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LESSON OBJECTIVES &KDSWHU 11.1 Fundamental Theorem of Counting • Use the Fundamental Theorem of Counting to find probabilities. 11.2 Compound Events • Find the probability of compound events. 11.3 Collecting Data • Analyze methods of gathering information. • Design an appropriate study to gather information. 11.4 Probability Models • Use probability to model random events. • Use expected value to make decisions. &RQWHQWV 11.5 The Normal Distribution • Use the Empirical Rule to find probabilities. • Find z-scores and use the standard normal distribution. )XQGDPHQWDO7KHRUHP RI&RXQWLQJ &RPSRXQG(YHQWV &ROOHFWLQJ'DWD 3UREDELOLW\'DWD 7KH1RUPDO'LVWULEXWLRQ 0DWK/DEV 6SUHDGRIDQ,QIHFWLRQ 5RFN3DSHU6FLVVRUV 0DWK$SSOLFDWLRQV &KDSWHU5HYLHZ &KDSWHU$VVHVVPHQW &KDSWHU3UREDELOLW\DQG6WDWLVWLFV Mental Math Skills Review Multiply. Simplify each expression. 1. 5 + 11 8 18 18 9 2. 1 + 1 − 1 7 4 2 20 10 3 3. • 8 6 4 25 25 7 1 45 4. 28 4 45 1.4 • 5 • 2 40 2.2 • 3 • 6 36 3.4 • 7 • 2 • 3 168 4.5 • 8 • 3 • 9 1,080 506 Chapter 11 Probability and Statistics 3UREDELOLW\DQG6WDWLVWLFV :K\VKRXOG,OHDUQWKLV" : 3URE 3UREDELOLW\LVDUDWLRWKDWTXDQWL¿HVWKHOLNHOLKRRGRIDSDUWLFXODUHYHQW ,QWHUSUHWLQJSUREDELOLWLHVKHOSVWRSUHSDUHLQGLYLGXDOVIRUGDLO\OLIHHYHQWVVXFKDV ,QWHU WKHZHDWKHULQYHVWPHQWRSSRUWXQLWLHVDQGFRPSHWLWLRQV WKH $QDO\]LQJVWDWLVWLFVDQGSUREDELOLWLHVDQGPDNLQJLQIRUPHGGHFLVLRQVDUHDPRQJ $Q WKHUHVSRQVLELOLWLHVRISURIHVVLRQDOVWKDWKDQGOHGDWD WKH &ULPH6FHQH,QYHVWLJDWRUVSURFHVVGDWDJDWKHUHGWRFRPSXWH SUREDELOLWLHVWKDWFDQOHDGWRVROYLQJFULPHV 0DUNHWHUVDQG5HWDLOHUVXVHVWDWLVWLFVWRGHVLJQVWRUHGLVSOD\V &KHPLVWVXVHSUREDELOLWLHVWRFRPSDUHGLIIHUHQWFRPSRXQGV ,QWKLVFKDSWHU\RXZLOOFRPSXWHWKHSUREDELOLW\RIFRPSRXQGHYHQWVDQG XVHVWDWLVWLFVWRVROYHSUREOHPVLQYROYLQJSUREDELOLW\ 3URMHFW,GHD0DNHD)DVKLRQ6WDWHPHQW 0RVWFORWKLQJLWHPVLQ\RXUFORVHWFDQEHZRUQDVSDUWRIPRUHWKDQ RQHRXWILW)DVKLRQFRQVXOWDQWVUHFRPPHQGWKDWHYHU\RQHKDYHDIHZ FRUHLWHPVWKDWFDQEHZRUQDVSDUWRIPDQ\RXWILWV7KHPRUHEDVLF FRORUV\RXKDYHLQ\RXUZDUGUREHWKHPRUHYHUVDWLOH\RXUZDUGUREH %XLOGDVDPSOHYHUVDWLOHZDUGUREHXVLQJFORWKHV\RXDOUHDG\RZQDQG DIHZQHZRQHV6HOHFWILYHDUWLFOHVRIFORWKLQJIURP\RXUFORVHW<RX FDQMXVWOLVWWKHFORWKHVZLWKGHVFULSWLRQV)RUWKHQHZFORWKHVXVH FDWDORJVDQGVDOHIO\HUVWRLGHQWLI\WKUHHDUWLFOHVRIFORWKLQJWKDWFRXOG EHSXUFKDVHGIRUDWRWDORIRUOHVVWRDGGWR\RXUZDUGUREH8VH FRPELQDWLRQVWRGHWHUPLQHWKHQXPEHURIGLIIHUHQWRXWILWV\RXUHLJKW DUWLFOHVRIFORWKLQJFDQPDNH3ODQ\RXUGDLO\DWWLUHIRUDVPDQ\GD\V DV\RXFDQEHIRUH\RXKDYHWRUHSHDWDQH[DFWRXWILW &KDSWHU3UREDELOLW\DQG6WDWLVWLFV Look to Your Future Actuaries collect and scrutinize data to approximate the probability of an event. Actuaries work for the insurance industry where probabilities of events, such as illness, disability, or loss of property are calculated as a risk. The probability or risk of such an event determines what policies the company can issue and at what cost in order to remain profitable. PLANNING THE CHAPTER Math Labs, pp. 538–541 Data Sheet (Lab Data Sheets) Math Applications, pp. 542–549 Chapter Review, pp. 550–551 Chapter Test, p. 552 Software Generated Assessment Standardized Test Practice, p. 553 Grid Response Form (CRB) Chapter Resource Book (CRB) Reteaching, pp. 393, 397, 403, 407 Extra Practice, pp. 395, 399, 405, 409 Enrichment, pp. 401, 411 Standardized Test Response Form, pp. 413, 414 Standardized Test Answers, p. 415 Classroom/Journal Topics What’s Ahead? In this chapter, students will learn how to calculate the probability of compound events. The Fundamental Theorem of Counting is used in many real-life situations to find probabilities. Students should become familiar with probability models and the normal distribution. Chapter 11 Probability and Statistics 507 LESSON PLANNING /HVVRQ )XQGDPHQWDO7KHRUHP / RI&RXQWLQJ 2EMHFWLYHV 2 Vocabulary Fundamental Theorem of Counting independent events dependent events 8VHWKH)XQGDPHQWDO 7KHRUHPRI&RXQWLQJWR ILQGSUREDELOLWLHV Extra Resources Reteaching 11.1 Extra Practice 11.1 $ODQLVSOD\LQJDFDUGJDPHZLWK DIULHQG+HSLFNVRQHFDUGIURP DGHFNRIFDUGVDQGWKHQSLFNV DQRWKHUFDUGZLWKRXWUHSODFLQJWKH ¿UVW$ODQQHHGVWRSLFNWZRKHDUWV LQDURZWRZLQWKHJDPH:KDWLV WKHSUREDELOLW\WKDW$ODQZLOOZLQ WKHJDPH" )XQGDPHQWDO7KHRUHPRI&RXQWLQJ Assignment )XQGDPHQWDO7KHRUHPRI&RXQWLQJ ,IRQHHYHQWFDQRFFXULQPGLIIHUHQWZD\VDQGDQRWKHU HYHQWFDQRFFXULQQGLIIHUHQWZD\VWKHQWKHUHDUH PQGLIIHUHQWZD\VIRUERWKHYHQWVWRRFFXU In-class practice: 1–5 Homework: 6–26 Math Applications 7KH)XQGDPHQWDO7KHRUHPRI&RXQWLQJFDQEHXVHGWR¿QGWKH QXPEHURISRVVLEOHRXWFRPHVRIGLIIHUHQWHYHQWV Exercises 4, 6, and 9 from pages 542–549 ([DPSOH 8VLQJWKH)XQGDPHQWDO7KHRUHP RI&RXQWLQJ START UP 7KHFDIHWHULDDWDQRI¿FHFRPSOH[VHUYHVWKUHHGLIIHUHQWW\SHVRI VDODGVWKUHHGLIIHUHQWW\SHVRIVRXSVDQGVL[GLIIHUHQWW\SHVRIVDODG GUHVVLQJV+RZPDQ\GLIIHUHQWRUGHUVXVLQJRQHVRXSRQHW\SHRI VDODGDQGRQHW\SHRIVDODGGUHVVLQJDUHSRVVLEOH" Remind students that tree diagrams and organized lists are good strategies for problem solving. Ask students to make either a tree diagram or an organized list to represent all of the possible orders for Example 1. 6ROXWLRQ 8VHWKH)XQGDPHQWDO7KHRUHPRI&RXQWLQJWR¿QGWKHWRWDOQXPEHU RIRUGHUV 7KHUHDUHGLIIHUHQWRUGHUVRIVRXSDQGVDODGZLWKRQHW\SHRI VDODGGUHVVLQJ INSTRUCTION Use the students’ tree diagrams and organized lists to help students develop the Fundamental Counting Principal. After the students have expressed the mechanics of the Fundamental Counting Principal in their own words, present the formal notation. Example 1 Extend the problem by modifying it to include three choices of beverages. Emphasize to students that they can use the answer to the original problem to obtain the answer to the modified problem. &KDSWHU3UREDELOLW\DQG6WDWLVWLFV R.E.A.C.T. Strategy Cooperating Instruct students to write a Fundamental Counting Principal problem. Select a partner for each student. Students should solve their partner’s problem. Then have students write a second problem that is a modification of the problem they just solved (their partner’s problem). After the second problems are written, ask students to exchange papers and solve the problem written by their partner. 508 Chapter 11 Probability and Statistics INSTRUCTION 2QJRLQJ$VVHVVPHQW Ask students to provide examples of other events which would be considered independent. In contrast, ask students to provide examples of events which would not be considered independent. $QXUVHU\VHOOVGLIIHUHQWW\SHV RIÀRZHUVGLIIHUHQWW\SHV RIEXVKHVDQGGLIIHUHQWW\SHV RIWUHHV+RZPDQ\GLIIHUHQW ZD\VFDQDFXVWRPHURUGHU ÀRZHUEXVKDQGWUHH" 'HSHQGHQWDQG,QGHSHQGHQW(YHQWV Tell students that experimental probability is the surveyed chance of an event occurring, while theoretical probability is the mathematical or calculated chance of an event occurring. Point out that there is no way to calculate a theoretical probability concerning free throw attempts. 0DUWLQDUROOHGDQXPEHUFXEHDQGÀLSSHGDFRLQ%HFDXVHWKHRXWFRPHVRI HDFKHYHQWDUHQRWLQÀ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§ 7KHSUREDELOLW\WKDW0ROO\ZLOOVXFFHVVIXOO\PDNHHDFKRIKHUQH[WWKUHHIUHH WKURZVLVDERXWRU 2QJRLQJ$VVHVVPHQW $WDFDUQLYDOGXFNSRQGJDPHWKHUHDUHUXEEHUGXFNVÀRDWLQJRQ WKHZDWHU7ZHQW\¿YHRIWKHGXFNVDUHPDUNHG³:LQQHU´RQWKHERWWRP $SOD\HUUDQGRPO\VHOHFWVDGXFNDQGFKHFNVWRVHHLIKHRUVKHZLQVDSUL]H 7KHQWKHGXFNLVUHWXUQHGWRWKHZDWHU:KDWLVWKHSUREDELOLW\WKDWWKHQH[W WZRSOD\HUVZLOOZLQDSUL]H"([SUHVVWKHDQVZHUDVDIUDFWLRQ 1 36 )XQGDPHQWDO7KHRUHPRI&RXQWLQJ Chalkboard Examples Ask students to name two events A and B, such that: 1. P(A and B) = 1 sample answer: Event A: a card is pulled from a standard deck and the card is either red or black; Event B: a standard six-sided die is rolled and a number less than 7 lands face up 2. P(A and B) = 0 sample answer: Event A: a card is pulled from a standard deck and the card is either red or black; Event B: a standard six-sided die is rolled and a number greater than 6 lands face up 3. P(A and B) = 0.25 sample answer: Event A: a card is pulled from a standard deck and the card is red; Event B: a standard six-sided die is rolled and an even number lands face up 11.1 Fundamental Theorem of Counting 509 INSTRUCTION :KHQWKHRXWFRPHRIRQHHYHQWDIIHFWVWKHRXWFRPHRIDQRWKHUHYHQWWKH HYHQWVDUHGHSHQGHQWHYHQWV)RUH[DPSOHWKHUHDUHFDUGVLQDVWDFN QXPEHUHGIURPWR-XOLRVHOHFWVDFDUGDQGGRHVQRWSXWWKHFDUGEDFNLQ WKHVWDFN7KHQ*LQDVHOHFWVDFDUG7KHRXWFRPHRI*LQD¶VVHOHFWLRQZLOOEH LQÀXHQFHGE\WKHRXWFRPHRI-XOLR¶VVHOHFWLRQ Have students provide examples of other events which would be considered dependent. To reinforce vocabulary comprehension, students should justify why the events are not independent. 3UREDELOLW\RI'HSHQGHQW(YHQWV ,I$DQG%DUHGHSHQGHQWHYHQWVDQG%IROORZV$WKHQ WKHSUREDELOLW\WKDWERWK$DQG%ZLOORFFXULVJLYHQE\ WKHIRUPXOD Example 3 An alternate way 3$DQG% 3$3%_$ to explain this problem stresses the use of the Fundamental Counting Principal. Explain that: 3%_$LVWKHSUREDELOLW\RI%JLYHQWKDW$KDVRFFXUUHG ([DPSOH 3UREDELOLW\RI'HSHQGHQW(YHQWV P(both spades) = (number of ways two spades can be chosen from a deck) divided by (number of ways two cards can be chosen from a deck). • P(both spades) = 13 12 = 1 52 • 51 17 $WWKHEHJLQQLQJRIWKLVOHVVRQWKHUHZDVDGHVFULSWLRQRIDFDUGJDPH :KDWLVWKHSUREDELOLW\WKDW$ODQZLOOZLQWKHJDPH"([SUHVVWKHSUREDELOLW\ DVDIUDFWLRQ 6ROXWLRQ $VWDQGDUGGHFNRISOD\LQJFDUGVKDVKHDUWVVSDGHVGLDPRQGV DQGFOXEV%HFDXVHWKH¿UVWFDUGLVQRWUHSODFHGWKHRXWFRPHRIWKH VHFRQGFDUGLVGHSHQGHQWRQWKHRXWFRPHRIWKH¿UVWFDUG 3KHDUWRQ¿UVWSLFN 13 = 1 52 4 $VVXPLQJWKDWDKHDUWLVVHOHFWHGRQWKH¿UVWSLFNWKHUHZLOOEHKHDUWVOHIW LQWKHUHPDLQLQJFDUGV Reteaching 11.1 (CRB) 3KHDUWRQVHFRQGSLFNDIWHUKHDUWLVSLFNHG 12 51 1 12 3ERWKKHDUWV 4 51 3ERWKKHDUWV 3 = 1 51 17 7KHSUREDELOLW\WKDW$ODQZLOOZLQWKHJDPHLV 1 17 2QJRLQJ$VVHVVPHQW 7KHUHDUHUHGPDUEOHVEOXHPDUEOHVJUHHQPDUEOHVDQG\HOORZ PDUEOHVLQDMDU&DUOWRQVHOHFWVDPDUEOHDWUDQGRPDQGGRHVQRWUHSODFHWKH PDUEOH7KHQKHVHOHFWVDQRWKHUPDUEOH:KDWLVWKHSUREDELOLW\WKDW&DUOWRQ ZLOOVHOHFWUHGPDUEOHV"([SUHVVWKHDQVZHUDVDIUDFWLRQ 3 95 &KDSWHU3UREDELOLW\DQG6WDWLVWLFV Diversity in the Classroom Kinesthetic Learner Divide students into groups to play the card game in Example 3. Each group should play the game 100 times and record the results of each game. Each group should use these results to calculate the experimental probability that Alan will win the game. Finally, pool the results of each group in the class in order to calculate the experimental probability that Alan will win the game. Ask students to explain why the pooled results are closer to the theoretical probability than the results of each individual group. 510 Chapter 11 Probability and Statistics Problem Solving Have students use pencil and paper or Geometer’s Sketchpad to draw a scale picture of the dartboard. VHHPDUJLQ $FLUFXODUGDUWERDUGKDVDGLDPHWHURILQFKHV DQGDFLUFXODUEXOO·VH\HZLWKDGLDPHWHURI 3 4 LQFK,I5HJJLHWKURZVDGDUWDQGLWLVHTXDOO\ OLNHO\WRODQGDQ\ZKHUHRQWKHERDUGZKDWLV WKHSUREDELOLW\RIKLWWLQJWKHEXOO·VH\H" 6WHS&DUU\2XWWKH3ODQ 8VHWKHIRUPXODIRUWKHDUHDRIDFLUFOH$ ʌU &RPSDUHWKHDUHDRIWKHEXOO·VH\HWRWKHDUHD RIWKHFLUFXODUGDUWERDUG([SUHVVWKHSUREDELOLW\ DVDSHUFHQW 6WHS8QGHUVWDQGWKH3UREOHP 'HVFULEHWKHDVSHFWVRIWKHSUREOHPVLWXDWLRQ :KDWDUH\RXEHLQJDVNHGWRILQG" 6WHS&KHFNWKH5HVXOWV 'RHV\RXUDQVZHUVHHPUHDVRQDEOH"$ERXW KRZPDQ\WLPHVODUJHUWKDQWKHEXOO·VH\HLVWKH HQWLUHGDUWERDUG" 6WHS'HYHORSD3ODQ 3UREOHPVROYLQJVWUDWHJ\8VHDIRUPXOD 7KHSUREDELOLW\RIDQHYHQWLVWKHTXRWLHQWRIWKH QXPEHURISRVVLEOHVXFFHVVHVDQGWKHQXPEHU RISRVVLEOHRXWFRPHV+RZFDQ\RXXVHDQDUHD PRGHOWRILQGWKHSUREDELOLW\RIWKHGDUWODQGLQJ RQWKHEXOO·VH\H" Ask students if the answer to the problem would change if the 3 -inch circle were not in the 4 center of the dartboard. Ask students to calculate the probability of not hitting the bull’s-eye of the dartboard. Emphasize that there are two different ways of performing this calculation. Understand the Problem You are given the size of a dartboard and its bull’s-eye. You want to find the probability that a randomly thrown dart will land on the bull’s-eye. /HVVRQ$VVHVVPHQW 7KLQNDQG'LVFXVVVHHPDUJLQ ([SODLQWKH)XQGDPHQWDO7KHRUHPRI&RXQWLQJ :KDWGRHVLWPHDQIRUWZRHYHQWVWREHGHSHQGHQW" :KDWGRHVLWPHDQIRUWZRHYHQWVWREHLQGHSHQGHQW" +RZFDQ\RXILQGWKHSUREDELOLW\WKDWERWKHYHQWV$DQG%ZLOO RFFXULIWKHHYHQWVDUHLQGHSHQGHQW" +RZFDQ\RXILQGWKHSUREDELOLW\WKDWERWKHYHQWV$DQG%ZLOO RFFXULIWKHRXWFRPHRI%GHSHQGVRQWKHRXWFRPHRI$" )XQGDPHQWDO7KHRUHPRI&RXQWLQJ Think and Discuss Answers Develop a Plan Divide the area of the bull’s-eye by the area of the dartboard. Carry Out the Plan 2 • P(bull’s-eye) = π 7.5 2 π • 0.375 = 0.0025 = 0.25% Check the Results Check: the answer seems reasonable. The dartboard is a few hundred times the size of the bull’s-eye. 1. Answers will vary. Sample answer: if one event can occur m ways and another event can occur n ways, then both events can occur in m × n ways. 2. the outcome of one event affects the outcome of the other event 3. the outcomes of the events have no effect on each other 4. multiply the probabilities 5. multiply the probability of A by the probability of B given that A has already occurred 11.1 Fundamental Theorem of Counting 511 WRAP UP 3UDFWLFHDQG3UREOHP6ROYLQJ To ensure mastery of objectives, students should be able to: • Use the Fundamental Counting Principal to find the number of outcomes in each situation. • Use the Fundamental Counting Principal to calculate probabilities. 8VHWKH)XQGDPHQWDO7KHRUHPRI&RXQWLQJWRILQGWKHQXPEHU RIRXWFRPHVLQHDFKVLWXDWLRQ Assignment In-class practice: 1–5 Homework: 6–26 Math Applications Exercises 4, 6, and 9 from pages 542–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xtra Practice 11.1 (CRB) 1 3DQG& 40 1 3HYHQQXPEHUDQG$ 10 3 3RGGQXPEHUDQGYRZHO 51 3QXPEHUOHVVWKDQDQG' 20 9 3QXPEHUJUHDWHUWKDQDQGFRQVRQDQW 40 &KDSWHU3UREDELOLW\DQG6WDWLVWLFV R.E.A.C.T. Strategy Relating Ask students to design a 15-inch dartboard such that the probability of hitting the bull’s-eye would be equal to the probability of not hitting the bull’s-eye. Encourage students to write an equation as part of the solution process rather than using guess and check. 512 Chapter 11 Probability and Statistics Mixed Review Additional Answers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nswers to Math Applications (DFKTXHVWLRQRQDPXOWLSOHFKRLFHWHVWKDVIRXUDQVZHUFKRLFHV :KDWLVWKHSUREDELOLW\RIJHWWLQJWKHILUVWTXHVWLRQVFRUUHFWLI 1 \RXJXHVVWKHDQVZHUV"([SUHVVWKHDQVZHUDVDIUDFWLRQ 64 0L[HG5HYLHZ )LQGDOOWKHUHDOIRXUWKURRWVRIHDFKQXPEHU *UDSKWKHWULJRQRPHWULFIXQFWLRQV\ VLQșDQG\ FRV θ − π 2 RYHUWKHLQWHUYDO²ʌșʌRQDJUDSKLQJFDOFXODWRU:KDWGR \RXQRWLFHDERXWWKHJUDSKV"([SODLQVHHPDUJLQ ( ) )XQGDPHQWDO7KHRUHPRI&RXQWLQJ 9. a.(0.28)3 = 2.2% b.1 – 0.28 = 0.72 c.0.28 • 0.28 • 0.72 = 5.6% d.0.72 • 0.28 • 0.72 = 14.5% Sample answer: the graphs are the same because the sine and cosine functions are π 2 radians out of phase. Math Applications for this chapter are on pages 542–549. The notes and solutions shown below accompany the suggested applications to assign with this lesson. 4. a. 725 = 29 3, 000 120 b. 725 − 1 = 724 3, 000 − 1 2, 999 c. 29 • 724 ≈ 0.058 120 2, 999 ≈ 5.8% 6. a.Because each click does not depend on any other event, they are independent events. b.Because each click is independent, the probability remains the same: 0.14. c.Because each click is independent, the probability remains the same: 0.14. d.0.14 • 0.14 • 0.14 • 0.14 ≈ 0.0004 ≈ 0.04% 11.1 Fundamental Theorem of Counting 513 LESSON PLANNING /HVVRQ &RPSRXQG(YHQWV / Vocabulary 2EMHFWLYHV 2 )LQGWKHSUREDELOLW\RI FRPSRXQGHYHQWV compound event mutually exclusive inclusive events conditional probability 6LJPXQGSLFNHGDSDLURIVRFNVRXWRIKLVGUDZHU ZLWKRXWORRNLQJLQWKHGUDZHU6LJPXQGKDVSDLUVRI EODFNGUHVVVRFNVSDLUVRIEURZQGUHVVVRFNVSDLUV RIEODFNVSRUWVRFNVDQGSDLUVRIZKLWHVSRUWVRFNV :KDWLVWKHSUREDELOLW\WKDW6LJPXQGZLOOVHOHFWDSDLU RIEODFNVRFNVRUDSDLURIVSRUWVRFNV" 0XWXDOO\([FOXVLYH(YHQWV Extra Resources $FRPSRXQGHYHQWLQYROYHVWZRRUPRUHHYHQWVVXFKDVWRVVLQJWKUHH QXPEHUFXEHVRUFKRRVLQJWZRSHRSOHWRVHUYHRQDFRPPLWWHH:KHQ WZRHYHQWVFDQQRWRFFXUDWWKHVDPHWLPHWKHHYHQWVDUHPXWXDOO\ H[FOXVLYH Reteaching 11.2 Extra Practice 11.2 Enrichment 11.2 3UREDELOLW\RI0XWXDOO\([FOXVLYH(YHQWV Assignment ,I$DQG%DUHPXWXDOO\H[FOXVLYHHYHQWVWKHQWKH SUREDELOLW\WKDW$RU%ZLOORFFXULVJLYHQE\WKHIRUPXOD In-class practice: 1–5 Homework: 6–37 3$RU% 3$3% ,I$DQG%DUHPXWXDOO\H[FOXVLYHHYHQWVWKHQ3$DQG% VLQFH $DQG%FDQQRWRFFXUDWWKHVDPHWLPH Math Applications ([DPSOH )LQGLQJ3UREDELOLW\RI 0XWXDOO\([FOXVLYH(YHQWV Exercises 1, 3, 7, 10, and 12 from pages 542–549 $IUXLWEDVNHWFRQWDLQVSHDUVUHGDSSOHVJUHHQDSSOHVDQG RUDQJHV,I.DUHQUHDFKHVLQWRWKHEDVNHWDQGVHOHFWVDSLHFHRIIUXLW DWUDQGRPZKDWLVWKHSUREDELOLW\WKDWLWLVHLWKHUUHGRURUDQJH" START UP 6ROXWLRQ Demonstrate the concepts of mutual exclusion and inclusion in the following way. First, have all female students sit on the left side of the room and all male students sit on the right side of the room. Second, have all students sit on the left side of the room and all male students sit on the right side of the room. 2QO\WKHUHGDSSOHVDUHUHGDQGRQO\WKHRUDQJHVDUHRUDQJH6RWKH HYHQWVRISLFNLQJDUHGSLHFHRIIUXLWRUDQRUDQJHSLHFHRIIUXLWDUH PXWXDOO\H[FOXVLYH$SLHFHRIIUXLWFDQQRWEHUHGDQGRUDQJHDWWKH VDPHWLPH 3UHGRURUDQJH 3UHG3RUDQJH 3UHG 2 12 3RUDQJH 3 12 3UHGRURUDQJH 2 + 3 5 12 12 12 7KHSUREDELOLW\WKDWWKHSLHFHRIIUXLWLVHLWKHUUHGRURUDQJHLV 5 12 &KDSWHU3UREDELOLW\DQG6WDWLVWLFV INSTRUCTION Example 1 Ask students to calculate the probability that the fruit picked is not red or orange once they know the answer to the original problem. Emphasize that P(A and B) = 0 when A and B are mutually exclusive events by asking students to calculate the probability P(apple and orange). Diversity in the Classroom English Language Learner Some students will struggle with new vocabulary in this lesson. Students probably computed problems such as Example 1 in previous courses without using the formula. Remind them that Example 1 could have been calculated in the following way. P(orange or red) = number of winners number of possible P(oraange or red) = 5 12 514 Chapter 11 Probability and Statistics INSTRUCTION 2QJRLQJ$VVHVVPHQW 7KHUHDUHQRYHOVELRJUDSKLHVSRHWU\ERRNVDQG VFLHQFHERRNVRQDVKHOI,I3DXODVHOHFWVDERRNDW UDQGRPZKDWLVWKHSUREDELOLW\WKDWLWLVDELRJUDSK\RUD SRHWU\ERRN"([SUHVV\RXUDQVZHUDVDIUDFWLRQ 1 3 ,QFOXVLYH(YHQWV (YHQWVWKDWDUHQRWPXWXDOO\H[FOXVLYHDUHLQFOXVLYHHYHQWV7KHVHDUH HYHQWVWKDWFDQKDSSHQDWWKHVDPHWLPH)RUH[DPSOHZKHQFDOFXODWLQJWKH SUREDELOLW\RIVHOHFWLQJDGLDPRQGRUDIDFHFDUGWKHUHDUHWKUHHPXWXDOO\ LQFOXVLYHHYHQWVSLFNLQJDMDFNRIGLDPRQGVDTXHHQRIGLDPRQGVDQGD NLQJRIGLDPRQGV 3UREDELOLW\RI,QFOXVLYH(YHQWV ,I$DQG%DUHLQFOXVLYHHYHQWVWKHQWKHSUREDELOLW\WKDW $RU%ZLOORFFXULVJLYHQE\WKHIRUPXOD 3$RU% 3$3%²3$DQG% Example 2 Show students a Venn diagram which represents the socks in Sigmund’s drawer. This diagram will provide students of a visual representation of why P(A and B) must be subtracted. Ask students to explain why the selection of the socks from the drawer might not be a completely random event. Tell students that another term used to describe inclusive events is “non-mutually exclusive” events. ([DPSOH )LQGLQJ3UREDELOLW\RI,QFOXVLYH(YHQWV $WWKHEHJLQQLQJRIWKLVOHVVRQDVLWXDWLRQZDVSUHVHQWHGLQZKLFK6LJPXQG ZDVVHOHFWLQJVRFNVIURPDGUDZHUZLWKRXWORRNLQJ:KDWLVWKHSUREDELOLW\ WKDW6LJPXQGZLOOVHOHFWDSDLURIEODFNVRFNVRUDSDLURIVSRUWVRFNV" 6ROXWLRQ 7KHHYHQWVDUHLQFOXVLYHHYHQWVEHFDXVHDSDLURIVRFNVFDQEHERWKDVSRUW VRFNDQGEODFN 3EODFNRUVSRUW 3EODFN3VSRUW±3EODFNDQGVSRUW 6 3EODFN 15 8 3VSRUW 15 3 3EODFNDQGVSRUW 15 11 3EODFNRUVSRUW 6 + 8 − 3 15 15 15 15 7KHSUREDELOLW\WKDW6LJPXQGVHOHFWVDSDLURIEODFNVRFNVRUDSDLURIVSRUW VRFNVLV 11 15 &RPSRXQG(YHQWV Enriching the Lesson Ask students to justify why the formula for calculating the P(A or B) when A and B are inclusive events can also be used when calculating the P(A or B) when A and B are mutually exclusive events. Tell students to use examples in their justification. 11.2 Compound Events 515 INSTRUCTION 2QJRLQJ$VVHVVPHQW Explain to students that conditional probabilities are those in which the sample space for an event is restricted. Remind students that a sample space is the set of all possible outcomes of a probability event. 7KHUHDUHJLUOVDQGER\VLQ0LVV5HDGLQJ¶VKRPHURRP)LYHRIWKHJLUOV SOD\VSRUWVDQGGRQRWSOD\VSRUWV(LJKWRIWKHER\VSOD\VSRUWVDQGGR QRWSOD\VSRUWV,IDVWXGHQWLVVHOHFWHGDWUDQGRPZKDWLVWKHSUREDELOLW\WKDW WKHVWXGHQWLVDER\RUSOD\VVSRUWV"([SUHVV\RXUDQVZHUDVDIUDFWLRQ 17 20 &RQGLWLRQDO(YHQWV $FRQGLWLRQDOSUREDELOLW\LVWKHSUREDELOLW\WKDWRQHHYHQWZLOORFFXUJLYHQ WKDWDQRWKHUHYHQWKDVDOUHDG\RFFXUUHG7KLVFRQFHSWZDVLQWURGXFHGZLWK GHSHQGHQWHYHQWV Tell students to think of conditional probability in terms of deciding what kind of car you buy. The model of car is going to depend on the make of car that is available at the car dealership where you are shopping. 3UREDELOLW\RI&RQGLWLRQDO(YHQWV 7KHFRQGLWLRQDOSUREDELOLW\RIHYHQW$JLYHQHYHQW%KDV DOUHDG\RFFXUUHGLVJLYHQE\WKHIRUPXOD P( A and B) 3$_% ZKHUH3% P ( B) ([DPSOH )LQGLQJ3UREDELOLW\RI&RQGLWLRQDO(YHQWV 7KHWDEOHEHORZVKRZVWKHUHVXOWVRIDFODVVVXUYH\)LQGWKHFRQGLWLRQDO SUREDELOLW\WKDWDVWXGHQWGLGPRUHWKDQKRXUVRIKRPHZRUNODVWQLJKW JLYHQWKDWWKHVWXGHQWLVDIHPDOH 'LG\RXGRPRUHWKDQKRXUV RIKRPHZRUNODVWQLJKW" 0DOH )HPDOH <HV 1R 6ROXWLRQ 7KHUHDUHPDOHVDQGIHPDOHVIRUDWRWDORIVWXGHQWV$WRWDORI VWXGHQWVGLGPRUHWKDQKRXUVRIKRPHZRUNZKLOHVWXGHQWVGLGQRW 3 (more than 2 hours and female) 3 (female) 10 10 32 3PRUHWKDQKRXUV_IHPDOH 18 = 18 5 9 32 7KHSUREDELOLW\WKDWDVWXGHQWGLGPRUHWKDQKRXUVRIKRPHZRUNODVWQLJKW JLYHQWKDWWKHVWXGHQWLVDIHPDOHLV 5 9 3PRUHWKDQKRXUV_IHPDOH &KDSWHU3UREDELOLW\DQG6WDWLVWLFV Diversity in the Classroom Visual Learner Help students to understand conditional probabilities better by constructing tree diagrams. Each first branch in the tree diagram will represent a simple probability. Each second branch in the tree diagram will represent a conditional probability. 516 Chapter 11 Probability and Statistics $FWLYLW\ ACTIVE LEARNING 5HFHVVLYH7UDLWV ,QWKHVWXG\RIKXPDQJHQHWLFVDGRPLQDQWJHQHLVSDUWRIDSDLU RIWUDLWVVXFKDVH\HFRORUWKDWLVSDVVHGRQWRRIIVSULQJ7KHRWKHU SDUWRIWKHSDLURIWUDLWVLVWKHUHFHVVLYHJHQH$GRPLQDQWJHQHZLOO ´RYHUSRZHUµ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iscuss with students other inherited traits and whether these traits are dominant or recessive. Use the Internet for research purposes. Calculate the experimental probability of a student in your class being left handed. Compare this experimental probability to the theoretical probability calculated in the activity. /HVVRQ$VVHVVPHQW 7KLQNDQG'LVFXVVVHHPDUJLQ :KDWLVDFRPSRXQGHYHQW"*LYHDQH[DPSOH :KDWGRHVLWPHDQIRUWZRHYHQWVWREHPXWXDOO\H[FOXVLYH" :KDWGRHVLWPHDQIRUWZRHYHQWVWREHLQFOXVLYH" +RZFDQ\RXGHWHUPLQHWKHSUREDELOLW\WKDWHLWKHURQHRIWZR LQFOXVLYHHYHQWV$DQG%ZLOORFFXU" 'HVFULEHDFRQGLWLRQDOSUREDELOLW\LQ\RXURZQZRUGV &RPSRXQG(YHQWV Think and Discuss Answers 1. A compound event involves two or more events. Answers will vary. Sample answer: rolling a number cube and flipping a coin. 2. The events cannot occur at the same time. 3. The events can occur at the same time. 4. P(A or B) = P(A) + P(B) – P(A and B) 5. Answers will vary. Sample answer: the probability that one event will occur given that another event has already occurred. 11.2 Compound Events 517 WRAP UP To ensure mastery of objectives, students should be able to: • Determine the probability of compound events which are mutually exclusive. • Determine the probability of compound events which are inclusive. • Determine the probability of compound events which are conditional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ssignment In-class practice: 1–5 Homework: 6–37 3RGGRU 21 3RGGRUJUHDWHUWKDQ 3HYHQRUSULPH 65 3RGGRUSULPH 23 3RUHYHQ 23 3OHVVWKDQRUJUHDWHUWKDQ 5 Math Applications 8VHWKHWDEOHEHORZWRILQGHDFKFRQGLWLRQDOSUREDELOLW\ ([SUHVVWKHDQVZHUVDVGHFLPDOVURXQGHGWRWKHQHDUHVW WKRXVDQGWKLIQHFHVVDU\ Exercises 1, 3, 7, 10, and 12 from pages 542–549 Reteaching 11.2 (CRB) NAME CLASS DATE RETEACHING 14.2 COMPOU A compound even ND EVENTS choosing two peo t involves two or more even ts such as tossing ple to serve on time, the events a three number cub are mutually excl committee. When two events cannot occur at es or usive. If A and B are the same mut given by the form ually exclusive events, then the events, then P(A ula: P(A or B) = P(A) + P(B). probability that A or B will occu If A and B are and B) = 0 sinc r is mut e A uall and B cannot occu Events that are y exclusive not r at the same time happen at the sam mutually exclusive are inclusiv . e time. e events. These are events that If A and B are can incl the formula: P(A usive events, then the probabil ity that A or B or B) = P(A) + will occur is give P(B) – P(A and A conditional B). n by pro event has already bability is the probability that one event will occurred. occur given that The conditional another probability of even t A, given even t B has already the formula: P(A occurred, is give |B) = P( A and B) n by , where P(B) ≠ P( B) 0. EXAMPLE 1 Suppose Jamal rolls a standard an even number number cube. Find or a 6. the probability SOLUTION of rolling The events are inclusive since rolling a 6 is an even number. P(rolling an even 3 number) = = 1 ; P(rolling a 6) = 1 6 2 6 P(rolling an even number and a 6) 1 = 6 P(rolling an even number or a 6) 1 1 1 1 = + – 2 6 6 = EXERCISES 2 Two number cub es mutually exclusiv are tossed. State whether or not the two even 1. the sum rolle e. ts are d is odd; the num bers are both odd mutually excl usive is less than 7; both numbers are less than 4 not mutually excl is even; the num usive bers are both even not mutually excl 4. the sum rolle usive d is 12; at least one number is odd mutually excl usive 398 2. the sum rolle d 3. the sum rolle d >Algebra 2 Chap ter Resource Copyright © CORD Book 518 Chapter 11 Probability and Statistics 6 %OXH(\HV *UHHQ(\HV %URZQ(\HV 0DOH )HPDOH 3JUHHQH\HV_IHPDOH 3EOXHH\HV_PDOH 3EURZQH\HV_PDOH 3JUHHQRUEURZQH\HV_IHPDOH 3EOXHRUJUHHQH\HV_PDOH 3IHPDOH_EURZQH\HV &KDSWHU3UREDELOLW\DQG6WDWLVWLFV Enrichment 11.2 (CRB) Extra Practice 11.2 (CRB) 7KHUHDUH'9'VYLGHRJDPHV&'VDQGYLGHRWDSHVRQ -DLPH·VEHGURRPVKHOI,I-DLPHVHOHFWVDQLWHPDWUDQGRPIURP WKHVKHOIZKDWLVWKHSUREDELOLW\WKDWLWLVD'9'RUYLGHRWDSH" ([SUHVVWKHDQVZHUDVDIUDFWLRQ 5 12 .DUHQ·VERRNEDJFRQWDLQVQRYHOVELRJUDSK\DQGVFLHQFH ERRN0DQQ\·VERRNEDJFRQWDLQVPDWKERRNVFLHQFHERRNV DQGSRHWU\ERRN(DFKVWXGHQWVHOHFWVDERRNDWUDQGRPIURP KLVRUKHUEDJ:KDWLVWKHSUREDELOLW\WKDWHLWKHU.DUHQ·VERRNLV DQRYHORU0DQQ\·VERRNLVDPDWKERRN"([SUHVVWKHDQVZHUDV DGHFLPDO $VXUYH\RIGRZQWRZQZRUNHUVUHYHDOHGWKDWIHPDOHVULGH WKHWUDLQIHPDOHVULGHWKHEXVDQGIHPDOHVFDUSRRO7KH VXUYH\DOVRIRXQGWKDWPDOHVULGHWKHWUDLQPDOHVULGHWKH EXVDQGPDOHVFDUSRRO D &RPSOHWHWKHWDEOHEHORZWRRUJDQL]HWKHVXUYH\UHVXOWV +RZGR\RXJHWWRZRUNHDFKGD\" 0DOH )HPDOH 7UDLQ %XV ND EVENTS 14.2 COMPOU ENRICHMENT DATE CLASS NAME EXTRA PRACTICE ND EVENTS 14.2 COMPOU (A or B). events. Find P P(A) = 35%, P(B) = 42% ually exclusive 2. A and B are mut ) = 0.15 0.7 , P(B 1. P(A) = 0.32 1 (B) = 1 ,P 6 3. P(A) = 2 7 in a time to take a turn ner is spun one Suppose the spin whether events A and B are e board game. Stat e. Then find P(A or B). mutually exclusiv ter than 5; B: an grea 5. A: a number 6 sible by 4; B: a divi 8. A: a number 1 8 3 mutually excl 0.35 0.685 42 5 8 usive; usive; Democrat Supports Issue 5 Does not 5 Support Issue No opinion nal find each conditio as Use the table to ress the answers probability. Exp to the nearest decimals rounded ssary. thousandth, if nece support) 3 not mutually excl by number divisible t) Book> Reso terport issue | Democra s urce Algebra P(sup 9.2 Chap 2 4 5 mutually excl ber B: an even num t | does not 11. P(Democra 0.33 not mutually excl even number ber 8; B: an odd num 6. A: a 4 or an number; 7. A: a prime 77% , P(B) = 0.22 4. P(A) = 0.11 1 2 Copyright © CORD &DU3RRO DATE CLASS NAME 3 4 usive; usive; 1 2 Republican 82 29 63 9 15 s issue 10. P(support 7 8 403 0.683 | Republican) ort or P(does not supp 12. 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The notes and solutions shown below accompany the suggested applications to assign with this lesson. 7KHUHDUHJLUOVDQGER\VLQ0U'DGH·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a.The options are inclusive because he could have a red, long-sleeved shirt. b.There are 4 red longsleeved shirts and 23 possible shirts, so the probability is 4 . 23 c.There are 11 long-sleeved shirts and 5 other red shirts, so the probability is 16 . 23 d.There are 23 shirts, 16 pants, and 25 shoes, so there are 23 • 16 • 25 = 9,200 different ways Marcellus can dress his avatar. 0L[HG5HYLHZ 6ROYHHDFKHTXDWLRQLQWKHJLYHQLQWHUYDO5RXQG\RXUDQVZHUV WRWKHQHDUHVWKXQGUHGWKUDGLDQRUWHQWKGHJUHH 3,514 • 3,513 3. a. 1− 1− 12,578 12 578 , ≈ 0.519 ≈ 51.9% FRVș șʌ VLQșVLQșFRVș șʌ FRVș²FRVș² ș ²FRVș șʌ $ELF\FOHPDQXIDFWXUHUPDNHVURDGELNHVDQGPRXQWDLQELNHV (DFKW\SHRIELNHFDQEHPDGHZLWKRQHRIIRXUGLIIHUHQW VSHHGV(DFKELNHFDQEHPDGHZLWKDKDUGVHDWRUDFXVKLRQHG VHDW+DQGOHEDUVIRUWKHELNHVPD\EHDQJOHGIRUWRXULQJRUIRU UDFLQJ+RZPDQ\GLIIHUHQWFRPELQDWLRQVRIELF\FOHVGRHVWKH PDQXIDFWXUHUPDNH" b. 3, 278 + 3,113 • 3, 278 + 3,112 12,578 12,578 ≈ 0.258 ≈ 25.8% c. 274 ≈ 0.010 ≈ 1.0% 2, 674 7. a.The events are inclusive, so P(type B or positive) = P(type B) + P(positive) – P(type B and positive) = 22 + 23 − 20 = 25 38 38 38 38 b.P(type A or negative) = P(type A) + P(negative) – P(type A and negative) = 16 + 15 − 13 = 18 = 9 38 38 38 38 19 &KDSWHU3UREDELOLW\DQG6WDWLVWLFV 10. a.P(no seatbelt or no license) = P(no seatbelt) + P(no license) – P(no seatbelt and no license) = 0.005 + 0.056 – 0.003 = 0.058 b.P(no seatbelt and no license) = 0.003 • 2,564 ≈ 8 520 Chapter 11 Probability and Statistics 12. a.44 + 25 + 27 + 31 = 127; 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Sample answer: Correlation; While studying more would likely improve grades, there could be other factors involved. For example, those students might have gotten better sleep the night before, have been better test takers, or have been more relaxed. 'HWHUPLQHZKHWKHUHDFKRIWKHIROORZLQJVLWXDWLRQVUHSUHVHQWV FRUUHODWLRQRUFDXVDWLRQ([SODLQ\RXUUHDVRQLQJ 2. Sample answer: Correlation; While exercising would likely help people lose weight, there could be other factors involved. For example, those in the program might have been eating healthier. 1LNNLDVNVWKHVWXGHQWVLQKHUFODVVKRZPXFKWLPHWKH\VSHQW VWXG\LQJIRUILQDOH[DPV6KHFRQFOXGHVWKDWVWXGHQWVZKR VWXGLHGPRUHHDUQHGEHWWHUJUDGHVVHHPDUJLQ ,QDQH[SHULPHQWRISDUWLFLSDQWVLQDQH[HUFLVHSURJUDPORVW ZHLJKWZKLOHRQO\RIVWXG\SDUWLFLSDQWVZKRZHUHQRWLQWKH SURJUDPORVWZHLJKWVHHPDUJLQ 0DOFROPREVHUYHGWKDWZKHQWKHZDWHUWHPSHUDWXUHRIWKHODNH GURSVEHORZ)WKHODNHEHJLQVWRIUHH]HVHHPDUJLQ 7KLQNDQG'LVFXVVVHHPDUJLQ Think and Discuss Answers 2. Answers will vary. Sample answer: Are you fed up with the incompetence of our country’s leadership? (biased); How many times per day do you brush your teeth? (unbiased) /HVVRQ$VVHVVPHQW 3. Sample answer: Causation; There are no other possible alternatives that could cause the lake water to freeze other than the water temperature reaching the freezing point. 1. Answers will vary. Sample answer: See if there is a relationship between the number of years of service of teachers and their students’ scores on standardized tests. &RUUHODWLRQDQG&DXVDWLRQ *LYHDQH[DPSOHRIDUHDOZRUOGVLWXDWLRQLQZKLFK\RXPLJKW ZDQWWRFRQGXFWDQREVHUYDWLRQDOVWXG\ *LYHDQH[DPSOHRIDELDVHGVXUYH\TXHVWLRQDQGDQH[DPSOHRI DQXQELDVHGVXUYH\TXHVWLRQ ([SODLQKRZWRFRQGXFWDQXQELDVHGH[SHULPHQW 'HVFULEHZKDWDUDQGRPVDPSOHPHDQVLQ\RXURZQZRUGV &KDSWHU3UREDELOLW\DQG6WDWLVWLFV 3. Sample answer: Divide participants into 2 groups. Perform a treatment on half of the participants and give a placebo to the other half so that they do not know which group they are in. Compare the results of the two groups to see if the treatment had an effect. 4. Sample answer: A random sample is a sample of members of a population that is representative of the entire population. No part of the population is more represented than any other part. 5. Sample answer: Correlation does not imply causation. Two events can show a relationship without one event having caused the other. 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The company wants to know if using their plant food has an appreciable effect on the growth of flowering plants. (b) An experiment would be most appropriate because the company will treat some of the plants with their food. (c) The company can take a sample of 100 flowering plants and divide them into 2 groups. Treat one half of the plants with the new plant food and treat the other half of the plants with a regular plant food. Give each plant the same amount of water and sunlight so that growing conditions are identical. Measure the growth of the plants in each group after several weeks and compare the results. The experiment should be unbiased since a plant could be assigned to either group and will receive the exact same growing conditions. Practice and Problem Solving Answers 6. Observational study; no treatment applied to either group of parents; no attempt to influence the results. 7. Experiment; teacher is applying a treatment (playing music during test taking) for one group but not the other. 8. Survey; a question is being asked to gather information. 9. Unbiased; does not favor a particular response. 10. Biased; the word thrilling could sway the respondent to favor action movies. 11. Likely biased; employees of a certain make of car might be inclined to say that it is their favorite make. 12. Likely unbiased; each city resident has an equal chance of being mailed the survey. 13. Answers will vary. Sample answer: (a) The population is all students at Karen’s school, and the sample is the students whose opinions Karen will gather. She wants to determine whether the students favor a school dress code. (b) A survey would be most appropriate because Karen wants to gather opinions on an issue. (c) Karen can select a random sample of 50 students and ask them, “Do you support or oppose having a school dress code?” The survey results should be unbiased because the sample is representative of the population and the survey question is unbiased. 14. Answers will vary. 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Answers will vary. Sample answer: Random numbers can be used to represent each outcome of an event with no bias toward any particular outcome. For example, you can use random numbers to decide who has to wash the dishes after dinner. 'HVFULEHKRZ\RXFDQXVHUDQGRPQXPEHUVWRPDNHDIDLU GHFLVLRQ*LYHDQH[DPSOH ([SODLQZKDWLWPHDQVIRUDQHYHQWWREHUDQGRPLQ\RXURZQZRUGV 'HVFULEHWKHDGYDQWDJHVRIFRQGXFWLQJDVLPXODWLRQDQGKRZD VLPXODWLRQFDQEHXVHGWRKHOS\RXPDNHDSUHGLFWLRQ :LOOWKHH[SHFWHGYDOXHRIDQHYHQWEHWKHVDPHDVWKHDFWXDO YDOXH"([SODLQ ([SODLQKRZFDOFXODWLQJH[SHFWHGYDOXHFDQKHOS\RXPDNHD GHFLVLRQ 3UDFWLFHDQG3UREOHP6ROYLQJVHHPDUJLQ 'HVFULEHKRZ\RXFDQXVHUDQGRPQHVVWRPDNHDIDLUGHFLVLRQ LQHDFKVLWXDWLRQ 2. Answers will vary. Sample answer: For an event to be random it means that there is no bias or leaning toward any particular outcome. Each outcome is equally likely to occur. 3. Answers will vary. Sample answer: A simulation can allow you to see what is likely to happen in an experiment without actually conducting the experiment. It may be costly or impractical to conduct the experiment. The results of the simulation can give you insight into what will likely happen. You can use this information to make predictions about the event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nswers will vary. Sample answer: The expected value is not likely to be the same as the actual value of an event, but it is possible that they are the same. If the same experiment is conducted many times, the average results will get closer and closer to the expected value. 5. Answers will vary. Sample answer: Calculating expected value allows you to predict what is likely to happen in an event based on probability. 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Sample answer: No, the company should not invest in the skate park because the expected value of the investment is a loss of $900. 14. Sample answer: The expected value of developing Product A is $115,000, and the expected value of developing Product B is $88,000. The owner should invest in Product A because it has a higher expected value of earnings. Practice and Problem Solving Answers 6. Answers will vary. Sample answer: Roll a number cube. Let 1 or 2 represent Terrell, 3 or 4 represent Carmen, and 5 or 6 represent Alicia. 7. Answers will vary. Sample answer: Assign a number from 1 to 12 to each salesperson and use a random number generator to generate 2 integers from 1 to 12. 8. Answers will vary. Sample answer: Assign a number from 1 to 32 to each player and use a random number generator to generate 3 integers from 1 to 32. 9. Answers will vary. Sample answer: Create a spinner with 4 equal spaces that represent each special offer. Spin the spinner until each special offer has been spun at least once, and stop as soon as all 4 special offers have been spun. Keep track of the results in a frequency table. Conduct several trials of the simulation and find the average number of purchases needed to get all 4 special offers. 10. Answers will vary. Sample answer: Use a random number generator to generate integers from 1 through 100. Let 1 through 35 represent an ace and let 36 through 100 represent a non-ace. Generate numbers until there are 3 aces in a row and count how many “serves” it took. 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UHSUHVHQWWKHDPRXQWRISUHFLSLWDWLRQ&HGDU*URYHUHFHLYHVHDFK\HDU:KDW LVWKHSUREDELOLW\WKDWWKHFLW\ZLOOKDYHEHWZHHQDQGLQFKHVRI SUHFLSLWDWLRQLQDIXWXUH\HDU" 6ROXWLRQ )LQGWKHFRUUHVSRQGLQJ]VFRUHV ]1 = 48.68 − 42.6 37.64 − 42.6 = 1.9 = −1.55 ]2 = 3.2 3.2 8VHDJUDSKLQJFDOFXODWRUWR¿QGWKHDUHDXQGHUWKHVWDQGDUGQRUPDOFXUYH EHWZHHQ] ±DQG] 7KHQRUPDOFGIFRPPDQGFDQEHXVHGWR ¿QGWKLVDUHD PLACEHOLDER 6RWKHSUREDELOLW\WKDWDUDQGRPO\VHOHFWHGIXWXUH\HDULQ&HGDU*URYHZLOO KDYHEHWZHHQDQGLQFKHVRISUHFLSLWDWLRQLVDERXW 7RHVWLPDWHWKHDUHDWRWKHOHIWRUULJKWRID]VFRUH\RXFDQXVH±DQG DVWKHORZHUDQGXSSHUOLPLWUHVSHFWLYHO\7KLVLVEHFDXVHWKHUHLVVROLWWOH DUHDXQGHUWKHVWDQGDUGQRUPDOFXUYHEH\RQGVWDQGDUGGHYLDWLRQVWKDWLW LVQHJOLJLEOH)RUH[DPSOHQRUPDOFGI±±FRXOGEHXVHGWRHVWLPDWH 3=± 7KH1RUPDO'LVWULEXWLRQ 11.5 The Normal Distribution 535 Think and Discuss Answers 1. Sample answer: The area under the normal curve represents the probability that a random variable falls in the range of values on the x-axis under the curve. 2. Sample answer: If a z-score is negative, the x-value is less than the mean. If it is positive, the x-value is greater than the mean. 3. Subtract the mean from the x-value and divide the result by the standard deviation. 4. Sample answer: The z-score tells you how many standard deviations above or below the mean the x-value lies. For example a z-score of –2.2 tells you that the x-value lies 2.2 standard deviations below the mean. 5. Sample answer: You can use standard normal tables, calculators, and spreadsheet programs to find probabilities for the standard normal curve. Practice and Problem Solving Answers 6.81.5% 7.95% 8.49.85% 9.15.85% 10.97.35% 11.–2.5 12.1.125 13.1.8 14.–3.1 536 Chapter 11 Probability and Statistics 2QJRLQJ$VVHVVPHQW :KDWLVWKHSUREDELOLW\WKDWDUDQGRPO\VHOHFWHGIXWXUH\HDUZLOOKDYHOHVV WKDQLQFKHVRISUHFLSLWDWLRQ"DERXW /HVVRQ$VVHVVPHQW 7KLQNDQG'LVFXVVVHHPDUJLQ :KDWLVWKHUHODWLRQVKLSEHWZHHQWKHDUHDXQGHUDQRUPDOFXUYH DQGSUREDELOLW\" :KDWGRHVWKHVLJQRID]VFRUHWHOO\RXDERXWDQ[YDOXHDQGWKH PHDQ" 'HVFULEHKRZWRWUDQVIRUP[YDOXHVIURPDUDQGRPYDULDEOHZLWK PHDQDQGVWDQGDUGGHYLDWLRQıLQWR]VFRUHV :KDWGRHVWKH]VFRUHWHOO\RXDERXWUHODWLRQVKLSEHWZHHQDQ [YDOXHDQGWKHPHDQRIDQRUPDOO\GLVWULEXWHGGDWDVHW"*LYHDQ H[DPSOH +RZGRHVWUDQVIRUPLQJ[YDOXHVLQWR]VFRUHVKHOS\RXFDOFXODWH SUREDELOLWLHV" 3UDFWLFHDQG3UREOHP6ROYLQJVHHPDUJLQ 8VHWKH(PSLULFDO5XOHWRHVWLPDWHHDFKSUREDELOLW\ SRXQGVı SRXQGV3OE;OE ı 3; PHWHUVı PHWHUV3P;P VDOHVı VDOHV3VDOHV;VDOHV 6XSSRVHWKHKHLJKWVRIWKHIHPDOHVWXGHQWVDW3OHDVDQW5LGJH +LJK6FKRRODUHQRUPDOO\GLVWULEXWHGZLWKDPHDQRILQFKHV DQGDVWDQGDUGGHYLDWLRQRILQFKHV:KDWLVWKHSUREDELOLW\ WKDWDUDQGRPO\VHOHFWHGIHPDOHVWXGHQWIURPWKHVFKRROLV EHWZHHQDQGLQFKHV" &RQYHUWHDFK[YDOXHWRD]VFRUH [ ı [ ı [ ı [ ı &KDSWHU3UREDELOLW\DQG6WDWLVWLFV 8VHWKHVWDQGDUGQRUPDOGLVWULEXWLRQWRVROYHHDFKSUREOHP 7KHSULFHVRIODUJHVFUHHQ79VDWDQHOHFWURQLFVUHWDLOVWRUHDUH QRUPDOO\GLVWULEXWHGZLWKDPHDQRIDQGDVWDQGDUG GHYLDWLRQRI6XSSRVHDODUJHVFUHHQ79LVVHOHFWHGDW UDQGRP:KDWLVWKHSUREDELOLW\WKDWWKHSULFHRIWKH79LV EHWZHHQDQG" 7KHILQDOH[DPVFRUHVLQ0U+HU]RJ·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ractice and Problem Solving Answers 15. about 85.4% 16. about 60.1% 17. about 65.5% 18. Brand B; The z-scores for each box are 1.45 for Brand A and –1.8 for Brand B. The Brand A box is 1.45 standard deviations above the mean and the Brand B box is 1.8 standard deviations below the mean. VWDQGDUGGHYLDWLRQ 6XSSRVHDER[RI%UDQG$FHUHDOZHLJKVRXQFHVDQGD ER[RIEUDQG%FHUHDOZHLJKVRXQFHV:KLFKER[ZHLJKWLV IDUWKHUIURPWKHPHDQLQWHUPVRIVWDQGDUGGHYLDWLRQV"([SODLQ KRZ\RXIRXQG\RXUDQVZHU 7KH1RUPDO'LVWULEXWLRQ 11.5 The Normal Distribution 537 MATH LAB Activity 1 0DWK/DEV PREPARE • Remind students of the meaning of a simulation. Explain that this lab is a simulation for how influenza may spread throughout your school or community. • Discuss with students that in a real situation of an infectious disease, even though contact might be made with an infected person that is not a guarantee of getting the infection. TEACH • This lab is a whole class activity. • In a typical classroom setting, the teacher will be activity administrator and each student will receive one numbered note card. • Keep the “infected” number secret, as well as the identity of the students that receives it. • If the class is too small for this simulation as it is described, give each student two numbered note cards. Instruct the students to mentally assign one of the numbers to the right hand and the other number to the left hand. When two people come in immediate contact, before rolling the dice, have each person in the pair choose right or left for the other person. Each person then identifies the ID number that has been associated with the right or left hand. • The activity administrator should be sure to change the infected ID number and the condition for being contagious with each round. These changes are necessary to get the simulation as random as possible. $FWLYLW\6SUHDGRIDQ,QIHFWLRQ 3UREOHP6WDWHPHQW (TXLSPHQW ,PDJLQHWKHUHLVDQRXWEUHDNRIWKHIOXDW \RXUVFKRRO+RZOLNHO\LVLWWKDW\RXZLOO EHLQIHFWHGE\WKHYLUXV"8VHDVLPXODWLRQ WRPRGHOWKHVSUHDGRIWKHLQIHFWLRQ7KHQ FDOFXODWHWKHH[SHULPHQWDOSUREDELOLW\RI EHLQJLQIHFWHGE\WKHYLUXV 1RWHFDUGV 1XPEHUFXEHV 3URFHGXUH 7KHDFWLYLW\DGPLQLVWUDWRUGLVWULEXWHVDQRWHFDUGWRHDFK VWXGHQWZLWKDQ,'QXPEHUVXFKDV«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• Discuss the conditions used for contagious and list all possible rolls that satisfied the condition. Determine if the condition chosen resulted in a greater probability of being infected. Using all possible rolls, describe a condition where very few people would likely become infected and a condition where a lot of people would likely become infected. Math Lab Solutions and Notes 2. The probability will vary depending on the number of students in the simulation; 1 where n is the number of students in the simulation. n 538 Chapter 11 Probability and Statistics 8VHDWDEOHVLPLODUWRWKHRQHVKRZQEHORZWRUHFRUGHDFK SHUVRQDOHQFRXQWHU6WXGHQWVPRYHDURXQGWKHFODVVURRP DQGPHHWZLWKRWKHUVWXGHQWV)RUHDFKHQFRXQWHUUHFRUG WKHVWXGHQW·V,'QXPEHU7KHQHDFKVWXGHQWLQWKHSDLUUROOVD QXPEHUFXEHDQGUHFRUGVWKHVXP&RQWLQXHPHHWLQJVWXGHQWV XQWLOHDFKVWXGHQWKDVPHWZLWKILYHRWKHUVWXGHQWV%HVXUH QRWWRHQFRXQWHUWKHVDPHVWXGHQWWZLFH 5RXQG ,' 6XP $WWKHFRQFOXVLRQRI5RXQGWKHDFWLYLW\DGPLQLVWUDWRUUHYHDOV WKH,'QXPEHURIWKHVWXGHQWRULJLQDOO\LQIHFWHGZLWKWKHIOX YLUXV7KRVHVWXGHQWVZKRFDPHLQFRQWDFWZLWKWKLVVWXGHQW VKRXOGFKHFNWKHLUVXPVIURPWKHHQFRXQWHUV,IWKHVXPRI \RXUUROOZLWKWKH´LQIHFWHGµVWXGHQWZDVRUWKHQ \RXZHUHJLYHQWKHIOX,I\RXUVXPZDVDQ\RWKHUQXPEHU \RXGLGQRWFRQWUDFWWKHIOX&RXQWWKHQXPEHURILQIHFWHG VWXGHQWVDWWKHFRQFOXVLRQRI5RXQG 'HWHUPLQHWKHWKHRUHWLFDOSUREDELOLW\RIFRQWUDFWLQJWKHIOX RQFH\RXHQFRXQWHUWKH´LQIHFWHGVWXGHQWµ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infected (from Round 1) encounter no previously infected. 3 × 5 = 15 possible newly-infected (7 or 8 expected to be newly-infected) 2 of 3 infected (from Round 1) encounter each other, and third infected encounters no previously infected. 5 + 4 + 4 = 13 potential newly-infected (6 or 7 expected to be newly-infected) 1 of 3 infected (from Round 1) encounters both other infected people and other 2 infected people do not encounter each other. 3 + 4 + 4 = 11 potential newly-infected people (5 or 6 expected to be newly-infected) 3 infected (from Round 1) each encounter other 2 infected. 3 + 3 + 3 = 9 potential newly-infected (4 or 5 expected to be newly-infected) Math Lab Solutions and Notes 6. In Round 1, each person encounters 5, so infected person has possibility of in fecting 5. Using theoretical probability 1 , 5 • 1 = 2.5, 2 2 2 or 3 new people are expected to be infected. Expected number infected after Round 1 is 1 + (2 or 3) = 3 or 4. 1 7. ; 36 different outcomes 2 when two number cubes rolled. Of those, 18 rolls have sum of 7, 8, 9, or 10. Table at bottom of following page shows possible dice combinations and probabilities used. You can alter the lab by using different probabilities, such as 1 or 1 . An example of 3 4 each is given; use the table to find other conditions that will result in desired probability. 1 = 12 condition with a 3 36 total of 12: 3, 4, 6, or 11 1 = 9 condition with a 4 36 total of 9: 2, 3, 4, or 10 8. During Round 2, if an infected person encounters another infected person, the expected number of newly-infected people will be affected. Described at the right are the 4 possible scenarios based on 3 people infected in Round 1. Note that between 1 and 6 people may be infected going into Round 2, so the example only considers one possible result of Round 1 going into Round 2. Math Labs 539 MATH LAB $FWLYLW\5RFN3DSHU6FLVVRUV Activity 2 PREPARE • Introduce the game Rock, Paper, Scissors. Explain that rock wins over scissors because the rock can crush the scissors. Scissors wins over paper because scissors can cut paper. Paper wins over rock because paper can cover rock. • Invite one student to come to front with you and play a coupled game. This will give students a chance to better understand the game before beginning the lab. 5RFN (TXLSPHQW )ROORZWKHVWHSVEHORZWRFRPSDUHWKH WKHRUHWLFDOSUREDELOLW\RIZLQQLQJDGRXEOH HOLPLQDWLRQWRXUQDPHQWWRWKHH[SHULPHQWDO SUREDELOLWLHV &KDONERDUG 3DSHUDQG SHQFLO 'LYLGHLQWRJURXSVRIIRXUVWXGHQWV'UDZDGRXEOHHOLPLQDWLRQ WRXUQDPHQWEUDFNHWOLNHWKHRQHVKRZQEHORZRQDVKHHWRI SDSHURURQWKHFKDONERDUG5DQGRPO\DVVLJQSRVLWLRQVIRU HDFKSOD\HUIRUWKHILUVWWZRPDWFKHVLQWKH:LQQHU·V%UDFNHW :ULWHWKHLQLWLDOVRIHDFKVWXGHQWRQWKHDSSURSULDWHOLQHRQ WKHEUDFNHW 6FLVVRUV FOLLOW-UP 540 Chapter 11 Probability and Statistics 3UREOHP6WDWHPHQW 3URFHGXUH 3DSHU TEACH • Students should work in groups of 4 students. • Each student will take on the roles of participate and recorder. • Review the difference between experimental probability and theoretical probability. • If the class size does not accommodate groups of four, ask students to consider a difference format for playing the tournament. • Discuss how the activity changes if it is a singleelimination tournament. Talk about advantages and disadvantages to playing double elimination compare to single elimination. • Brainstorm different types of tournaments and classify each as more likely to be a single elimination or double elimination tournament. 5RFN3DSHU6FLVVRUVLVDJDPHSOD\HGEHWZHHQWZRSHRSOH,WLVRIWHQ XVHGDVDGHFLVLRQPDNLQJJDPHVLPLODUWRIOLSSLQJDFRLQ2QWKH FRXQWRIWKUHHHDFKSOD\HU´WKURZVµDURFNSDSHURUVFLVVRUVXVLQJ KLVRUKHUKDQG$URFNLVIRUPHGE\PDNLQJDILVWSDSHUE\DIODW SDOPDQGVFLVVRUVE\KROGLQJRXWWZRILQJHUV,QSOD\LQJWKHJDPH 5RFNEHDWV6FLVVRUV6FLVVRUVEHDWV3DSHUDQG3DSHUEHDWV5RFN,IWKH SOD\HUVERWKWKURZWKHVDPHLWHPWKHJDPHLVUHSHDWHGXQWLOWKHUHLV RQHZLQQHU &KDSWHU3UREDELOLW\DQG6WDWLVWLFV Sum of dice 2 1+1 3 4 1 + 2; 1 + 3; 2 + 1; 3 + 1; 2 + 2; 5 6 7 8 9 10 11 12 1 + 4; 4 + 1; 3 + 2; 2 +3 1 + 5; 5 + 1; 2 + 4; 4 + 2; 3+3 1 + 6; 6 + 1; 2 + 5; 5 + 2; 3 + 4; 4+3 2 + 6; 6 + 2; 3 + 5; 5 +3; 4+4 4 + 5; 5 +4; 3 + 6; 6+3 4 + 6; 6 + 4; 5+5 5 + 6; 6+5 6+6 Number of Rolls 1 2 3 4 5 6 5 4 3 2 1 Probability 1/36 1/18 1/12 1/9 5/36 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The survey would produce unbiased results because the sample is a random sample and there is no bias in the survey question. 3,514 • 3,513 1− 3. a. 1− 12,578 12,578 ≈ 0.519 ≈ 51.9% b. 3, 278 + 3,113 • 3, 278 + 3,112 12,578 12,578 ≈ 0.258 ≈ 25.8% c. 274 ≈ 0.010 ≈ 1.0% 2, 674 Math Applications 543 Math Applications Solutions and Notes 4. a. 725 = 29 3, 000 120 b. 725 − 1 = 724 3, 000 − 1 2, 999 c. 29 • 724 ≈ 0.058 120 2, 999 ≈ 5.8% D :KDWLVWKHSUREDELOLW\WKDWWKHILUVWZLQQLQJVWXGHQWFKRVHQLV DVRSKRPRUH" 29 120 E $VVXPLQJWKDWWKHILUVWZLQQLQJVWXGHQWFKRVHQZDVDVRSKRPRUHZKDWLV WKHSUREDELOLW\WKDWWKHVHFRQGZLQQLQJVWXGHQWLVDOVRDVRSKRPRUH" 724 2, 999 5. a.149 • 148 • 147 275 274 273 ≈ 0.158 ≈ 15.8% b.100% – 15.8% = 84.1% c.0.158 • 0.158 • 0.842 ≈ 0.021 ≈ 2.1% d.0.158 • 0.158 • 0.158 ≈ 0.004 ≈ 0.4% &DQGDFHLVDPHPEHURIKHUXQLYHUVLW\·VEDVNHWEDOOWHDP+HUWHDPDGYDQFHG WRWKH)LQDO)RXU%DVNHWEDOO7RXUQDPHQW7KHUHDUHQRWHQRXJKWLFNHWVDYDLODEOH IRUDOORIWKHVWXGHQWVWRDWWHQGWKH)LQDO)RXUVRWKHXQLYHUVLW\GHFLGHGWRKROG 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IUDXGXOHQW"5RXQGWRWKHQHDUHVWKXQGUHGWKSHUFHQW 0DUWLQDZRUNVIRUDFRPSDQ\WKDWPDNHVDQGVHOOVSHUVRQDOGLJLWDODVVLVWDQWV 3'$V0DUWLQDVDZWKHSURWRW\SHRIWKHFRPSDQ\·VODWHVWPRGHORI3'$ZKLFK LVH[WUHPHO\VORZ0DUWLQDILUVWHVWDEOLVKHVDIRFXVJURXSPDGHXSRIW\SH$DQG W\SH%SHUVRQDOLWLHV$SHUVRQZLWKDW\SH$SHUVRQDOLW\LVDOZD\VLQDUXVKDQG RIWHQPXOWLWDVNVXQGHUWLPHFRQVWUDLQWVZKLOHDSHUVRQZLWKDW\SH%SHUVRQDOLW\ LVPRUHUHOD[HGDQGSDWLHQW0DUWLQD·VIRFXVJURXSFRQVLVWVRIW\SH$ SHUVRQDOLWLHVDQGW\SH%SHUVRQDOLWLHV0DUWLQDKDVWKHPHPEHUVRIWKHIRFXV JURXSWU\RXWWKHSURWRW\SHRIWKHODWHVWPRGHORI3'$7KUHHRIWKHW\SH$ SHUVRQDOLWLHVJLYHLWDSRVLWLYHUHYLHZZKLOHRIWKHW\SH%SHUVRQDOLWLHVJLYHLW DSRVLWLYHUHYLHZ D :KDWLVWKHSUREDELOLW\WKDWDUDQGRPO\VHOHFWHGPHPEHURI0DUWLQD·VIRFXV JURXSLVDW\SH%SHUVRQDOLW\RUWKHSHUVRQJDYHWKHSURWRW\SHRIWKHODWHVW PRGHORI3'$DSRVLWLYHUHYLHZ" 25 Math Applications Solutions and Notes 6. a.Because each click does not depend on any other event, they are independent events. b.Because each click is independent, the probability remains the same: 0.14. c.Because each click is independent, the probability remains the same: 0.14. d.0.14 • 0.14 • 0.14 • 0.14 ≈ 0.0004 ≈ 0.04% 7. a.The events are inclusive, so P(type B or positive) = P(type B) + P(positive) – P(type B and positive) = 22 + 23 − 20 = 25 38 38 38 38 b.P(type A or negative) = P(type A) + P(negative) – P(type A and negative) = 16 + 15 − 13 = 18 = 9 38 38 38 38 19 38 E :KDWLVWKHSUREDELOLW\WKDWDUDQGRPO\VHOHFWHGPHPEHURI0DUWLQD·VIRFXV JURXSLVDW\SH$SHUVRQDOLW\RUWKHSHUVRQGLGQRWJLYHWKHSURWRW\SHRI WKHODWHVWPRGHORI3'$DSRVLWLYHUHYLHZ" 9 19 0DWK$SSOLFDWLRQV Math Applications 545 Math Applications Solutions and Notes 8. a.Sample answer: No, the sample is representative of the population since it is a random sample. $PDUNHWLQJFRPSDQ\FRQGXFWHGDVXUYH\RIDUDQGRPVDPSOHRIFRQVXPHUV DQGDVNHGWKHPWRJLYHWKHLURSLQLRQVRQDQHZEUDQGRIVKDPSRR7KH FRQVXPHUVZHUHDVNHGWRXVHWKHQHZEUDQGRIVKDPSRRIRUZHHNVDQG WKHQWHOOZKHWKHUWKH\SUHIHUWKHQHZEUDQGSUHIHUWKHLUROGEUDQGRUKDYHQR 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While increasing the sample size will produce more accurate results, it may not be feasible or cost effective to sample a very large portion of the population. 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Sample answer: The aggressive growth plan has a greater expected value, so Molly might consider investing in this plan if she still has several years before retirement. 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