184 Chapter Not binomial: Because the student receives instruction after incorrect answers, her probability of success is likely to increase.

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1 Chapter Chapter. Not binomial: There is not fixed number of trials n (i.e., there is no definite upper limit on the number of defects) and the different types of defects have different probabilities.. Yes: ) Success means person says Yes and failure means person says No. () We have a fixed number of observations (n = ). () It is reasonable to believe that each response is independent of the others. () It is reasonable to believe each response has the same probability of success (saying yes ) since the individuals are randomly chosen from a large city.. Yes: ) Success means reaching a live person and failure is any other outcome. () We have a fixed number of observations ( n = 5). () It is reasonable to believe that each call is independent of the others. () Each randomly-dialed number has chance p =. of reaching a live person.. Not binomial: There is no fixed number of attempts (n)..5 Not binomial: Because the student receives instruction after incorrect answers, her probability of success is likely to increase..6 The number who say they never have time to relax has (approximately) a binomial distribution with parameters n = 5 and p =. ) Success means the respondent never has time to relax and failure means the respondent has time to relax. (This is a good example to point out why success and failure should be referred to as labels.) ) We have a fixed number of observations (n = 5). ) It is reasonable to believe each response is independent of the others. ) The probability of success may vary from individual to individual (think about retired individuals versus parents versus students), but the opinion polls provide a reasonable approximation for the probability in the entire population..7 Let = the number of children with type O blood. is B(5,.5). 5 P ( = ) = (.5) (.75) (.5) (.75) = 7. Let = the number of broccoli plans that you lose. is B(, 5). P( ) = P ( = ) + P( = ) = (5) (.5) + (5) (.5) = (.5) ( )( ). Let = the number of children with blood type O. is B(5,.5). 5 P ( ) = P ( = ) = (.5) (.75) 5 (.75) 5 =.7 =.767. Let = the number of players who graduate. is B(,.).

2 The Binomial and Geometric Distributions 5 (a) ( ) (.) (.) P= = 7 (b) P ( = ) = (.) (.) 5 P ( < ) = P = 5 =.5 (c) ( ). Let = the number of Hispanics on the committee. is B(5,.). 5 (a) ( ) (.) (.7) 5 P= =.7 (b) P ( = ) = (.) (.7) 5 = 7. Let = the number of men called. is B(,.7). (a) P ( = ) = (.7) (.) 6 (b) P( st woman is the th call) = ( ) ( ).7... Let = the number of children with blood type O. is B(5,.5). (a) P ( =) = binompdf(5,.5, ) =.67. (b) A table with the values of, the pdf, and the cdf is shown below. 5 pdf P() cdf F() (c) The probabilities given in the table above for P() add to. (d) A probability histogram is shown below on the left.. Probability.. Count of children with type O blood 5 (e) See the table above for the cumulative probabilities. Cumulative distribution histograms are shown below for the number of children with type O blood (left) and the number of free throws made (right). Both cumulative distributions show bars that step up to one, but the bars in the cumulative histogram for the number of children with type O blood get taller sooner. That is, there are fewer steps and the steps are bigger.

3 6 Chapter.. Cumulative probability..6 Cumulative probability..6.. Count of children with type O blood Count of free throws made. Let = the number of correct answers. is B(5,.5). (a) P( 5) = P( ) = binomcdf (5,.5, ) =.556. (b) P( ) = P( ) = binomcdf (5,.5, ).7 =. (c) P( ) = P( ) = binomcdf (5,.5, ).675 = 5..5 (a) Let = the number of correct answers. is B(,.5). The probability of at least one correct answer is P( ) = P( = ) = binompdf (,.5,) 56 =.7. (b) Let = the number of correct answers. P( ) = P( = ). P( = ) is the probability of getting none of the questions correct, or every question wrong. Note that this is not a binomial random variable because each question has a different probability of a success. The probability of getting the first question wrong is /, the second question wrong is / and the third question wrong is /5. The probability of getting all of the questions wrong is P( = ) = (/) (/) (/5) =, because Erin is guessing so the responses to different questions are independent. Thus, P( ) = P( = ) = =.6..6 (a) Yes: ) Success means having an incarcerated parent and failure is not having an incarcerated parent. () We have a fixed number of observations ( n = ). () It is reasonable to believe that the responses of the children are independent. () Each randomly selected child has probability p = of having an incarcerated parent. (b) P( = ) is the probability that none of the selected children has an incarcerated parent. P( = ) = binompdf (,, ).6 and P(= ) = binompdf (,, ).77. (c) P( ) = P( ) = binomcdf (,, ) =.567. Alternatively, by the addition rule for mutually exclusive events, P( ) = (P( = ) + P( = )) = ( ) = = Let = the number of players who graduate. is B(,.). (a) P( =) = binompdf(,., ) 7. (b) P( = ) = binompdf(,., ) =. 5. (c) P( ) = P(=) 5 =.5.. (a) n = and p =.5. (b) P ( = ) = (.5) (.75) = binompdf(,.5, ).6. (c) P( ) = P = + P = + P = = binomcdf(,.5, ).556. ( ) ( ) ( )

4 The Binomial and Geometric Distributions 7. (a) np = 5.6 = 5, n( p) = 5 = ; both values are greater than, so the conditions are satisfied. (b) Let = the number of people in the sample who find shopping frustrating. is B(5,.6). Then P( 5) = P( 5) = binomcdf (5,.6, 5) =.766 =.7, which rounds to.. The probability correct to six decimal places is.. (c) P( 6) = binomcdf (5,.6, 6). Using the Normal approximation to the binomial, P( 6) 57, a difference of 7.. Let be the number of s and s; then has a binomial distribution with n = and p = 77 (in the absence of fraud). Using the calculator or software, we find P ( ) = binomcdf(, 77, ). Using the Normal approximation (the conditions are satisfied), we find a mean of. and standard deviation of σ = Therefore, ( ). P P Z = PZ (.) = 6. Either way, the.7 probability is quite small, so we have reason to be suspicious.. (a) The mean is µ = np =. = 6. (b) The standard deviation is σ = ()(.)(.) =..7. (c) If p =. then σ =...6, and if p =. then σ =. 5. As the probability of success gets closer to the standard deviation decreases. (Note that as p approaches, the probability histogram of the binomial distribution becomes increasingly skewed, and thus there is less and less chance of getting an observation an appreciable distance from the mean.). If H is the number of home runs, with a binomial(n = 5, p =.6) distribution, then H has mean µ = np = and standard deviation H σ = home runs. Therefore, H 7 5. PH ( 7) P Z = P( Z.5) = 6. Using a calculator or software, we 7.6 find that the exact value is binomcdf(5,.6, 6) = 7677 or about 76.. (a) Let = the number of people in the sample of adults from Richmond who approve of the President s response. The count is approximately binomial. ) Success means the respondent approves and failure means the respondent does not approve. ) We have a fixed number of observations (n = ). ) It is reasonable to believe each response is independent of the others. ) The probability of success may vary from individual to individual (think about people with affiliations in different political parties), but the national survey will provide a reasonable approximate probability for the entire nation. (b) P( 5) = binomcdf(,., 5). (c) The expect number of approvals is µ =. = 6 and the standard deviation is σ =. =. 5.5 approvals. (d) Using the Normal approximation, ( ) P P Z = P( Z.) =, a 5.5

5 Chapter difference of. The approximation is not very accurate, but note that p is close to so the exact distribution is skewed.. (a) The mean is µ = np = 5. = blacks and the standard deviation is σ = blacks. The Normal approximation is quite safe: n p = and n ( p) = are both more than. We compute 65 5 P( 65 5) P Z = P(. Z.).766. (Exact computation of this probability with a calculator or software gives.7.).5 The command cumsum (L ) L calculates and stores the values of P( x) for x =,,,,. The entries in L and the entries in L defined by binomcdf(,.75, L ) L are identical..6 (a) Answers will vary. The observations for one simulation are:,,,,,,,,, and, with a sum of 7. For these data, the average is x =.7. Continuing this simulation, sample means were obtained:.7,.6,.6,.,.,.5,.,.,., and.. The mean of these sample means is.7, which is close to, and the standard deviation of these means is.6, which is close to.7 /. (Note: Another simulation produced sample means of.,.,.5,.,.,.5,.6,.5,., and., which have an average of. and a standard deviation of 6. There is more variability in the standard deviation.) (b) For n = 5, one simulation produced sample means of.5,.,.,.,.,.7,.6,.7,., and.5, with a mean of. and a standard deviation of.56. For n = 5, one simulation produced sample means of., 5.5, 5.,.7, 5., 5.,.7,.,.7, and 6., with a mean of. and a standard deviation of.67. (c) As the number of switches increases from to 5 and then 5, the sample mean also increases from to.5 and then 5. As the sample size increases from to 5 and then from 5 to 5, the spread of x values increases. The number of simulated samples stays the same at, but σ changes from...7 to 5.. =.5 and then (a) Let S denote the number of contaminated eggs chosen by Sara. S has a binomial distribution with n = and p =.5; i.e., S is B(,.5) (b) Using the calculator and letting a contaminated egg and, or good egg, simulate choosing eggs by RandInt(,, ). Repeating this 5 times leads to occasions when at least one of the eggs is contaminated; P S = P S = P S = = binompdf(,.5, ) = (.75) ( ).6. (c) ( ) ( ) The value obtained by simulation is close to the exact probability; the difference is.. (a) We simulate 5 observations of = the number of students out of with a loan by using the command randbin (,.65, ) L: sum (L). Press ENTER 5 times. Then sort the list from largest to smallest using the command SortD(L) (this command is found on the TI / under Stat EDIT :SortD) and then look to see how many values are greater than. Only one of the simulated values was greater than, so the estimated probability is /5 =.

6 The Binomial and Geometric Distributions (b) Using the calculator, we find P( ) P( ) > = = binomcdf(,.65, ).767 =. (c) Using the Normal approximation, we find.5 P( > ) P Z > = P( Z >.7) = 7. The Normal approximation is not very.65 good in this situation, because n ( p) =.5 is very close to the cutoff for our rule of thumb. The difference between the two probabilities in (b) and (c) is. Note that the simulation provides a better approximation than the Normal distribution.. Let = the number of s among n random digits. is B(n,.). (a) When n =, P( = 5 ) = binompdf(,., ) 5. (b) When n = 5, P( ) = P( = ) = (.).55 = 5.. (a) The probability of drawing a white chip is 5/5 =.. The number of white chips in 5 draws is B(5,.). Therefore, the expected number of white chips is 5. = 7.5. (b) The probability of drawing a blue chip is /5 =.. The number of blue chips in 5 draws is B(5,.). Therefore, the standard deviation of the number of blue chips is 5.. = blue chips. (c) Let the digits,,,, fl red chip, 5, 6, 7 fl white chip, and, flblue chip. Draw 5 random digits from Table B and record the number of times that you get chips of various colors. Using the calculator, you can draw 5 random digits using the command randint (,, 5) L. Repeat this process 5 times (or however many times you like) to simulate multiple draws of 5 chips. A sample simulation of a single 5-chip draw using the TI- yielded the following result: Digit Frequency 6 This corresponds to drawing red chips, white chips, and 7 blue chips. (d) The expected number of blue chips is 5. = 5, and the standard deviation is. It is very likely that you will draw or fewer blue chips. The actual probability is binomcdf (5,., ).7. (e) You are almost certain to draw 5 or fewer blue chips; the probability is binomcdf (5,., 5)... (a) A binomial distribution is not an appropriate choice for field goals made by the National Football League player, because given the different situations the kicker faces, his probability of success is likely to change from one attempt to another. (b) It would be reasonable to use a binomial distribution for free throws made by the NBA player because we have n = 5 attempts, presumably independent (or at least approximately so), with chance of success p =. each time.. (a) Yes: ) Success means the adult approves and failure means the adult disapproved. ) We have a fixed number of observations (n = 55). ) It is reasonable to believe each response is independent of the others. ) The probability of success may vary from individual to individual, but a national survey will provide a reasonable approximate probability for the entire nation. (b) Not binomial: There are no separate trials or attempts being observed here. (c) Yes: Let = the number of wins in 5 weeks. ) Success means Joe wins and failure means Joe loses. ) We have a fixed number of observations (n = 5). )

7 Chapter The results from one week to another are independent. ) The probability of winning stays the same from week to week.. (a) Answers will vary. A table of counts is shown below. Line Number Count of zeros 5 6 A dotplot and a boxplot are shown below Count of zeros 7 Count of zeros 6 5 The sample mean for these lines is zeros and the standard deviation is about.6 zeros. The distribution is clearly skewed to the right. (b) The number of zeros is binomial because ) Success is a digit of zero and failure is any other digit. ) The number of digits on each line is n =. We are equally likely to get any of the digits in any position so ) the trials are independent and ) the probability of success is p =. for each digit examined. (c) As the number of lines used increases, the mean gets closer to, the standard deviation becomes closer to.7 and the shape will still be skewed to the right because we are simulating a binomial distribution with n = and p =.. (d) Dan is right, the number of zeros in digits will be approximately normal with a mean of and a standard deviation of 6. As n increases, n. > and n. > so the conditions are satisfied and we can use a Normal distribution with a mean of n. and a standard deviation of n.. to approximate the Binomial(n,.) distribution. However, the conditions are not satisfied for one line so the simulated distribution will not become approximately normal, no matter how many additional rows are examined.. (a) n = and p =.5. (b) The mean is µ =.5 = 5. (c) The probability of getting exactly five correct guesses is P ( = 5) = (.5) 5 (.75) Let = the number of correctly answered questions. is B(,.). (a) P( = ) 7. (b) P( = ).6. (c) P( = ). (d) P( ).7. (e) P( > ) = P( ).7 =. (f) P( = )! (g) A probability distribution table is shown below. P() 7.65

8 The Binomial and Geometric Distributions (h) The expected number of correct answers is.5 =.5..6 Let = the number of truthful persons classified as deceptive. is B(,.). (a) The probability that the lie detector classifies all as truthful is P ( = ) = (.) (.) 67, and the probability that at least one is classified as deceptive is P ( ) = P = 67 =.. (b) The mean number who will be ( ) classified as deceptive is. =. applicants and the standard deviation is...56 P (.) = P.55, using binomcdf(,., ). applicants. (c) ( ).7 In this case, n = and the probability that a randomly selected basketball player graduates is p =.. We will estimate P( ) by simulating observations of = number graduated and computing the relative frequency of observations that are or smaller. The sequence of calculator commands are: randbin(,.,) L : sum(l ) L (), where s represent players who graduated. Press Enter until numbers are obtained. The actual value of P( ) is binomcdf(,., ).. (a) ) Success is getting a response, and Failure is not getting a response. ) We have a fixed number of trials (n = 5). ) It is reasonable to believe that the business decisions to respond or not are independent. ) The probability of success (responding) is.5 for each business. (b) The mean is 5.5 = 75 responses. (c) The approximate probability is 7 75 P( 7) P Z = P( Z.) = 6; using unrounded values and software 6.7 yields 7. The exact probability is about.. (d) Use, since.5 =.. (a) Let = the number of auction site visitors. is B(,.5). (b) P( = ) = binompdf(,.5, ) ; P( ) = P( 7) = binomcdf(,.5,7).6 =... (a) Let = the number of units where antibodies are detected. is B(,.) (b) The probability that all contaminated units are detected is P ( = ) = (.) ().7, using binompdf(,., ), and the probability

9 Chapter that at least one unit is not detected is P P( ) ( < ) =., using binomcdf(,., ). (c) The mean is. units and the standard deviation is 5 units.. (a) Yes: ) success is getting a tail, failure is getting a head, and a trial is the flip of a coin, ) the probability of getting a tail on each flip is p =.5, ) the outcomes on each flip are independent, and ) we are waiting for the first tail. (b) No: The variable of interest is the number of times both shots are made, not the number of trials until the first success is obtained. (c) Yes: ) success is getting a jack, failure is getting something other than a jack, and a trial is drawing of a card, ) the probability of drawing a jack is p = /5 = 76, ) the observations are independent because the card is replaced each time, and ) we are waiting for the first jack. (d) Yes: ) success is matching all 6 numbers, failure is not matching all 6 numbers, and a trial is the Match 6 lottery game on a particular day, ) the probability of winning is p =, ) the observations are independent, and ) we are waiting for the 6 first win. (e) No: the probability of a success, getting a red marble, changes from trial to trial because the draws are made without replacement. Also, you are interested in getting successes, rather than just the first success.. (a) ) Success is rolling a prime number, failure is rolling number that is not prime, and a trial is rolling a die, ) the probability of rolling a prime is p =.5, ) the observations, outcomes from rolls, are independent, and ) we are waiting for the first prime number. (b) The first five possible values of the random variable and their corresponding probabilities are shown in the table below 5 P() F() (c) A probability histogram is shown below (left). (d) The cumulative probabilities are shown in the third row of the table above, and a cumulative probability histogram is shown below (right)..5. Probability.. Cumulative probability Number of rolls Number of rolls 7 i.5.5 = =..5 (e) ( ) i=

10 The Binomial and Geometric Distributions. (a) ) Success is a defective hard drive, failure is a working hard drive, and a trial is a test of a hard drive, ) the probability of a defective hard drive is p =, ) the observations, results of the tests on different hard drives, are independent, and ) we are waiting for the first defective hard drive. The random variable of interest is = number of hard drives tested in order to find the first defective. (b) ( ) ( ) 5 P = 5 = 66 (c) The first four entries in the table for the pdf of are shown below.; P ( = x) for p = and =,, and =,, and 7. P(). 7. The probability of getting the first success on the fourth trial for parts (a), (c), and (d) in Exercise. is: (a) and (d) P( ) ( ) ( ) P = =.5.5 = 65, (c) P( = ) = 65, 5 5 = = (a) The random variable of interest is = number of flips required in order to get the first head. is a geometric random variable with p =.5. (b) The first five possible values of the random variable and their corresponding probabilities are shown in the table below 5 P() F() A probability histogram is shown below (left). (d) The cumulative probabilities are shown in the third row of the table above, and a cumulative probability histogram is shown below (right)..5. Probability.. Cumulative probability Number of rolls Number of rolls 7 > =. 5 =..6 (a) P( ). (b) P( ) P( ) > = = geometcdf(/, ).7 (a) The cumulative probabilities for the first values of are:.66667, 5556, 6,.5777,.5,.665,.7,.767,.6, and.. A cumulative probability histogram is shown below.

11 Chapter.. Cumulative probability Number of rolls (b) P( >) = ( /6) = (5/6) =.65 or P( ).5 =.65. (c) Using the calculator, geometcdf(/6,5) =.57 and geometcdf(/6,6) =.65, so the smallest positive integer k for which P( k) >. is 6.. Let = the number of applicants who need to be interviewed in order to find one who is fluent in Farsi. is a geometric random variable with p = % =. (a) The expected number of interviews in order to obtain the first success (applicant fluent in Farsi) is µ = 5 p = =. (b) P( > 5) = ( ) 5 = (.6) 5.6; P( > ) = (.6).5.. (a) We must assume that the shots are independent, and the probability of success is the same for each shot. A success is a missed shot, so the p =.. (b) The first success (miss) is the sixth shot, so P( = 6) = ( p) n- (p) = (.) 5. = 655. (c) P( 6) = P( > 6) = ( p ) 6 = (.) Using a calculator, P( 6) = geometcdf(., 6) (a) There are = possible outcomes, and only two of the possible outcomes (HHH and TTT) do not produce a winner. Thus, P(no winner) = / =.5. (b) P(winner) = P(no winner) =.5 =.75. (c) Let = number of rounds (tosses) until someone wins. ) Success is getting a winner, failure is not getting a winner, and a trial is one round (each person tosses a coin) of the game, ) the probability of success is.75, ) the observations are independent, and ) we are waiting for the first win. Thus, is a geometric random variable. (d) The first seven possible values of the random variable and their corresponding probabilities and cumulative probabilities are shown in the table below P() F() (e) P( ) = =.75. (f) P( > ) = (.5). (g) The expected number of rounds for someone to win is. p.75 tails, and enter the command randint (,, ) and press ENTER 5 times. In a simulation, we recorded the following frequencies: Freq. Relative Freq...

12 The Binomial and Geometric Distributions 5 The relative frequencies are not far away from the calculated probabilities of.75,.75, and 675 in part (d). Obviously, a larger number of trials would result in better agreement because the relative frequencies will converge to the corresponding probabilities..5 (a) Geometric: ) Success is selecting a red marble, failure is not selecting a red marble, and a trial is a selection from the jar, ) the probability of selecting a red marble is p = =.57, ) the observations, results of the marble selection, are independent because 5 7 the marble is placed back in the jar after the color is noted, and ) we are waiting for the first red marble. The random variable of interest is = number of marbles you must draw to find the first red marble. (b) The probability of getting a red marble on the second draw is P ( = ) = =. The probability of drawing a red marble by the second draw is 7 7 P ( ) = P ( = ) + P ( = ) = +.6. The probability that it takes more than draws to get a red marble is P ( > ) =.7. (c) Using TI- commands: 7 seq(,,,) L, geompdf(/7,l ) L and geomcdf(/7,l ) L [or cumsum(l ) L ]. The first ten possible values of the random variable and their corresponding probabilities and cumulative probabilities are shown in the table below P() F() (d) A probability histogram (left) and a cumulative probability histogram (right) are shown below..6. Probability.5.. Cumulative probability Number of marbles Number of marbles.5 (a) No. Since the marbles are being drawn without replacement and the population (the set of all marbles in the jar) is so small, the probability of getting a red marble is not independent from draw to draw. Also, a geometric variable measures the number of trials required to get the first success; here, we are looking for the number of trials required to get two successes. (b) No. Even though the results of the draws are now independent, the variable being measured is still not the geometric variable. This random variable has a distribution known as the negative binomial distribution. (c) The probability of getting a red marble on any draw is p = =.57. Let the digits,,, fl a red marble is drawn,, 5, 6 fl some other color 5 7

13 6 Chapter marble is drawn, and 7,, fl digit is disregarded. Start choosing random digits from Table B, or use the TI- command randint (,, ) repeatedly. After two digits in the set,,, have been chosen, stop the process and count the number of digits in the set {,,,,, 5, 6} that have been chosen up to that point; this count represents the observed value of. Repeat the process until the desired number of observations of has been obtained. Here are some sample simulations using the TI- (with R = red marble, O = other color marble, D = disregard): 7 = D R O R 6 = D R D O R 7 = etc. D D R R For repetitions, we recorded the following frequencies: Freq Relative Freq A simulated probability histogram for the repetitions is shown below. The simulated distribution is skewed to the right, just like the probability histogram in Exercise.5, but the two distributions are not the same. 6 5 Percent 5 6 Number of draws 7.5 Success is getting a correct answer. The random variable of interest is = number of questions Carla must answer until she gets one correct. The probability of success is p = /5 =. (all choices are equally likely to be selected). (b) P ( = 5) =... (c) ( ) (.) ( ) 5 P> = 6. (d) The first five possible values of the random variable and their corresponding probabilities are shown in the table below. 5 P()..6. (e) The expected number of questions Carla must answer to get one correct is µ = = (a) If success is having a son and the probability of success is p =.5, then the average number of children per family is µ = =. (b) The expected number of girls in this family is.5 µ =.5=. Alternatively, if the average number of children is, and the last child is a boy,

14 The Binomial and Geometric Distributions 7 then the average number of girls per family is =. (c) Let an even digit represent a boy, and an odd digit represent a girl. Read random digits until an even digit occurs. Count number of digits read. Repeat many times, and average the counts. Beginning on line in Table B and simulating 5 trials, the average number of children per family is.6, and the average number of girls is.6. These averages are very close to the expected values..55 Letting G = girl and B = boy, the outcomes are: {G, BG, BBG, BBBG, BBBB}. A success is having a girl. (b) The random variable can take on the values of,,, and. The multiplication rule for independent events can be used to obtain the probability distribution table for below. P() = = = = 6 6 Note that P ( ) =. (c) Let Y = number of children produced until first girl is born. Then Y is a geometric variable for Y = to but not for values greater than because the couple stops having children after. Note that BBBB is not included in the event Y=. The multiplication rule for independent events can be used to obtain the probability distribution table for Y below. Y P(Y) = = = 6 Note that this is not a valid distribution since PY ( ) <. The difficulty lies in the fact that one of the possible outcomes, BBBB, cannot be written in terms of Y. (d) If T is the total number of children in the family, then the probability distribution of T is shown in the table below. T P(T) = = + = = 6 The expected number of children for this couple is µ T = =.75. (e) P(T >.75) = P(T = ) =.5 or P(T >.75) = P(T = ) + P(T = ) + P(T = ) = =.5. (f) P(having a girl) = P(not having a girl) = P(BBBB) = 65 = Let,,,, fl Girl and 5, 6, 7,, fl Boy. Beginning with line in Table B, the simulated values are: B B G B G B G B G B B B G B G G B B G B B G G G G B G B B B B G G G B B G G B G B G B B G B B G G B G

15 Chapter The average number of children is 5/5 =., which is close to the expected value of Find the mean of 5 randomly generated observations of ; the number of children in the family. We can create a suitable string of random digits (say of length ) by using the command randint(,,) L. Let the digits to represent a boy and 5 to represent a girl. Scroll down the list and count the number of children until you get a girl number or boy numbers in a row, whichever comes first. The number you have counted is, the number of children in the family. Continue until you have 5 values of. The average for repetitions is 5 x = =., which is very close to the expected value of µ = (a) The table of probabilities p and expected values /p is shown below. : Y: (b) A scatterplot is shown below on the left... Expected value (Y) 6 Log(Y) Probability () Log() - -. p (c) From Chapter, the power law model is Y = a, we transform the data by taking the logarithm of both sides to obtain the linear model log( Y) = log( a) + plog( ). Thus, we need to compute the logarithms of and Y. (d) A scatterplot of the transformed data is shown above on the right. Notice, that the transformed data follow a linear pattern. (e) The correlation for the transformed data is approximately r =. (f) The estimated power function is Ŷ = =. (g) The power function illustrates the fact that the mean of a geometric random variable is equal to the reciprocal of the probability p of success: µ =. p CASE CLOSED! (a) The proportion of successes for all trials is / 56. The proportion of successes for high confidence is 55/65., for medium confidence is / =.5, and for low confidence is /75 =.. If Olof Jonsson is simply guessing, then we would expect his proportion of successes to be.5. Overall, he has done slightly better than expected, especially when he has high confidence. (b) Yes, this result does indicate evidence of psychic ability. Let = number of correct guesses in independent trials. is B(,.5). The probability of observing or more successes if Olof is simply guessing is P( ) = P( 7) = binomcdf(,.5, 7). =. There is a very small chance that Olof could

16 The Binomial and Geometric Distributions guess correctly or more times in trials. (Note: A formal hypothesis test of H : p=.5 versus H : p>.5 yields an approximate test statistic of Z =. and an approximate P-value of 5, so we have statistically significant evidence that Olof is doing better than guessing. Students will learn more about hypothesis testing procedures very soon.) (c) The probability of making a correct guess on the first draw is / =.5. If you are told that your first guess is incorrect, then the revised probability of making a correct guess on the second draw is / =.. If you are then told that your second guess is incorrect, then the revised probability of making a correct guess is / =.5. Finally, if you are told that your first three guesses are incorrect, then your revised probability of making a correct guess is / =. (d) The sum of the probabilities of guessing correctly on the first, second, third, and fourth draws without replacement is / + / + / + = / or about.. Thus, the proportion of correct guesses would be./ or about.5. The expected number of correct guesses without replacement is.5=5, which is much higher than the expected number of correct guesses without replacement,.5=7. If some of the computer selections were made without replacement, then the expected number of correct guesses would increase. (e) Let = number of runs in precognition mode. Since the computer randomly selected the mode (precognition or clairvoyance) with equal probability, is B(,.5). The probability of getting or fewer runs in precognition mode is P( ) = binomcdf(,.5, ) 7, somewhat unlikely..5 (a) Not binomial: the opinions of a husband and wife are not independent. (b) Not binomial: success is responding yes, failure is responding no, and a trial is the opinion expressed by the fraternity member selected. We have a fixed number of trail, n = 5, but these trial are not independent and the probability of success is not the same for each fraternity member..6 Let N be the number of households with or more cars. Then N has a binomial distribution with n = and p =.. (a) P( N = ) = (.) (.) = (.) 67. ( ) ( ) ( ) P N = P N = =... (b) The mean is µ = np =. =. and the standard deviation is np ( p) P( N ) P( N ) σ = =...56 households. (c) >. =.55 = 7..6 (a) The distribution of will be symmetric; the shape depends on the value of the probability of success. When p =.5, the distribution of will always be symmetric. (b) The values of and their corresponding probabilities and cumulative probabilities are shown in the table below P() F() A probability histogram (left) and a cumulative probability histogram (right) are shown below.

17 Chapter. Probability.5.5 Cumulative probability Number of brothers Number of brothers (c) P( = 7) = 75.6 Let = the result of the first toss ( = tail, = head) and Y = the number of trials to get the first head (). (a) A record of the 5 repetitions is shown below. Y Y Y Y Y 5 5 (b) Based on the results above, the probability of getting a head is /5 =.6, not a very good estimate. (c) A frequency table for the number of tosses needed to get the first head is shown below. Based on the frequency table, the probability that the first head appears on an oddnumbered toss is ( + + )/5 =.7. This estimate is not bad, since the theoretical probability is / or about Y Count Percent N= 5.6 Let = number of southerners out of who believe they have been healed by prayer. Then is a binomial random variable with n = and p = 6. (a) P( = ) = binompdf((, 6, ).65. (b) P(< <5) = P( ) = binomcdf(, 6, ) binomcdf(, 6, ).7.7 =.7 or P( 5) = binomcdf(, 6, 5) binomcdf(, 6, ) =, depending on your interpretation of between being exclusive or inclusive. (c) P( >5) = P( 5) =.. (d) P( <) = P( 7) = binomcdf(, 6, 7).

18 The Binomial and Geometric Distributions.6 Let = the number of schools out of who say they have a soft drink contract. is binomial with n = and p =.6. (a) P( = ) = binompdf(,.6, ) (b) P( ) = binomcdf(,.6, ) (c) P( ) = P( ) =. (d) P( ) = P( ) P( ) = binomcdf(,.6, ) binomcdf(,.6, ).57 =.576. (e) The random variable of interest is defined at the beginning of the solution and the probability distribution table is shown below. P() P() (f) The probability histogram for is shown below..5 Probability Number of schools with soft drink contracts.65 (a), the number of positive tests, has a binomial distribution with parameters n = and p =. (b) µ = np = = positive tests. (c) To use the Normal approximation, we need np and n ( p) both bigger than, and as we saw in (b), np =..66 Let = the number of customers who purchase strawberry frozen yogurt. Then is a binomial random variable with n = and p =.. The probability of observing 5 or more orders for strawberry frozen yogurt among customers is P( 5) = P( ) = binomcdf(,., ).67 =. Even though there is only about a.% chance of observing 5 or more customers who purchase strawberry frozen yogurt, these rare events do occur in the real world and in our simulations. The moral of the story is that the regularity that helps us to understand probability comes with a large number of repetitions and a large number of trials.

19 Chapter.67 is geometric with p =.5. (a) P( =) =.5. (b) P( ) = P( = ) + P( = ) + P( = ) =.5+(.5) - (.5) + (.5) - (.5).65. Alternatively, P( ) = P( > ) = (c) P( > ) = (.5) 76. (d) The expected number of times Roberto will have to go to the plate to get his first hit is µ = =.76, p.5 or just over at bats. (e) Use the commands: seq(,,,) L, geompdf(.5,l ) L and geomcdf(.5,l ) L. (f) A probability histrogram (left) and a cumulative probability histogram (right) are shown below..5.. Probability.5.5 Cumulative probability Number of at-bats Number of at -bats.6 (a) By the rule, the probability of any one observation falling within the interval µ σ to µ + σ is about.6. Let = the number of observations out of 5 that fall within this interval. Assuming that the observations are independent, is B(5,.6). Thus, P( = ) = binompdf (5,.6, ). (b) By the rule, 5% of all observations fall within the interval µ σ to µ + σ. Thus,.5% (half of 5%) of all observations will fall above µ + σ. Let = the number of observations that must be taken before we observe one falling above µ + σ. Then is geometric with p = 5. Thus, P( = ) = ( 5) 5 = (.75) 5.