Test 1 (ch. 5) Practice questions

(5.2#13)Consider a binomial experiment with n=6 trials where the probability of success on a single trial is p=0.85. (a)Find P(r< or=1). (b)InterpretationIf you conducted the experiment and got fewer than 2 successes, would you be surprised? Why?

(a) 0.000. (b) Yes, the probability of 0 or 1 success is 0.000 to three places after the decimal. It would be a very rare event to get fewer than 2 successes when the probability of success on a single trial is so high.

(5.1#17)Jim is a 60-year-old Anglo male in reasonably good health. He wants to take out a $50,000 term (i.e., straight death benefit) life insurance policy until he is 65. The policy will expire on his 65th birthday. The probability of death in a given year is provided by the Vital Statistics Section of the Statistical Abstract of the United States (116th edition) (LOOK AT TEST 1 PICTURE REFERENCE) 5.1 #17 Jim is applying to Big Rock Insurance Company for his term insurance policy. (a)What is the probability that Jim will die in his 60th year? Using this prob-ability and the $50,000 death benefit, what is the expected cost to Big Rock Insurance? (b)Repeat part (a) for years 61, 62, 63, and 64. What would be the total expected cost to Big Rock Insurance over the years 60 through 64? (c)InterpretationIf Big Rock Insurance wants to make a profit of $700 above the expected total cost paid out for Jim's death, how much should it charge for the policy? (d)InterpretationIf Big Rock Insurance Company charges $5000 for the policy, how much profit does the company expect to make?

(a) 0.01191; $595.50. (b) $646; $698; $751.50; $806.50; $3497.50 total. (c) $4197.50. (d) $1502.50.

(5.2#11)Consider a binomial experiment with n=7 trials where the probability of success on a single trial is p=0.30. (a)Find P(r=0). (b)Find P(r > or =1) by using the complement rule

(a) 0.082. (b) 0.918.

(5.1#15)The college hiking club is having a fundraiser to buy new equipment for fall and winter outings. The club is selling Chinese fortune cookies at a price of $1 per cookie. Each cookie contains a piece of paper with a different number written on it. A random drawing will determine which number is the winner of a dinner for two at a local Chinese restaurant. The dinner is valued at $35. Since the fortune cookies were donated to the club, we can ignore the cost of the cookies. The club sold 719 cookies before the drawing. (a)Lisa bought 15 cookies. What is the probability she will win the dinner for two? What is the probability she will not win?(b)InterpretationLisa's expected earnings can be found by multiplying the value of the dinner by the probability that she will win. What are Lisa's expected earnings? How much did she effectively contribute to the hiking club?

(a) 15/719; 704/719. (b) $0.73; $14.27.

(5.2#9)According to the college registrar's office, 40% of stu-dents enrolled in an introductory statistics class this semester are freshmen, 25% are sophomores, 15% are juniors, and 20% are seniors. You want to determine the probability that in a random sample of five students enrolled in introductory statistics this semester, exactly two are freshmen. (a)Describe a trial. Can we model a trial as having only two outcomes? If so, what is success? What is failure? What is the probability of success? (b)We are sampling without replacement. If only 30 students are enrolled in introductory statistics this semester, is it appropriate to model 5 trials as inde-pendent, with the same probability of success on each trial? Explain. What other probability distribution would be more appropriate in this setting?

(a) A trial consists of looking at the class status of a student enrolled in introductory statistics. Two outcomes are "fresh-man" and "not freshman." Success is freshman status; failure is any other class status. P(success) =0.40. (b) Trials are not independent. With a population of only 30 students, in 5 trials without replacement, the probability of success rounded to the nearest hundredth changes for the later trials. Use the hypergeometric distribution for this situation.

(5.1#1)Which of the following are continuous variables, and which are discrete? (a)Number of traffic fatalities per year in the state of Florida (b)Distance a golf ball travels after being hit with a driver (c)Time required to drive from home to college on any given day (d)Number of ships in Pearl Harbor on any given day (e)Your weight before breakfast each morning

(a) Discrete. (b) Continuous. (c) Continuous. (d) Discrete. (e) Continuous

(5.1#13)The following data are based on information taken from Daily Creel Summary, published by the Paiute Indian Nation, Pyramid Lake, Nevada. Movie stars and U.S. presidents have fished Pyramid Lake. It is one of the best places in the lower 48 states to catch trophy cutthroat trout. In this table, x5number of fish caught in a 6-hour period. The percentage data are the percentages of fishermen who catch x fish in a 6-hour period while fishing from shore. (LOOK AT TEST 1 PICTURE REFERENCE) 5.1 #13 (a)Convert the percentages to probabilities and make a histogram of the probability distribution. (b)Find the probability that a fisherman selected at random fishing from shore catches one or more fish in a 6-hour period. (c)Find the probability that a fisherman selected at random fishing from shore catches two or more fish in a 6-hour period. (d)Compute m, the expected value of the number of fish caught per fisher-man in a 6-hour period (round 4 or more to 4). (e)Compute s, the standard deviation of the number of fish caught per fisher-man in a 6-hour period (round 4 or more to 4).

(a) LOOK AT BACK BOOK (b) 0.56. (c) 0.20. (d) 0.82. (e) 0.899

(5.2#7)In an experiment, there are n independent trials. For each trial, there are three outcomes, A, B, and C. For each trial, the probability of outcome A is 0.40; the probability of outcome B is 0.50; and the probability of outcome C is 0.10. Suppose there are 10 trials. (a)Can we use the binomial experiment model to determine the probability of four outcomes of type A, five of type B, and one of type C? Explain. (b)Can we use the binomial experiment model to determine the probabil-ity of four outcomes of type A and six outcomes that are not of type A? Explain. What is the probability of success on each trial?

(a) No. A binomial probability model applies to only two outcomes per trial. (b) Yes. Assign outcome A to "success" and outcomes B and C to "failure." p= 0.40

(5.3#25)Innocent until proven guilty? In Japanese criminal trials, about 95% of the defendants are found guilty. In the United States, about 60% of the defendants are found guilty in criminal trials (Source: The Book of Risks, by Larry Laudan, John Wiley and Sons). Suppose you are a news reporter following seven criminal trials. (a)If the trials were in Japan, what is the probability that all the defendants would be found guilty? What is this probability if the trials were in the United States? (b)Of the seven trials, what is the expected number of guilty verdicts in Japan? What is the expected number in the United States? What is the standard deviation in each case? (c)Quota Problem As a U.S. news reporter, how many trials n would you need to cover to be at least 99% sure of two or more convictions? How many trials n would you need if you covered trials in Japan?

(a) P(r=7 guilty in U.S.) = 0.028; P(r=7 guilty in Japan) =0.698. (b) For guilty in Japan, m= 6.65; s<0.58; for guilty in U.S., m= 4.2; s<1.30. (c) To be 99% sure of at least two guilty convictions in the U.S., look at n=8 trials. To be 99% sure of at least two guilty convictions in Japan, look at n=3 trials

(5.3#15)USA Today reported that about 20% of all people in the United States are illiterate. Suppose you interview seven people at random off a city street. (a)Make a histogram showing the probability distribution of the number of illiterate people out of the seven people in the sample. (b)Find the mean and standard deviation of this probability distribution. Find the expected number of people in this sample who are illiterate. (c)Quota Problem How many people would you need to interview to be 98% sure that at least seven of these people can read and write (are not illiterate)?

(a) SEE BACK BOOK (b) m= 1.4; s<1.058. (c) n=12. Note that n=12 gives P(r>or= 7) =0.98, where success =literate and p= 0.80.

(5.3#11)In Hawaii, January is a favorite month for surfing since 60% of the days have a surf of at least 6 feet (Reference: Hawaii Data Book, Robert C. Schmitt). You work day shifts in a Honolulu hospital emergency room. At the beginning of each month you select your days off, and you pick 7 days at random in January to go surfing. Let r be the number of days the surf is at least 6 feet. (a)Make a histogram of the probability distribution of r. (b)What is the probability of getting 5 or more days when the surf is at least 6 feet? (c)What is the probability of getting fewer than 3 days when the surf is at least 6 feet? (d)What is the expected number of days when the surf will be at least 6 feet? (e)What is the standard deviation of the r-probability distribution? (f)Interpretation Can you be fairly confident that the surf will be at least 6 feet high on one of your days off? Explain.

(a) SEE BACK OF BOOK (b) P(r > 5) = 0.420. (c) P(r < 3) = 0.096. (d) m = 4.2. (e) s ≈ 1.296. (f ) Yes, the probability of at least 1 day with surf of at least 6 feet is 0.002, and the expected number of such days is 4

(5.3#13)Old Friends Information Service is a California company that is in the business of finding addresses of long-lost friends. Old Friends claims to have a 70% success rate (Source: TheWall Street Journal). Suppose that you have the names of six friends for whom you have no addresses and decide to use Old Friends to track them. (a)Make a histogram showing the probability of r=0 to 6 friends for whom an address will be found. (b)Find the mean and standard deviation of this probability distribution. What is the expected number of friends for whom addresses will be found? (c)Quota Problem How many names would you have to submit to be 97% sure that at least two addresses will be found?

(a) SEE BACK OF BOOK (b) m= 4.2; s<1.122. (c) n=5. Note that n=5 gives P(r>OR= 2) = 0.97

(5.3#9)Critical ThinkingConsider a binomial distribution with n5 10 trials and the probability of success on a single trial p5 0.85. (a)Is the distribution skewed left, skewed right, or symmetric? (b)Compute the expected number of successes in 10 trials. (c)Given the high probability of success p on a single trial, would you expect P(r<or= 3) to be very high or very low? Explain. (d)Given the high probability of success p on a single trial, would you expect P(r>or= 8) to be very high or very low? Explain.

(a) Skewed left. (b) m= 8.5. (c) Very low; the expected number of successes in 10 trials is 8.5 and p is so high it would unusual to have so few successes in 10 trials. (d) Very high; the expected number of successes in 10 trials is 8.5 and p is so high it would be common to have 8 or more successes in 10 trials

(5.3#5) Consider a binomial distribution of 200 trials with ex-pected value 80 and standard deviation of about 6.9. Use the criterion that it is unusual to have data values more than 2.5 standard deviations above the mean or 2.5 standard deviations below the mean to answer the following questions. (a)Would it be unusual to have more than 120 successes out of 200 trials? Explain. (b)Would it be unusual to have fewer than 40 successes out of 200 trials? Explain. (c)Would it be unusual to have from 70 to 90 successes out of 200 trials? Explain.

(a) Yes, 120 is more than 2.5 standard deviations above the expected value. (b) Yes, 40 is less than 2.5 standard deviations below the expected value. (c) No, 70 to 90 successes is within 2.5 standard deviations of the expected value

(5.1#11)In the following table, income units are in thousands of dollars, and each interval goes up to but does not include the given high value. The midpoints are given to the nearest thousand dollars. (LOOK AT TEST 1 PICTURE REFERENCE) 5.1 #11 (a)Using the income midpoints x and the percent of super shoppers, do we have a valid probability distribution? Explain. (b)Use a histogram to graph the probability distribution of part (a). (c)Compute the expected income m of a super shopper. (d)Compute the standard deviation s for the income of super shoppers.

(a) Yes, events are distinct and probabilities total 1. (b) SEE ONLINE ANSWER (c) 32.3 thousand dollars. (d) 16.12 thousand dollars.

(5.1#3)Consider each distribution. Determine if it is a valid probability distribution or not, and explain your answer. (LOOK AT TEST 1 PICTURE REFERENCE) 5.1 #3

(a) Yes. (b) No; probabilities total more than 1.

(5.2#5)Suppose you are a hospital manager and have been told that there is no need to worry that respirator monitoring equipment might fail because the probability any one monitor will fail is only 0.01. The hospital has 20 such moni-tors and they work independently. Should you be more concerned about the prob-ability that exactly one of the 20 monitors fails, or that at least one fails? Explain.

Any monitor failure might endanger patient safely, so you should be concerned about the probability of at least one failure, not just exactly one failure.

(5.1#7)Consider the probability distribution shown in Problem 3(a). Compute the expected value and the standard deviation of the distribution (LOOK AT TEST 1 PICTURE REFERENCE) 5.1 #3

Expected value < 0.9. s< 0.6245.

(5.2#15)A fair quarter is flipped three times. For each of the following probabilities, use the formula for the binomial distribu-tion and a calculator to compute the requested probability. Next, look up the probability in Table 3 of Appendix II and compare the table result with the computed result. (a)Find the probability of getting exactly three heads. (b)Find the probability of getting exactly two heads. (c)Find the probability of getting two or more heads. (d)Find the probability of getting exactly three tails.

LOOK IN BACK OF BOOK FOR HOW (a) 0.125 (b) 0.375 (c) 0.500 (d) 0.125

(5.2#21)Aldrich Ames is a convicted traitor who leaked American secrets to a foreign power. Yet Ames took routine lie detec-tor tests and each time passed them. How can this be done? Recognizing control questions, employing unusual breathing patterns, biting one's tongue at the right time, pressing one's toes hard to the floor, and count-ing backward by 7 are countermeasures that are difficult to detect but can change the results of a polygraph examination (Source: Lies! Lies!! Lies!!! The Psychology of Deceit, by C. V. Ford, professor of psychiatry, University of Alabama). In fact, it is reported in Professor Ford's book that after only 20 minutes of instruction by "Buzz" Fay (a prison inmate), 85% of those trained were able to pass the polygraph examination even when guilty of a crime. Suppose that a random sample of nine students (in a psychology laboratory) are told a "secret" and then given instructions on how to pass the polygraph examination without revealing their knowledge of the secret. What is the probability that (a)all the students are able to pass the polygraph examination?(b)more than half the students are able to pass the polygraph examination? (c)no more than four of the students are able to pass the polygraph examination? (d)all the students fail the polygraph examination?

LOOK IN BACK OF BOOK FOR HOW (a) 0.232 (b) 0.995 (c) 0.005 or 0.006 (d) 0.000

(5.2#29)In the western United States, there are many dry-land wheat farms that depend on winter snow and spring rain to produce good crops. About 65% of the years, there is enough moisture to produce a good wheat crop, depending on the region (Reference: Agricultural Statistics, U.S. Department of Agriculture). (a)Let r be a random variable that represents the number of good wheat crops in n58 years. Suppose the Zimmer farm has reason to believe that at least 4 out of 8 years will be good. However, they need at least 6 good years out of 8 to survive financially. Compute the probability that the Zimmers will get at least 6 good years out of 8, given what they believe is true; that is, compute P(6<or=r | 4<or=r). See part (d) for a hint. (b)Let r be a random variable that represents the number of good wheat crops in n=10 years. Suppose the Montoya farm has reason to believe that at least 6 out of 10 years will be good. However, they need at least 8 good years out of 10 to survive financially. Compute the probability that the Montoyas will get at least 8 good years out of 10, given what they believe is true; that is, compute P(8<or=r | 6<or=r).

LOOK IN BACK OF BOOK FOR HOW (a) 0.478 (b) 0.348

(5.2#19)Sociologists say that 90% of married women claim that their husband's mother is the biggest bone of contention in their marriages (sex and money are lower-rated areas of contention). (See the source in Problem 18.) Suppose that six married women are having coffee together one morning. What is the probability that (a)all of them dislike their mother-in-law? (b)none of them dislike their mother-in-law? (c)at least four of them dislike their mother-in-law? (d)no more than three of them dislike their mother-in-law?

LOOK IN BACK OF BOOK FOR HOW (a) 0.531 (b) 0.000 (c) 0.983 (d) 0.017 or 0.016

(5.2#23)Approximately 75% of all marketing personnel are extroverts, whereas about 60% of all computer programmers are introverts (Source: A Guide to the Development and Use of the Myers-Briggs Type Indicator, by Myers and McCaulley). (a)At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts? What is the probability that 5 or more are extroverts? What is the probability that all are extroverts? (b)In a group of 5 computer programmers, what is the probability that none are introverts? What is the probability that 3 or more are introverts? What is the probability that all are introverts?

LOOK IN BACK OF BOOK FOR HOW (a) P(r>or=10)= 0.851 P(r>or=5)= 0.999 P(r=15)= 0.013 (b) P(r=0)= 0.010 P(r>or=3)= 0.683 P(r=5)= 0.078

(5.2#17)The following is based on information taken from The Wolf in the Southwest: The Making of an Endangered Species, edited by David Brown (University of Arizona Press). Before 1918, approximately 55% of the wolves in the New Mexico and Arizona region were male, and 45% were female. However, cattle ranchers in this area have made a determined effort to exterminate wolves. From 1918 to the present, approximately 70% of wolves in the region are male, and 30% are female. Biologists suspect that male wolves are more likely than females to return to an area where the population has been greatly reduced. (a)Before 1918, in a random sample of 12 wolves spotted in the region, what was the probability that 6 or more were male? What was the probability that 6 or more were female? What was the probability that fewer than 4 were female? (b)Answer part (a) for the period from 1918 to the present.

LOOK IN BACK OF BOOK FOR HOW (a) p(r>or=6)= 0.740 p(r<or=6)= 0.473 p(r.8)= 0.135 (b) p(r>or=6)= 0.961 p(r<or=6)= 0.117 p(r.8)= 0.493

(5.3#1)What does the expected value of a binomial distribution with n trials tell you?

The average number of successes.

(5.2#1)What does the random variable for a binomial experi-ment of n trials measure?

The random variable measures the number of successes out of n trials. This text uses the letter r for the random variable.

(5.2#3)For a binomial experiment, how many outcomes are possible for each trial? What are the possible outcomes?

Two outcomes, success or failure.

(5.3#3)Consider a binomial experiment with n58 trials and p50.20. (a)Find the expected value and the standard deviation of the distribution. (b)Interpretation Would it be unusual to obtain 5 or more successes? Explain. Confirm your answer by looking at the binomial probability distribution table

a) m= 1.6; s=1.13. (b) Yes, 5 successes is more than 2.5s above the expected value. P(r>or=5)= 0.010

(5.3#23)Does crime pay? The FBI Standard Survey of Crimes shows that for about 80% of all property crimes (burglary, larceny, car theft, etc.), the criminals are never found and the case is never solved (Source: TrueOdds, by James Walsh, Merrit Publishing). Suppose a neighborhood district in a large city suffers repeated property crimes, not always perpetrated by the same criminals. The police are investigating six property crime cases in this district. (a)What is the probability that none of the crimes will ever be solved? (b)What is the probability that at least one crime will be solved? (c)What is the expected number of crimes that will be solved? What is the standard deviation? (d)Quota Problem How many property crimes n must the police investigate before they can be at least 90% sure of solving one or more cases?

n= 6; p=0.80 do not solve; p=0.20 solve. (a) P(r=6 not solved) =0.262. (b) P(r>or= 1 solved)=0.738. (c) For solving crime, m= 1.2; s<0.98. (d) To be 90% sure of solving one or more crimes, investigate n=11 crimes