Chapter 17 Probability Models

Transcription

1 Chapter 17 Probability Models 241 Chapter 17 Probability Models 1 Bernoulli a) These are not Bernoulli trials The possible outcomes are 1, 2, 3, 4, 5, and There are more than two possible outcomes b) These may be considered Bernoulli trials There are only two possible outcomes, Type A and not Type A Assuming the donors are representative of the population, the probability of having Type A blood is 43% The trials are not independent, because the population is finite, but the donors represent less than 1% of all possible donors c) These are not Bernoulli trials The probability of getting a heart changes as cards are dealt without replacement d) These are not Bernoulli trials We are sampling without replacement, so the trials are not independent Samples without replacement may be considered Bernoulli trials if the sample size is less than 1% of the population, but 5 is more than 1% of 3 e) These may be considered Bernoulli trials There are only two possible outcomes, sealed properly and not sealed properly The probability that a package is unsealed is constant, at about 1%, as long as the packages checked are a representative sample of all packages Finally, the trials are not independent, since the total number of packages is finite, but the 24 packages checked probably represent less than 1% of the packages 2 Bernoulli 2 a) These may be considered Bernoulli trials There are only two possible outcomes, getting a and not getting a The probability of getting a is constant at 1/ The rolls are independent of one another, since the outcome of one die roll doesn t affect the other rolls b) These are not Bernoulli trials There are more than two possible outcomes for eye color c) These can be considered Bernoulli trials There are only two possible outcomes, properly attached buttons and improperly attached buttons As long as the button problem occurs randomly, the probability of a doll having improperly attached buttons is constant at about 3% The trails are not independent, since the total number of dolls is finite, but 37 dolls is probably less than 1% of all dolls d) These are not Bernoulli trials The trials are not independent, since the probability of picking a council member with a particular political affiliation changes depending on who has already been picked The 1% condition is not met, since the sample of size 4 is more than 1% of the population of 19 people e) These may be considered Bernoulli trials There are only two possible outcomes, cheating and not cheating Assuming that cheating patterns in this school are similar to the patterns in the nation, the probability that a student has cheated is constant, at 74% The trials are not independent, since the population of all students is finite, but 481 is less than 1% of all students

2 242 Part IV Randomness and Probability 3 Simulating the model a) Answers will vary A component is the simulation of the picture in one box of cereal One possible way to model this component is to generate random digits -9 Let and 1 represent Tiger Woods and 2-9 a picture of another sports star Each run will consist of generating random numbers until a or 1 is generated The response variable will be the number of digits generated until the first or 1 b) Answers will vary c) Answers will vary To construct your simulated probability model, start by calculating the simulated probability that you get a picture of Tiger Woods in the first box This is the number of trials in which a or 1 was generated first, divided by the total number of trials Perform similar calculations for the simulated probability that you have to wait until the second box, the third box, etc d) Let X = the number of boxes opened until the first Tiger Woods picture is found X )( 2) 8)( 2) 8)( 2) P(X) 2 = 1 = 8 = 124 e) Answers will vary 4 Simulation II a) Answers will vary A component is the simulation of one die roll One possible way to model this component is to generate random digits 1- Let 1 represent getting 1 (the roll you need and let 2- represent not getting the roll you need Each run will consist of generating random numbers until 1 is generated The response variable will be the number of digits generated until the first 1 b) Answers will vary c) Answers will vary To construct your simulated probability model, start by calculating the simulated probability that you roll a 1 on the first roll This is the number of trials in which a 1 was generated first divided by the total number of trials Perform similar calculations for the simulated probability that you have to wait until the second roll, the third roll, etc d) Let X = the number of rolls until the first 1 is rolled X P(X) e) Answers will vary

3 5 Tiger again Chapter 17 Probability Models 243 a) Answers will vary A component is the simulation of the picture in one box of cereal One possible way to model this component is to generate random digits -9 Let and 1 represent Tiger Woods and 2-9 a picture of another sports star Each run will consist of generating five random numbers The response variable will be the number of s and 1s in the five random numbers b) Answers will vary c) Answers will vary To construct your simulated probability model, start by calculating the simulated probability that you get no pictures of Tiger Woods in the five boxes This is the number of trials in which neither nor 1 were generated divided by the total number of trials Perform similar calculations for the simulated probability that you would get one picture, 2 pictures, etc d) Let X = the number of Tiger Woods pictures in 5 boxes X ( 2 ) ( 8 ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 2 ) ( 8 ) 4 P(X) e) Answers will vary Seatbelts a) Answers will vary A component is the simulation of one driver in a car One possible way to model this component is to generate pairs of random digits -99 Let 1-75 represent a driver wearing his or her seatbelt and let 7-99 and represent a driver not wearing his or her seatbelt Each run will consist of generating five pairs of random digits The response variable will be the number of pairs of digits that are -75 b) Answers will vary c) Answers will vary To construct your simulated probability model, start by calculating the simulated probability that none of the five drivers are wearing seatbelts This is the number of trials in which no pairs of digits were -75, divided by the total number of trials Perform similar calculations for the simulated probability that one driver is wearing his or her seatbelt, two drivers, etc d) Let X = the number of drivers wearing seatbelts in 5 cars X ( 75 ) ( 25 ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 75 ) ( 25 ) 4 P(X) e) Answers will vary

4 244 Part IV Randomness and Probability 7 On time These departures cannot be considered Bernoulli trials Departures from the same airport during a 2-hour period may not be independent They all might be affected by weather and delays 8 Lost luggage 9 Hoops The fate of these bags cannot be considered Bernoulli trials What happens to 22 pieces of luggage, all checked on the same flight probably aren t indpendent The player s shots may be considered Bernoulli trials There are only two possible outcomes (make or miss), the probability of making a shot is constant (8%), and the shots are independent of one another (making, or missing, a shot does not affect the probability of making the next) Let X = the number of shots until the first missed shot Let Y = the number of shots until the first made shot Since these problems deal with shooting until the first miss (or until the first made shot), a geometric model, either Geom( 8) or Geom( 2), is appropriate 4 a) Use Geom( 2 ) PX ( = 5) = ( 8 ) ( 2 ) = 8192 (Four shots made, followed by a miss) 3 b) Use Geom() 8 P( Y = 4) = ()() 2 8 = 4 (Three misses, then a made shot) c) Use Geom() 8 P( Y = 1) + P( Y = 2) + P( Y = 3) = () 8 + ()() ()() = Chips The selection of chips may be considered Bernoulli trials There are only two possible outcomes (fail testing and pass testing) Provided that the chips selected are a representative sample of all chips, the probability that a chip fails testing is constant at 2% The trials are not independent, since the population of chips is finite, but we won t need to sample more than 1% of all chips Let X = the number of chips required until the first bad chip The appropriate model is Geom( 2 ) 4 a) PX ( = 5) = ( 98 ) ( 2 ) 184 (Four good chips, then a bad one) b) P( 1 X 1) = ( 2 ) + ( 98 )( 2 ) + ( 98 ) 2 ( 2 ) + + ( 98 ) 9 ( 2 ) 183 (Use the geometric model on a calculator or computer for this one!) 11 More hoops As determined in a previous exercise, the shots can be considered Bernoulli trials, and since the player is shooting until the first miss, Geom( 2 ) is the appropriate model E( X) = 1 p = 1 2 = 5 shots The player is expected to take 5 shots until the first miss

5 Chips ahoy Chapter 17 Probability Models 245 As determined in a previous exercise, the selection of chips can be considered Bernoulli trials, and since the company is selecting until the first bad chip, Geom( 2 ) is the appropriate model E( X) = 1 p = 1 2 = 5 chips The first bad chip is expected to be the 5 th chip selected 13 Customer center operator The calls can be considered Bernoulli trials There are only two possible outcomes, taking the promotion, and not taking the promotion The probability of success is constant at 5% (5% of the 1% Platinum cardholders) The trials are not independent, since there are a finite number of cardholders, but this is a major credit card company, so we can assume we are selecting fewer than 1% of all cardholders Since we are calling people until the first success, the model Geom( 5 ) may be used 1 1 E( calls) = = = 2 calls We expect it to take 2 calls to find the first cardholder to take p 5 the double miles promotion 14 Cold calls 15 Blood The donor contacts can be considered Bernoulli trials There are only two possible outcomes, giving $1 or more, and not giving $1 or more The probability of success is constant at 1% (5% of the 2% of donors who will make a donation) The trials are not independent, since there are a finite number of potential donors, but we will assume that she is contacting less than 1% of all possible donors Since we are contacting people until the first success, the model Geom( 1 ) may be used 1 1 E( contacts) = = = 1 contacts We expect that Justine will have to contact 1 p 1 potential donors to find a $1 donor These may be considered Bernoulli trials There are only two possible outcomes, Type AB and not Type AB Provided that the donors are representative of the population, the probability of having Type AB blood is constant at 4% The trials are not independent, since the population is finite, but we are selecting fewer than 1% of all potential donors Since we are selecting people until the first success, the model Geom( 4 ) may be used Let X = the number of donors until the first Type AB donor is found a) E( X) = 1 p = 1 4 = 25 people We expect the 25 th person to be the first Type AB donor b) P( a Type AB donor among the first 5 people checked) = PX ( = 1) + PX ( = 2) + PX ( = 3) + PX ( = 4) + PX ( = 5) = ( 4 ) + ( 9 )( 4 ) + ( 9 ) ( 4 ) + ( 9 ) ( 4 ) + ( 9 ) ( 4 ) 185

6 24 Part IV Randomness and Probability c) P( a Type AB donor among the first people checked) = PX ( = 1) + PX ( = 2) + PX ( = 3) + PX ( = 4) + PX ( = 5) + PX ( = ) = ( 4 ) + ( 9 )( 4 ) + ( 9 ) ( 4 ) + ( 9 ) ( 4 ) + ( 9 ) ( 4 ) + ( 9 ) ( 4 ) d) P( no Type AB donor before the 1th person checked)= P( X > 9) = ( 9 ) 93 This one is a bit tricky There is no implication that we actually find a donor on the 1 th trial We only care that nine trials passed with no Type AB donor 1 Colorblindness These may be considered Bernoulli trials There are only two possible outcomes, colorblind and not colorblind As long as the men selected are representative of the population of all men, the probability of being colorblind is constant at about 8% Trials are not independent, since the population is finite, but we won t be sampling more than 1% of the population Let X = the number of people checked until the first colorblind man is found Since we are selecting people until the first success, the model Geom( 8 ), may be used 1 1 We expect to examine 5 people until finding the a) E( X) = = = 5 people p 8 first colorblind person b) P( no colorblind men among the first 4) = P( X > 4) = (92) c) P( first colorblind man is the sixth man checked)= P( X = ) = ( 92 ) ( 8 ) 527 d) P(she finds a colorblind man before the tenth man) = P( 1 X 9) 2 8 = ( 8 ) + ( 92 )( 8 ) + ( 92 ) ( 8 ) + + ( 92 ) ( 8 ) 528 (Use the geometric model on a calculator or computer for this one!) 17 Lefties These may be considered Bernoulli trials There are only two possible outcomes, lefthanded and not left-handed Since people are selected at random, the probability of being left-handed is constant at about 13% The trials are not independent, since the population is finite, but a sample of 5 people is certainly fewer than 1% of all people Let X = the number of people checked until the first lefty is discovered Let Y = the number of lefties among n = 5 a) Use Geom( 13 ) 4 P( first lefty is the fifth person)= P( X = 5) = ( 87 ) ( 13 ) 745

7 Chapter 17 Probability Models 247 b) Use Binom( 513, ) P( some lefties among the 5 people) = 1 P( no lefties among the first 5 people) = 1 PY ( = ) 5 = 1 ( 13 ) ( 87 ) 5 52 c) Use Geom( 13 ) 2 P( first lefty is second or third person)= P( X = 2) + P( X = 3) = ( 87 )( 13 ) + ( 87 ) ( 13 ) 211 d) Use Binom( 513, ) P( exactly 3 lefties in the group)= P( Y = ) = ( 13 ) ( 87 ) 1 3 e) Use Binom( 513, ) P( at least 3 lefties in the group)= P( Y = 3) + P( Y = 4) + P( Y = 5) ( 13 ) ( 87 ) + ( 13 ) ( 87 ) + ( 13 ) ( 87 ) f) Use Binom( 513, ) P( at most 3 lefties in the group)= P( Y = ) + P( Y = 1) + P( Y = 2) + P( Y = 3) 18 Arrows ( 13 ) ( 87 ) + ( 13 ) ( 87 ) ( 13 ) ( 87 ) + ( 13 ) ( ) These may be considered Bernoulli trials There are only two possible outcomes, hitting the bull s-eye and not hitting the bull s-eye The probability of hitting the bull s-eye is given, p = 8 The shots are assumed to be independent Let X = the number of shots until the first bull s-eye Let Y = the number of bull s-eyes in n = shots a) Use Geom( 8 ) 2 P(first bull s-eye is on the third shot) = PX ( = 3) = ( 2 ) ( 8 ) 32 b) Use Binom( 8, ) P( at least one miss out of shots) = 1 P( out of hits) = 1 PY ( = ) = 1 ( 8 ) ( 2 ) 738

8 248 Part IV Randomness and Probability c) Use Geom( 8 ) 3 4 P( first hit on fourth or fifth shot)= P( X = 4) + P( X = 5) = ( 2 ) ( 8 ) + ( 2 ) ( 8 ) = 78 d) Use Binom( 8, ) P( exactly four hits ) = P( Y = 4) = 4 2 ( 8 ) ( 2 ) 4 24 e) Use Binom( 8, ) P( at least four hits ) = P( Y = 4) + P( Y = 5) + P( Y = ) ( 8 ) ( 2 ) + ( 8 ) ( 2 ) + ( 8 ) ( 2 ) f) Use Binom( 8, ) P( at most four hits ) = P( Y = ) + P( Y = 1) + P( Y = 2) + P( Y = 3) + P( Y = 4) 19 Lefties redux ( 8 ) ( 2 ) + ( 8 ) ( 2 ) + ( 8 ) ( 2 ) ( 8 ) 3 ( 2 ) + ( 8 ) ( 2 ) a) In a previous exercise, we determined that the selection of lefties could be considered Bernoulli trials Since our group consists of 5 people, use Binom( 513, ) Let Y = the number of lefties among n = 5 EY ( ) = np= 5( 13 ) = 5 lefties b) SD( Y) = npq = 5( 13 )( 87 ) 75 lefties c) Use Geom( 13 ) Let X = the number of people checked until the first lefty is discovered 1 1 EX ( ) = = 79 people p 13 2 More arrows a) In a previous exercise, we determined that the shots could be considered Bernoulli trials Since the archer is shooting arrows, use Binom( 8, ) Let Y = the number of bull s-eyes in n = shots EY ( ) = np= 8 ( ) = 48 bull s-eyes b) SD( Y) = npq = ( 8 )( 2 ) 98 bull s-eyes

9 Chapter 17 Probability Models 249 c) Use Geom( 8 ) Let X = the number of arrows shot until the first bull s-eye 1 1 EX ( ) = = = 5 shots p 8 21 Still more lefties a) In a previous exercise, we determined that the selection of lefties (and also righties) could be considered Bernoulli trials Since our group consists of people, and now we are considering the righties, use Binom(, 87 ) b) i) Let Y = the number of righties among n = EY ( ) = np= ( 87 ) = 144 righties SD( Y) = npq = ( 87 )( 13 ) 1 1 righties P( not all righties) = 1 P( all righties) = 1 PY ( = ) ii) = 1 ( 87 ) ( 13 ) 8 P( no more than 1righties ) = P( Y 1) = PY ( = ) + PY ( = 1) + PY ( = 2) + + PY ( = 1) ( 87 ) ( 13 ) + ( 87 ) ( 13 ) + + ( 87 ) ( 13 ) iii) iv) P( exactly six of each ) = P( Y = ) ( 87 ) ( 13 ) 193 P( majority righties ) = P( Y 7) = PY ( = 7) + PY ( = 8) + PY ( = 9) + + PY ( = ) ( 87 ) ( 13 ) + ( 87 ) ( 13 ) + + ( 87 ) ( 13 )

10 25 Part IV Randomness and Probability 22 Still more arrows a) In a previous exercise, we determined that the archer s shots could be considered Bernoulli trials Since our archer is now shooting 1 arrows, use Binom( 18, ) b) i) Let Y = the number of bull s-eyes hit from n = 1 shots EY ( ) = np= 18 ( ) = 8 bull s-eyes hit SD( Y) = npq = 18 ( )( 2 ) 1 2 bull s-eyes hit P( no misses out of 1shots) = P( all hits out of 1shots) = PY ( = 1) ii) iii) 1 ( 8 ) ( 2 ) P( no more than 8 hits)= P( Y 8) = PY ( = ) + PY ( = 1) + PY ( = 2) + + PY ( = 8) 24 P( exactly 8 out of 1shots)= P( Y = 8) ( 8 ) ( 2 ) + ( 8 ) ( 2 ) + + ( 8 ) ( 2 ) Vision 1 ( 8 ) ( 2 ) iv) P( more hits than misses)= P( Y ) = PY ( = ) + PY ( = 7) + + PY ( = 1) 1 ( 8 ) ( 2 ) + ( 8 ) ( 2 ) + + ( 8 ) ( 2 ) The vision tests can be considered Bernoulli trials There are only two possible outcomes, nearsighted or not The probability of any child being nearsighted is given as p = Finally, since the population of children is finite, the trials are not independent However, 19 is certainly less than 1% of all children, and we will assume that the children in this district are representative of all children in relation to nearsightedness Use Binom( 19, ) µ = E( nearsighted) = np = 19( ) = 228 children σ = SD( nearsighted) = npq = 19( )( 88 ) 4 22 children

11 24 International students Chapter 17 Probability Models 251 The students can be considered Bernoulli trials There are only two possible outcomes, international or not The probability of any freshmen being an international student is given as p = Finally, since the population of freshmen is finite, the trials are not independent However, 4 is likely to be less than 1% of all students, and we are told that the freshmen in this college are randomly assigned to housing Use Binom( 4, ) µ = E( international) = np= 4 ( ) = 2 4 students σ = SD( nearsighted) = npq = 4 ( )( 94 ) 1 5 students 25 Tennis, anyone? The first serves can be considered Bernoulli trials There are only two possible outcomes, successful and unsuccessful The probability of any first serve being good is given as p = 7 Finally, we are assuming that each serve is independent of the others Since she is serving times, use Binom( 7, ) Let X = the number of successful serves in n = first serves a) b) P( all six serves in ) = P( X = ) P( exactly four serves in)= P( X = 4) c) ( 7 ) ( 3 ) 118 P( at least four serves in ) = P( X = 4) + P( X = 5) + P( X = ) ( 7 ) ( 3 ) d) ( 7 ) ( 3 ) + ( 7 ) ( 3 ) + ( 7 ) ( 3 ) P( no more than four serves in ) = P( X = ) + P( X = 1) + P( X = 2) + P( X = 3) + P( X = 4) 2 Frogs ( 7 ) ( 3 ) + ( 7 ) ( 3 ) + ( 7 ) ( 3 ) ( 7) ( 3 ) + ( 7 ) ( 3 ) The frog examinations can be considered Bernoulli trials There are only two possible outcomes, having the trait and not having the trait If the frequency of the trait has not changed, and the biologist collects a representative sample of frogs, then the probability of a frog having the trait is constant, at p = 5 The trials are not independent since the population of frogs is finite, but frogs is fewer than 1% of all frogs Since the biologist is collecting frogs, use Binom(, 5 ) Let X = the number of frogs with the trait, from n = frogs

12 252 Part IV Randomness and Probability a) b) c) d) P( no frogs have the trait)= P( X = ) ( 5 ) ( 875 ) 21 P( at least two frogs)= P( X 2) = PX ( = 2) + PX ( = 3) + + PX ( = ) = ( ) ( ) ( ) ( ) ( ) ( ) P( three or four frogs have trait)= P( X = 3) + P( X = 4) ( 5 ) ( 875 ) + ( 5 ) ( 875 ) P( no more than four)= P( X 4) = P( X = ) + P( X = 1) + + P( X = 4) 27 And more tennis ( 5 ) ( 875 ) + ( 5 ) ( 875 ) + + ( 5 ) ( 875 ) The first serves can be considered Bernoulli trials There are only two possible outcomes, successful and unsuccessful The probability of any first serve being good is given as p = 7 Finally, we are assuming that each serve is independent of the others Since she is serving 8 times, use Binom( 87, ) Let X = the number of successful serves in n = 8 first serves a) EX ( ) = np= 87 ( ) = 5 first serves in SD( X) = npq = 87 ( )( 3 ) 4 1 first serves in b) Since np = 5 and nq = 24 are both greater than 1, Binom( 87, ) may be approximated by the Normal model, N(5, 41) c) According to the Normal model, in matches with 8 serves, she is expected to make between 519 and 1 first serves approximately 8% of the time, between 478 and 42 first serves approximately 95% of the time, and between 437 and 83 first serves approximately 997% of the time

13 d) Using Binom(8, 7): P( at least 5 first serves)= P( X 5) = PX ( = 5) + PX ( = ) + + PX ( = 8) = 8 5 Chapter 17 Probability Models 253 ( ) ( ) + 8 ( ) ( ) ( 7 ) ( 3 ) According to the Binomial model, the probability that she makes at least 5 first serves out of 8 is approximately 11 Using N(5, 41): x z = µ σ 5 5 z = 41 z According to the Normal model, the probability that she makes at least 5 first serves out of 8 is approximately 141 PX ( 5) Pz ( > 2 195) More arrows These may be considered Bernoulli trials There are only two possible outcomes, hitting the bull s-eye and not hitting the bull s-eye The probability of hitting the bull s-eye is given, p = 8 The shots are assumed to be independent Since she will be shooting 2 arrows, use Binom(2, 8) Let Y = the number of bull s-eyes in n = 2 shots a) EY ( ) = np= 2( 8) = 1 bull s-eyes SD( Y) = npq = 28 ( )( 2 ) 5 bull s-eyes b) Since np = 1 and nq = 4 are both greater than 1, Binom( 2, 8 ) may be approximated by the Normal model, N(1, 5) c) According to the Normal model, in matches with 2 arrows, she is expected to get between and 15 bull s-eyes approximately 8% of the time, between 1488 and bull s-eyes approximately 95% of the time, and between 1432 and 1798 bull s-eyes approximately 997% of the time

14 254 Part IV Randomness and Probability d) Using Binom(2, 8): P( at most 14hits)= P( Y 14) = PY ( = 1 ) + PY ( = ) + + PY ( = 14) = 2 5 ( ) ( ) + 2 ( ) ( ) ( 8 ) ( 7 ) According to the Binomial model, the probability that she makes at most 14 bull s-eyes out of 2 is approximately 5 Using N(1, 5): y According to the Normal z = µ model, the probability that she σ hits at most 14 bull s-eyes out 141 z = of 2 is approximately 2 5 z PY ( 2) Pz ( < 3 534) 2 Using either model, it is apparent that it is very unlikely that the archer would hit only 14 bull s-eyes out of 2 29 Apples a) A binomial model and a normal model are both appropriate for modeling the number of cider apples that may come from the tree Let X = the number of cider apples found in the n = 3 apples from the tree The quality of the apples may be considered Bernoulli trials There are only two possible outcomes, cider apple or not a cider apple The probability that an apple must be used for a cider apple is constant, given as p = The trials are not independent, since the population of apples is finite, but the apples on the tree are undoubtedly less than 1% of all the apples that the farmer has ever produced, so model with Binom(3, ) EX ( ) = np= 3( ) = 18 cider apples SD( X) = npq = 3( )( 94) 4 11 cider apples Since np = 18 and nq = 282 are both greater than 1, Binom( 3, ) may be approximated by the Normal model, N(18, 411)

15 b) Using Binom(3, ): P( at most cider apples)= P( X ) = PX ( = ) + + PX ( = ) = 3 85 ( ) ( ) ( ) ( 94 ) Chapter 17 Probability Models According to the Binomial model, the probability that no more than cider apples come from the tree is approximately 85 Using N(18, 411): x z = µ σ 18 z = 411 z = 1 4 PX ( ) Pz ( < ) According to the Normal model, the probability that no more than apples out of 3 are cider apples is approximately 72 c) It is extremely unlikely that the tree will bear more than 5 cider apples Using the Normal model, N(18, 411), 5 cider apples is about 78 standard deviations above the mean 3 Frogs, part II The frog examinations can be considered Bernoulli trials There are only two possible outcomes, having the trait and not having the trait If the frequency of the trait has not changed, and the biologist collects a representative sample of frogs, then the probability of a frog having the trait is constant, at p = 5 The trials are not independent since the population of frogs is finite, but 15 frogs is fewer than 1% of all frogs Since the biologist is collecting 15 frogs, use Binom( 155, ) Let X = the number of frogs with the trait, from n = 15 frogs a) EX ( ) = np= 155 ( ) = frogs SD( X) = npq = 155 ( )( 875 ) 4 5 frogs b) Since np = 1875 and nq = 135 are both greater than 1, Binom( 2, 5 ) may be approximated by the Normal model, N(1875, 45) c) Using Binom(15, 5): P( at least 22 frogs)= P( X 22) = PX ( = 22) + + PX ( = 15) = ( ) ( ) ( 5 ) ( 875 ) According to the Binomial model, the probability that at least 22 frogs out of 15 have the trait is approximately 2433

16 25 Part IV Randomness and Probability Using N(1875, 45): x z = µ σ z = 45 z 82 PX ( 22) Pz ( > 82 ) 2111 According to the Normal model, the probability that at least 22 frogs out of 15 have the trait is approximately 2111 Using either model, the probability that the biologist discovers 22 of 15 frogs with the trait simply as a result of natural variability is quite high This doesn t prove that the trait has become more common 31 Lefties again Let X = the number of righties among a class of n = 188 students Using Binom(188, 87): These may be considered Bernoulli trials There are only two possible outcomes, righthanded and not right-handed The probability of being right-handed is assumed to be constant at about 87% The trials are not independent, since the population is finite, but a sample of 188 students is certainly fewer than 1% of all people Therefore, the number of righties in a class of 188 students may be modeled by Binom(188, 87) If there are 171 or more righties in the class, some righties have to use a left-handed desk P( at least 171 righties)= P( X 171) = PX ( = 171) + + PX ( = 188) = ( ) ( ) ( 87 ) ( 13 ) According to the binomial model, the probability that a right-handed student has to use a left-handed desk is approximately 1 Using N(135, 41): EX ( ) = np= 188( 87 ) = 13 5 righties SD( X) = npq = 188( 87 )( 13 ) 4 1 righties Since np = 135 and nq = 2444 are both greater than 1, Binom( 188, 87 ) may be approximated by the Normal model, N(135, 41)

17 Chapter 17 Probability Models 257 x z = µ σ z = 41 z 1 14 PX ( 171) Pz ( > 1 14) 53 According to the Normal model, the probability that there are at least 171 righties in the class of 188 is approximately No-shows Let X = the number of passengers that show up for the flight of n = 275 passengers Using Binom(275, 95): These may be considered Bernoulli trials There are only two possible outcomes, showing up and not showing up The airlines believe the probability of showing up is constant at about 95% The trials are not independent, since the population is finite, but a sample of 275 passengers is certainly fewer than 1% of all passengers Therefore, the number of passengers who show up for a flight of 275 may be modeled by Binom(275, 95) If 2 or more passengers show up, someone has to get bumped off the flight P( at least 2 passengers)= P( X 2) = PX ( = 2) + + PX ( = 275) = ( ) ( ) ( 95 ) ( 5 ) According to the binomial model, the probability someone on the flight must be bumped is approximately 11 Using N(25, 31): EX ( ) = np= 275( 95 ) = passengers SD( X) = npq = 275( 95 )( 5 ) 3 1 passengers Since np = 25 and nq = 1375 are both greater than 1, Binom( 275, 95 ) may be approximated by the Normal model, N(25, 31) x z = µ PX ( 2) Pz ( > 1 31) 941 σ z = 31 z 1 31 According to the Normal model, the probability that at least 2 passengers show up is approximately 941

18 258 Part IV Randomness and Probability 33 Annoying phone calls Let X = the number of sales made after making n = 2 calls Using Binom(2, ): These may be considered Bernoulli trials There are only two possible outcomes, making a sale and not making a sale The telemarketer was told that the probability of making a sale is constant at about p = The trials are not independent, since the population is finite, but 2 calls is fewer than 1% of all calls Therefore, the number of sales made after making 2 calls may be modeled by Binom(2, ) P( at most 1)= P( X 1) = PX ( = 1 ) + + PX ( = ) = 2 ( ) ( ) ( ) ( 88 ) According to the Binomial model, the probability that the telemarketer would make at most 1 sales is approximately Using N(24, 4): EX ( ) = np= 2( ) = 24 sales SD( X) = npq = 2 ( )( 88 ) 4 sales Since np = 24 and nq = 17 are both greater than 1, Binom( 2, ) may be approximated by the Normal model, N(24, 4) x z = µ σ 124 z = 4 z 343 PX ( 1) Pz ( < 3 43) According to the Normal model, the probability that the telemarketer would make at most 1 sales is approximately Since the probability that the telemarketer made 1 sales, given that the % of calls result in sales is so low, it is likely that he was misled about the true success rate 34 The euro Let X = the number of heads after spinning a Belgian euro n = 25 times Using Binom(25, 5): These may be considered Bernoulli trials There are only two possible outcomes, heads and tails The probability that a fair Belgian euro lands heads is p = 5 The trials are independent, since the outcome of a spin does not affect other spins Therefore, Binom(25, 5) may be used to model the number of heads after spinning a Belgian euro 25 times

19 P( at least 14)= P( X 14) = PX ( = 1425 ) + + PX ( = ) = ( ) ( ) ( 5) ( 5) Chapter 17 Probability Models According to the Binomial model, the probability that a fair Belgian euro comes up heads at least 14 times is 332 Using N(5, 791): EX ( ) = np= 255 ( ) = 5 heads SD( X) = npq = 255 ( )( 5 ) 7 91 heads Since np = 5 and nq = 5 are both greater than 1, Binom( 255, ) may be approximated by the Normal model, N(5, 791) x z = µ σ 145 z = 791 z 1 89 PX ( 141 ) Pz ( > 89) 29 According to the Normal model, the probability that a fair Belgian euro lands heads at least 14 out of 25 spins is approximately 29 Since the probability that a fair Belgian euro lands heads at least 14 out of 25 spins is low, it is unlikely that the euro spins fairly However, the probability is not extremely low, and we aren t sure of the source of the data, so it might be a good idea to spin it some more 35 Seatbelts II These stops may be considered Bernoulli trials There are only two possible outcomes, belted or not belted Police estimate that the probability that a driver is buckled is 8% (The probability of not being buckled is therefore 2%) Provided the drivers stopped are representative of all drivers, we can consider the probability constant The trials are not independent, since the population of drivers is finite, but the police will not stop more than 1% of all drivers a) Let X = the number of cars stopped before finding a driver whose seat belt is not buckled Use Geom(2) to model the situation 1 1 E( X) = = = 5 cars p 2 5 b) P( First unbelted driver is in the sixth car)= P( X = ) = ( 8 ) ( 2 ) c) P( The first ten drivers are wearing seatbelts) = (8) 1 17

20 2 Part IV Randomness and Probability d) Let Y = the number of drivers wearing their seatbelts in 3 cars Use Binom(3, 8) EY ( ) = np= 38 ( ) = 24 drivers SD( Y) = npq = 38 ( )( 2 ) 2 19 drivers e) Let W = the number of drivers not wearing their seatbelts in cars Using Binom(, 2): P( at least 2)= P( W 2) = PW ( = 2 ) + + PW ( = ) = Using N(24, 438): EW ( ) = np= 2 ( ) = 24 drivers ( ) ( ) ( 2) ( 8) 21 SD( W ) = npq = 2 ( )( 8 ) 4 38 drivers According to the Binomial model, the probability that at least 2 out of drivers are not wearing their seatbelts is approximately 848 Since np = 24 and nq = 9 are both greater than 1, Binom( 2, ) may be approximated by the Normal model, N(24, 438) w µ z = σ 224 z = 438 z 913 PW ( ) Pz ( > 913 ) 8194 According to the Normal model, the probability that at least 2 out of drivers stopped are not wearing their seatbelts is approximately Rickets The selection of these children may be considered Bernoulli trials There are only two possible outcomes, vitamin D deficient or not vitamin D deficient Recent research indicates that 2% of British children are vitamin D deficient (The probability of not being vitamin D deficient is therefore 8%) Provided the students at this school are representative of all British children, we can consider the probability constant The trials are not independent, since the population of British children is finite, but the children at this school represent fewer than 1% of all British children a) Let X = the number of students tested before finding a student who is vitamin D deficient Use Geom(2) to model the situation 7 P( First vitamin D deficient child is the eighth one tested)= P( X = 8) = ( 8 ) ( 2 ) 42 b) P( The first ten children tested are okay) = (8) 1 17

21 1 1 c) E( X) = = = 5 kids p 2 Chapter 17 Probability Models 21 d) Let Y = the number of children who are vitamin D deficient out of 5 children Use Binom(5, 2) EY ( ) = np= 52 ( ) = 1 children SD( Y) = npq = 52 ( )( 8 ) 2 83 children e) Using Binom(32, 2): P( no more than 5children have the deficiency)= P( X 5) = PX ( = 5 ) + + PX ( = ) = ( ) ( ) ( 2) ( 8) According to the Binomial model, the probability that no more than 5 of the 32 children have the vitamin D deficiency is approximately 27 Using N(4, 71): EY ( ) = np= 322 ( ) = 4 children SD( Y) = npq = 322 ( )( 8 ) 7 1 children Since np = 4 and nq = 25 are both greater than 1, Binom( 322, ) may be approximated by the Normal model, N(4, 71) y z = µ σ 54 z = 71 z PY ( 51 ) Pz ( < 955) 253 According to the Normal model, the probability that no more than 5 out of 32 children have the vitamin D deficiency is approximately ESP Choosing symbols may be considered Bernoulli trials There are only two possible outcomes, correct or incorrect Assuming that ESP does not exist, the probability of a correct identification from a randomized deck is constant, at p = 2 The trials are independent, as long as the deck is shuffled after each attempt Since 1 trials will be performed, use Binom(1, 2) Let X = the number of symbols identified correctly out of 1 cards E( X) = np= 1( 2) = 2 correct identifications SD( X) = npq = 12 ( )( 8 ) = 4 correct identifications Answers may vary In order be convincing, the mind reader would have to identify at least 32 out of 1 cards correctly, since 32 is three standard deviations above the mean Identifying fewer cards than 32 could happen too often, simply due to chance

22 22 Part IV Randomness and Probability 38 True-False Guessing at answers may be considered Bernoulli trials There are only two possible outcomes, correct or incorrect If the student was guessing, the probability of a correct response is constant, at p = 5 The trials are independent, since the answer to one question should not have any bearing on the answer to the next Since 5 questions are on the test use Binom(5, 5) Let X = the number of questions answered correctly out of 5 questions E( X) = np= 55 ( ) = 25 correct answers SD( X) = npq = 55 ( )( 5 ) 3 54 correct answers Answers may vary In order be convincing, the student would have to answer at least 3 out of 5 questions correctly, since 3 is approximately three standard deviations above the mean Answering fewer than 3 questions correctly could happen too often, simply due to chance 39 Hot hand A streak like this is not unusual The probability that he makes 4 in a row with a 55% free throw percentage is ( 55 )( 55 )( 55 )( 55 ) 9 We can expect this to happen nearly one in ten times for every set of 4 shots that he makes One out of ten times is not that unusual 4 New bow A streak like this is not unusual The probability that she makes consecutive bulls-eyes with an 8% bulls-eye percentage is ( 8)( 8)( 8)( 8)( 8)( 8) 2 If she were to shoot several flights of arrows, she is expected to get bulls-eyes about 2% of the time An event that happens due to chance about one out of four times is not that unusual 41 Hotter hand The shots may be considered Bernoulli trials There are only two possible outcomes, make or miss The probability of success is constant at 55%, and the shots are independent of one another Therefore, we can model this situation with Binom(32, 55) Let X = the number of free throws made out of 4 E( X) = np= 455 ( ) = 22 free throws made SD( X) = npq = 455 ( )( 45 ) 3 15 free throws Answers may vary The player s performance seems to have increased 32 made free throws is ( 32 22)/ standard deviations above the mean, an extraordinary feat, unless his free throw percentage has increased This does NOT mean that the sneakers are responsible for the increase in free throw percentage Some other variable may account for the increase The player would need to set up a controlled experiment in order to determine what effect, if any, the sneakers had on his free throw percentage

23 42 New bow, again Chapter 17 Probability Models 23 The shots may be considered Bernoulli trials There are only two possible outcomes, hit or miss the bulls-eye The probability of success is constant at 8%, and the shots are independent of one another Therefore, we can model this situation with Binom(5, 8) Let X = the number of bulls-eyes hit out of 5 E( X) = np= 58 ( ) = 4 bulls-eyes hit SD( X) = npq = 58 ( )( 2 ) 2 83 bulls-eyes Answers may vary The archer s performance doesn t seem to have increased 45 bulls-eyes is ( )/ standard deviations above the mean This isn t unusual for an archer of her skill level