mean. e) This analysis depends on each of the policies being independent from each other. This assumption of independence may be violated if there are many fire insurance claims as a result of a forest fire, or other natural disaster. 3. Casino. The standard deviation of the slot machine payouts is large because most players will lose their dollar, but a few large payouts are epected. The payouts are highly variable. µ = E( profit from 5 plays) = 5( E( profit from one play)) = 5( 0. 08) = $ 0. 40 = SD( profit from 5 plays) = 5( Var( profit from one play)) = 5( 10 ) $ 68. 33 µ = E( profit from 1000 plays) = 1000( E( profit from one play)) = 1000( 0. 08) = $ 80 = SD( profit from 1000 plays) = 1000( Var( profit from one play)) = 1000( 10 ) $ 3, 794. 73 d) If the machine is played only 1000 times a day, the chance of being profitable will be slightly more than 50%, since $80 is approimately 0.0 standard deviations above 0. But if the casino has many slot machines, the chances of being profitable will go up. 33. Cereal. E( large bowl small bowl) = E( large bowl ) E(small bowl) = 5. 15. = 1ounce = SD( large bowl small bowl) = Var( large) + Var( small) = 04. + 03. = 05. ounces 0 z = 1 = 05. The small bowl will contain more cereal than the large bowl when the difference between the amounts is less than 0. According to the Normal model, the probability of this occurring is approimately 0.03. d) µ = E( large bowl + small bowl) = E( large bowl ) + E(small bowl) = 5. + 15. = 4ounce = SD( large bowl + small bowl) = Var( large) + Var( small) = 04. + 03. = 05. ounces

e) 45. 4 z = = 1 05. Chapter 16 Random Variables 59 According to the Normal model, the probability that the total weight of cereal in the two bowls is more than 4.5 ounces is approimately 0.159. f) µ = E( bo large small) = E( bo) E( large) E( small ) = 16. 3. 5 1. 5 = 1. 3 ounces = SD( bo large small) = Var( bo) + Var( large) + Var(small) = 0. + 03. + 04. 054. ounces 34. Pets. µ = E( dogs cats) = E( dogs) E( cats) = 100 10 = $ 0 = SD( dogs cats) = Var( dogs) + Var( cats) = 30 + 35 $ 46. 10 35. More cereal. µ = E( bo large small) = E( bo) E( large) E( small ) = 16.. 5 1. 5 = 1. ounces 0 z = ( 0) 46. 10 z = 0. 4338 = SD( bo large small) = Var( bo) + Var( large) + Var(small) The epected cost of the dog is greater than that of the cat when the difference in cost is positive (greater than 0). According to the Normal model, the probability of this occurring is about 0.33. 13 1. z = 051. z = 157. = 01. + 03. + 04. 051ounces. According to the Normal model, the probability that the bo contains more than 13 ounces is about 0.058.

60 Part IV Randomness and Probability 36. More pets. Let X = cost for a dog, and let Y = cost for a cat. Total cost = X + X + Y µ = E( X + X + Y) = E( X) + E( X) + E( Y) = 100 + 100 + 10 = $ 30 = SD( X + X + Y) = Var( X) + Var( X) + Var(Y) = 30 + 30 + 35 = $ 55 Since the models for individual pets are Normal, the model for total costs is Normal with mean $30 and standard deviation $55. 400 30 z = 55 z = 145. According to the Normal model, the probability that the total cost of two dogs and a cat is more than $400 is approimately 0.073. 37. Medley. µ = E( # 1+ # + # 3+ # 4) = E( # 1) + E( # ) + E( # 3) + E(# 4) = 50. 7 + 55. 51 + 49. 43 + 44. 91 = 00. 57 seconds = SD( # 1+ # + # 3+ # 4) = Var( # 1) + Var( # ) + Var(#3) + Var(# 4) = 0. 4 + 0. + 0. 5 + 0. 1 0. 46 seconds 199. 48 00. 57 z = 0. 461 z = 36. The team is not likely to swim faster than their best time. According to the Normal model, they are only epected to swim that fast or faster about 0.9% of the time.

Chapter 16 Random Variables 61 38. Bikes. µ = E( unpack + assembly + tuning) = E( unpack) + E( assembly) + E( tuning) = 3. 5 + 1. 8 + 1. 3 = 37. 6 minutes = SD( unpack + assembly + tuning) = Var( unpack) + Var( assembly) + Var(tuning) 30 37. 6 z = 368. z = 07. = 07. + 4. + 7. 37. minutes The bike is not likely to be ready on time. According to the Normal model, the probability that an assembly is completed in under 30 minutes is about 0.019. 39. Farmer s market. Let A = price of a pound of apples, and let P = price of a pound of potatoes. Profit = 100A+ 50P µ = E( 100A+ 50P ) = 100( E( A)) + 50( E( P)) = 100( 0. 5) + 50( 0. 3) = $ 63 = SD( 100A + 50P ) = 100 ( Var( A)) + 50 ( Var( P)) = 100 ( 0. ) + 50 ( 0. 1 ) $ 0. 6 d) No assumptions are necessary to compute the mean. To compute the standard deviation, independent market prices must be assumed. 40. Bike sale. Let B = number of basic bikes sold, and let D =number of delue bikes sold. Net Profit = 10B+ 150D 00 µ = E( 10B+ 150D 00) = 10( E( B)) + 150( E( D)) 00 = 10( 5. 4) + 150( 3. ) 00 = $ 98 = SD( 10B + 150D 00) = 10 ( Var( B)) + 150 ( Var( D)) = 10 (. 1 ) + 150 (. 0 8 ) $ 187. 45 d) No assumptions are necessary to compute the mean. To compute the standard deviation, independent sales must be assumed.