AP STAT -- unit 4
A nonprofit organization plans to hold a raffle to raise funds for its operations. A total of 1,000 raffle tickets will be sold for $1.00 each. After all the tickets are sold, one ticket will be selected at random and its owner will receive $50.00. The expected value for the net gain for each ticket is -$0.95. What is the meaning of the expected value in this context? A The ticket owners lose an average of $0.05 per raffle ticket. B The ticket owners lose an average of $0.95 per raffle ticket. C Each ticket owner will lose $0.95 per raffle ticket. D A ticket owner would have to purchase 19 more tickets for the expected value of his or her net gain to increase to $0.00. E A ticket owner has a 95 percent chance of having a ticket that is not selected.
Answer B
A certain factory that manufactures office chairs has a quality control process to identify defective chairs. The binomial random variable DD represents the number of chairs in a sample of chairs that are defective. The mean of DD is 10 chairs and the standard deviation is 3 chairs. Based on the distribution of DD, which of the following would be an accurate interpretation of the value 0.1 ? A The total number of defective chairs made B The total number of non-defective chairs made C The relative frequency of the sample size to the population of chairs D The probability of identifying a non-defective chair E The probability of identifying a defective chair
Answer E Correct. The mean of 10 indicates np=10np=10, and the standard deviation of 3 indicates np(1−p)−−−−−−−−√=10(1−p)−−−−−−−−√=3np(1−p)=10(1−p)=3. Solving for pp gives p=0.1p=0.1, which is the probability of success, that is, identifying a defective chair.
At a sporting event, cheerleaders will throw 50 bundled T-shirts into the crowd. The T-shirt sizes consist of 10 small, 15 medium, and the remainder either large or extra large. Suppose Ana catches a T-shirt. What is the probability that she will catch a T-shirt that is not a size small? A 0.10 B 0.20 C 0.50 D 0.67 E 0.80
Answer E Correct. There are 50 total T-shirts and 10 are small, so 40 are not small. The probability is thus 4050=0.804050=0.80.
In a certain school, 17 percent of the students are enrolled in a psychology course, 28 percent are enrolled in a foreign language course, and 32 percent are enrolled in either a psychology course or a foreign language course or both. What is the probability that a student chosen at random from this school will be enrolled in both a foreign language course and a psychology course? A 0.45 B 0.32 C 0.20 D 0.13 E 0.05
Correct answer is D
FRQ check term - upcoming concert
For an upcoming concert, each customer may purchase up to 3 child tickets and 3 adult tickets. Let C be the number of child tickets purchased by a single customer. The probability distribution of the number of child tickets purchased by a single customer is given in the table below. Compute the mean and the standard deviation of C. Suppose the mean and the standard deviation for the number of adult tickets purchased by a single customer are 2 and 1.2, respectively. Assume that the numbers of child tickets and adult tickets purchased are independent random variables. Compute the mean and the standard deviation of the total number of adult and child tickets purchased by a single customer. Suppose each child ticket costs $15 and each adult ticket costs $25. Compute the mean and the standard deviation of the total amount spent per purchase. Overall [0-4] 0-4 points Solution Part (a): The mean of C is 0 x 0.4 + 1 x 0.3 + 2 x 0.2 + 3 x 0.1 = 1. The standard deviation of C is (0−1)2×0.4+(1−1)2×0.3+(2−1)2×0.2(3−1)2×0.1−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√=1(0−1)2×0.4+(1−1)2×0.3+(2−1)2×0.2(3−1)2×0.1=1 Part (b): Let T = C + A, where A is the total number of adult tickets purchased by a single customer, denote the total number of tickets purchased by a single customer. The mean of T is μT = μC + μA = 1 + 2 = 3 The standard deviation of T is σT=σ2C+σ2A−−−−−−−√=12+1.22−−−−−−−√=2.44−−−−√=1.562σT=σC2+σA2=12+1.22=2.44=1.562 Part (c): Let M = 15 × C + 25 × A denote the total amount of money spent per purchase. The mean of M is μM = 15μC + 25μA = 15 x 1 + 25 x 2 = $65. The standard deviation of M is σM=152σ2C+252σ2A−−−−−−−−−−−−√=225×12+625×1.22−−−−−−−−−−−−−−−−−√=1125−−−−√=$33.54
As a promotion, the first 50 customers who entered a certain store at a mall were asked to choose from one of two discounts. The first discount choice was 20% off all purchases made that day. The second discount choice was 10% off all purchases for the week. Of those who received the discounts, 28 chose the first discount and 22 chose the second discount. One customer will be selected at random from those who received a discount. Let F represent the event that the selected person chose the first discount, and let S represent the event that the selected person chose the second discount. Are F and S mutually exclusive events? A Yes, because P(F∩S)=0. B Yes, because P(F∩S)=0.12. C Yes, because P(F∩S)=1. D No, because P(F∩S)=0. E No, because P(F∩S)=1.
Answer A Correct. A person could choose only one discount. Because the events cannot occur at the same time, the events are mutually exclusive.
FRQ rocket super igniter failure rate
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FRQ antibiotic plans for ear infection
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FRQ car model customer satisfaction
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In the 1830s, land surveyors began to survey the land acquired in the Louisiana Purchase. Part of their task was to note the sizes of trees they encountered in their surveying. The table of data below is for bur oak trees measured during the survey. Which of the following differences in cumulative relative frequencies gives the proportion of trees that are 12 inches to 16 inches, inclusive, in diameter? A 0.615 - 0.325 B 0.615 - 0.473 C 0.726 - 0.325 D 0.726 - 0.473 E 0.731 - 0.325
Correct answer is C
According to 2015 census data, 42.7 percent of Colorado residents were born in Colorado. If a sample of 250 Colorado residents is selected at random, what is the standard deviation of the number of residents in the sample who were born in Colorado? A 6.75 B 7.82 C 10.33 D 11.97 E 61.17
Answer B Correct. Let random variable CC represent the number of people from a sample of 250 who were born in Colorado. The random variable has a binomial distribution. The standard deviation of a binomial random variable is np(1−p)−−−−−−−−√=250(0.427)(0.573)−−−−−−−−−−−−−−√≈7.82
Let SS represent the number of randomly selected adults in a community surveyed to find someone with a certain genetic trait. The random variable SS follows a geometric distribution with mean 4.66. Which of the following is a correct interpretation of the mean? A A value randomly selected from the distribution of SS is expected to be 4.66. B In repeated sampling from the distribution of SS, the average of the values will approach 4.66. C For a sample of values randomly selected from the distribution of SS, the average of the sample will be 4.66. D The probability is 0.66 that a value randomly selected from the distribution of SS will be close to the mean. E For a sample of values randomly selected from the distribution of SS, the average of the sample will vary from the population mean by no more than 4.66.
Answer B Correct. The mean of a probability distribution is the long-run average of the values selected from many repetitions.
A middle school chess club has 5 members: Adam, Bradley, Carol, Dave, and Ella. Two students from the club will be selected at random to participate in the county chess tournament. What is the probability that Adam and Ella will be selected? A 120120 B 110110 C 1818 D 1717 E 14
Answer B (1/10) Correct. The two players selected could be ABAB, ACAC, ADAD, AEAE, BCBC, BDBD, BEBE, CDCD, CECE, or DEDE, so there are 10 equally likely possible outcomes. Alternatively, there are (52)=10(52)=10outcomes when 2 people are chosen from 5 people. One of these outcomes is AEAE, so the probability is 110110.
One student from a high school will be selected at random. Let A be the event that the selected student is a student athlete, and let B be the event that the selected student drives to school. If P(A∩B)=0.08 and P(B|A)=0.25, what is the probability that the selected student will be a student athlete? A 0.02 B 0.17 C 0.32 D 0.33 E 3.13
Answer C Correct. By the multiplication rule, P(A∩B)=P(A)⋅P(B|A)P(A∩B)=P(A)⋅P(B|A), so 0.08=P(A)⋅0.250.08=P(A)⋅0.25. Solving for P(A)P(A) gives 0.32.
A fair die with its faces numbered from 1 to 6 will be rolled. Which of the following is the best interpretation of the probability that the number landing face up will be less than 3 ? A For many rolls of the die, the long-run relative frequency of a number less than 3 landing face up is 1313. B For many rolls of the die, the long-run relative frequency of a number less than 3 landing face up is 1212. C For many rolls of the die, the long-run relative frequency of a number less than 3 landing face up is 2323. D For three rolls of the die, a number less than 3 will land face up one time. E It will take three rolls of the die before a number less than 3 lands face up for the first time.
Answer A Correct. A roll of a fair die can result in 6 possible outcomes. There are 2 outcomes with a number less than 3 (1 and 2). In the long run, the relative frequency of rolling a number less than 3 is 26=1326=13.
The table shows data that were collected from people who attended a certain high school basketball game and indicates the team each person rooted for and whether each of these people purchased food during the game. A person who attended the game will be selected at random. Which of the following correctly interprets mutually exclusive events represented by the table? A Rooting for the home team and rooting for the away team B Rooting for the home team and purchasing food during the game C Rooting for the away team and purchasing food during the game D Rooting for the home team and not purchasing food during the game E Not rooting for the home team and not purchasing food during the game
Answer A Correct. Based on the table, each person either rooted for the home team or rooted for the away team. Because the events cannot occur at the same time, the events are mutually exclusive.
According to a recent survey, 31 percent of the residents of a certain state who are age 25 years or older have a bachelor's degree. A random sample of 50 residents of the state, age 25 years or older, will be selected. Let the random variable BB represent the number in the sample who have a bachelor's degree. What is the probability that BB will equal 40 ? A (5040)(0.31)40(0.69)10(5040)(0.31)40(0.69)10 B (5040)(0.69)40(0.31)10(5040)(0.69)40(0.31)10 C (4010)(0.31)40(0.69)10(4010)(0.31)40(0.69)10 D (4010)(0.69)40(0.31)10(4010)(0.69)40(0.31)10 E 40(0.31)50
Answer A Correct. Define success as selecting a person with the degree and failure as selecting a person without the degree. Random variable BB has a binomial distribution, with probability of success of 0.31 and probability of failure of 0.69. The probability of 40 out of 50 people having a degree is given by (5040)(0.31)40(0.69)10(5040)(0.31)40(0.69)10.
In a certain region, 94 percent of the people have a certain characteristic in their blood. Suppose a group of 45 people from the region are selected at random. Let the random variable BB represent the number of people in the sample without the characteristic. Random variable BB follows a binomial distribution with a mean of 2.7 people. Which of the following is the best interpretation of the mean? A For all groups of 45 people, the average number of people without the characteristic is 2.7. B Every group of 45 people will have 2.7 people with the characteristic. C Every group of 45 people will have 2.7 people without the characteristic. D On average, 2.7 people are selected until finding someone with the characteristic. E On average, 2.7 people are selected until finding someone without the characteristic.
Answer A Correct. For the random variable BB, the mean number of people without the characteristic is equal to 2.7 for all possible groups of 45 people.
Let the random variable Q represent the number of students who go to a certain teacher's office hour each day. The standard deviation of Q is 2.2. Which of the following is the best interpretation of the standard deviation? A On average, the number of students going to an office hour varies from the mean by about 2.2 students. B For a randomly selected office hour, the number of students who will go is 2.2. C For a randomly selected office hour, the number of students expected to go will vary from the mean by 2.2 students. D For a random selection of office hours, the average number of students expected to go is 2.2. E For a random selection of office hours, the average number of students expected to go will vary from the mean by 2.2 students.
Answer A Correct. The standard deviation is, roughly speaking, the average or typical amount of variability from the mean of the population
A high school science teacher has 78 students. Of those students, 35 are in the band and 32 are on a sports team. There are 16 students who are not in the band or on a sports team. One student from the 78 students will be selected at random. Let event B represent the event of selecting a student in the band, and let event S represent the event of selecting a student on a sports team. Are B and S mutually exclusive events? A No, because P(B∩S)=578P(B∩S)=578. B No, because P(B∩S)=4878P(B∩S)=4878. C No, because P(B∩S)=6278P(B∩S)=6278. D Yes, because P(B∩S)=578P(B∩S)=578. E Yes, because P(B∩S)=6278P(B∩S)=6278.
Answer A Correct. The sum of 35, 32, and 16 is 83, which is 5 greater than 78. So there must be 5 students who are both in the band and on a sports team. Because the events can occur at the same time, they are not mutually exclusive.
A student is applying to two different agencies for scholarships. Based on the student's academic record, the probability that the student will be awarded a scholarship from Agency A is 0.55 and the probability that the student will be awarded a scholarship from Agency B is 0.40. Furthermore, if the student is awarded a scholarship from Agency A, the probability that the student will be awarded a scholarship from Agency B is 0.60. What is the probability that the student will be awarded at least one of the two scholarships? A 0.60 B 0.62 C 0.71 D 0.73 E 0.95
Answer B Correct. The probability that a student will be awarded at least one of the two scholarships is P(A∪B)=P(A)+P(B)−P(A∩B)P(A∪B)=P(A)+P(B)−P(A∩B). To find P(A∩B)P(A∩B), use the formula for conditional probability, P(B∣∣A)=P(A∩B)P(A)P(B|A)=P(A∩B)P(A), then solve P(A∩B)=[P(A)⋅P(B|A)]=(0.55)(0.6)=0.33P(A∩B)=[P(A)⋅P(B|A)]=(0.55)(0.6)=0.33. Then P(A∪B)=0.55+0.40−0.33=0.62P(A∪B)=0.55+0.40−0.33=0.62.
For a certain dog breed, the number of puppies in a litter typically varies from 2 to 6. The following table shows the probability distribution of the random variable NN, where NN represents the number of puppies in a litter. Also shown are the squared deviations, or distances, from the expected value of 4.5 for the distribution. Number of puppies23456Squared deviation6.252.250.250.252.25Probability0.050.150.250.350.20 What is the variance of the distribution? A 1.12 B 1.25 C 1.41 D 1.58 E 2.25
Answer B Correct. The variance is the sum of the products of the squared deviations and the respective probabilities. In this case, 6.25(0.05)+2.25(0.15)+0.25(0.25)+0.25(0.35)+2.25(0.20)=1.256.25(0.05)+2.25(0.15)+0.25(0.25)+0.25(0.35)+2.25(0.20)=1.25.
At a large regional collegiate women's swim meet, an official records the time it takes each swimmer to swim 100 meters for all swimmers who compete in only one stroke category. The following table shows the mean times and corresponding standard deviations for the collegiate women at the swim meet for each of the four stroke categories. Stroke CategoryMean 100 meter TimeStandard DeviationBackstroke55.6 seconds0.70 secondsBreaststroke63.3 seconds0.92 secondsButterfly54.4 seconds0.94 secondsFreestyle50.2 seconds0.76 seconds For each of the 4 stroke categories, consider a random variable representing the time of a randomly selected swimmer in that category. What is the standard deviation of the sum of the 4 random variables? A 0.83 seconds B 1.67 seconds C 2.80 seconds D 3.32 seconds E 3.76 seconds
Answer B Correct. To find the standard deviation the variances are added. σ2X+Y+Z+W=σ2X+σ2Y+σ2Z+σ2W=0.702+0.922+0.942+0.762=2.7976σX+Y+Z+W2=σX2+σY2+σZ2+σW2=0.702+0.922+0.942+0.762=2.7976, so σ=2.7976−−−−−√≈1.67σ=2.7976≈1.67.
At a certain bakery, the price of each doughnut is $1.50. Let the random variable DD represent the number of doughnuts a typical customer purchases each day. The expected value and variance of the probability distribution of DD are 2.6 doughnuts and 3.6 (doughnuts)2(doughnuts)2, respectively. Let the random variable PPrepresent the price of the doughnuts that a typical customer purchases each day. Which of the following is the standard deviation, in dollars, of the probability distribution of PP ? A 1.5(3.6)1.5(3.6) B 1.53.6−−−√1.53.6 C 1.5(3.6)−−−−−−√1.5(3.6) D 1.5(2.6)1.5(2.6) E 1.52.6−−−√
Answer B (1.5sqr3.6) Correct. This is the price of each doughnut, $1.50, multiplied by the standard deviation of the number of doughnuts 3.6−−−√3.6.
For which of the following probability assignments are events A and B independent? A P(A∩Bc)=0.3P(A∩Bc)=0.3, P(A∩B)=0.12P(A∩B)=0.12, and P(Ac∩B)=0.4P(Ac∩B)=0.4. B P(A∩Bc)=0.3P(A∩Bc)=0.3, P(A∩B)=0.3P(A∩B)=0.3, and P(Ac∩B)=0.3P(Ac∩B)=0.3. C P(A∩Bc)=0.1P(A∩Bc)=0.1, P(A∩B)=0.1P(A∩B)=0.1, and P(Ac∩B)=0.4P(Ac∩B)=0.4. D P(A∩Bc)=0.3P(A∩Bc)=0.3, P(A∩B)=0.0P(A∩B)=0.0, and P(Ac∩B)=0.2P(Ac∩B)=0.2. E P(A∩Bc)=0.5P(A∩Bc)=0.5, P(A∩B)=0.1P(A∩B)=0.1, and P(Ac∩B)=0.4P(Ac∩B)=0.4.
Answer C Correct. Events AA and BB are independent if P(A∩B)=[P(A)][P(B)]P(A∩B)=[P(A)][P(B)]. Here, P(A)=P(A∩B)+P(A∩Bc)=0.1+0.1=0.2P(A)=P(A∩B)+P(A∩Bc)=0.1+0.1=0.2 and P(B)=P(B∩A)+P(B∩Ac)=0.1+0.4=0.5P(B)=P(B∩A)+P(B∩Ac)=0.1+0.4=0.5. Therefore, [P(A)][P(B)]=(0.2)(0.5)=0.1[P(A)][P(B)]=(0.2)(0.5)=0.1 and 0.1=P(A∩B)0.1=P(A∩B).
Let random variable U represent the field goal percentage (percentage of shots made) for players in a basketball league. The following table shows the probability distribution of the random variable U. Field Goal PercentageProbability0.30.100.40.450.50.300.60.100.70.05 Fatima claims that the distribution of U is uniform with a median of 0.4 field goal percentage. Is Fatima's claim supported by the table? A Yes, the distribution is uniform with a median of 0.4 field goal percentage. B No, the distribution is uniform with a median of 0.5 field goal percentage. C No, the distribution is skewed to the right with a median of 0.4 field goal percentage. D No, the distribution is skewed to the right with a median of 0.5 field goal percentage. E No, the distribution is skewed to the left with a median of 0.4 field goal percentage.
Answer C Correct. Fatima's claim is not supported by the table. A relative frequency histogram of the data will display most of the weight over the bars associated with a field goal percentage of 0.3 and 0.4 with a tail from 0.5 to 0.7, indicating a right skew. The median is equal to 0.4 field goal percentage, because 0.4 is the middle value of the distribution.
A player pays $15 to play a game in which a chip is randomly selected from a bag of chips. The bag contains 10 red chips, 4 blue chips, and 6 yellow chips. The player wins $5 if a red chip is selected, $10 if a blue chip is selected, and $20 if a yellow chip is selected. Let the random variable XX represent the amount won from the selection of the chip, and let the random variable WW represent the total amount won, where W=X−15W=X−15. What is the mean of WW ? A $10.50 B $4.50 C −$4.50−$4.50 D −$6.50−$6.50 E −$10.50
Answer C Correct. The expected payout is E(X)=1020($5)+420($10)+620($20)=$10.50E(X)=1020($5)+420($10)+620($20)=$10.50. After subtracting the $15 to play the game, the result is -$4.50 expected winnings.
The following table shows the probability distribution for the number of books a student typically buys at the annual book fair held at an elementary school. Number of Books01234567Probability0.350.200.150.100.070.080.040.01 Let the random variable BB represent the number of books a student buys at the next book fair. What is the expected value of BB ? A 0 B 1.00 C 1.79 D 3.50 E 28
Answer C Correct. The expected value of BB is the sum of the products of the number of books and their respective probabilities. In this case, 0(0.35)+1(0.20)+2(0.15)+3(0.10)+4(0.07)+5(0.08)+6(0.04)+7(0.01)=1.790(0.35)+1(0.20)+2(0.15)+3(0.10)+4(0.07)+5(0.08)+6(0.04)+7(0.01)=1.79.
Joslyn performed an experiment using a die with its faces numbered from 1 to 6. She rolled the die and recorded whether the 5 landed face up. She repeated the process many times and kept a cumulative record of the total number of rolls and the total number of 5s landing face up. The following table shows part of her record. Total Number of RollsTotal Number of 5s51102152203 Suppose Joslyn could roll the die 10,000 times and keep a record of the total number of 5s landing face up in the 10,000 rolls. What would such a record illustrate? A The conditional probability rule B The multiplication rule C The addition rule D The law of large numbers E The property of mutually exclusive events
Answer D Correct. As the number of repetitions increases, the relative frequencies of the 5 landing face up out of the total number of rolls will approach the theoretical probability of 1616.
A company is considering purchasing the mineral rights to two different mountains. The probability that it will purchase the mineral rights to the first mountain is 0.55. The probability that it will purchase the mineral rights to the second mountain is 0.4. Assuming the decisions to purchase the mineral rights to each mountain are made independently, what is the probability that it will purchase the mineral rights to exactly one of the two mountains? A 0.18 B 0.22 C 0.33 D 0.51 E 0.95
Answer D Correct. If FF represents the event that the mineral rights to the first mountain are purchased and SS represents the event that the mineral rights to the second mountain are purchased, then P(F∩Sc)+P(Fc∩S)=[P(F)][P(Sc)]+[P(Fc)][P(S)]=(0.55)(0.6)+(0.45)(0.4)=0.51P(F∩Sc)+P(Fc∩S)=[P(F)][P(Sc)]+[P(Fc)][P(S)]=(0.55)(0.6)+(0.45)(0.4)=0.51.
Let random variable Y represent the number of interviews conducted for job openings at a certain company. The following table shows the cumulative probability distribution of the discrete random variable Y. yP(Y≤y)5060.270.480.690.8101.0 Khaleed claims that the distribution of Y is skewed to the left with mean equal to 8 interviews. Is Khaleed's claim correct? A Yes, the distribution is skewed to the left with mean equal to 8 interviews. B No, the distribution is skewed to the left with mean greater than 8 interviews. C No, the distribution is skewed to the right with mean equal to 8 interviews. D No, the distribution is uniform with mean equal to 8 interviews. E No, the distribution is uniform with mean greater than 8 interviews.
Answer D Correct. Khaleed's claim is not correct. The probabilities shown in the table are cumulative; the individual probabilities for 6, 7, 8, 9, and 10 interviews are each 0.2, indicating a uniform distribution. Each value has the same weight, so the mean is the value in the center, which is 8 interviews.
According to a recent survey, 81 percent of adults in a certain state have graduated from high school. If 15 adults from the state are selected at random, what is the probability that 5 of them have not graduated from high school? A (2015)(0.19)15(0.81)5(2015)(0.19)15(0.81)5 B (105)(0.19)15(0.81)15(105)(0.19)15(0.81)15 C (105)(0.81)5(0.19)10(105)(0.81)5(0.19)10 D (155)(0.19)5(0.81)10(155)(0.19)5(0.81)10 E (155)(0.81)5(0.19)10
Answer D Correct. Let XX represent the number of adults selected who have not graduated from high school. Random variable XX has a binomial distribution. The probability of "success" is the probability that a student has not graduated from high school, or 0.19. So the probability that exactly 5 adults have not graduated from high school is P(X=5)=(155)(0.19)5(0.81)10
In a certain board game, a player rolls two fair six-sided dice until the player rolls doubles (where the value on each die is the same). The probability of rolling doubles with one roll of two fair six-sided dice is 1616. What is the probability that it takes three rolls until the player rolls doubles? A (16)3(16)3 B (56)3(56)3 C (16)(56)3(16)(56)3 D (16)(56)2(16)(56)2 E (56)(16)2(56)(16)2
Answer D Correct. Let the discrete random variable XX represent the number of rolls it takes until the player rolls doubles. The random variable has a geometric distribution, where P(X=3)=(16)(56)2P(X=3)=(16)(56)2.
In 2014, 85 percent of households in the United States had a computer. For a randomly selected sample of 200 households in 2014, let the random variable CC represent the number of households in the sample that had a computer. What are the mean and standard deviation of CC ? A The mean is 85 households, and the standard deviation is 0.36 household. B The mean is 144.5 households, and the standard deviation is 5.05 households. C The mean is 144.5 households, and the standard deviation is 13.04 households. D The mean is 170 households, and the standard deviation is 5.05 households. E The mean is 170 households, and the standard deviation is 0.36 household.
Answer D Correct. Random variable CC has a binomial distribution. The mean of the random variable is np=200(0.85)np=200(0.85), and the standard deviation of the random variable is np(1−p)−−−−−−−−√=200(0.85)(0.15)−−−−−−−−−−−−−√np(1−p)=200(0.85)(0.15).
Data were collected on the number of days per week that members visit a certain fitness center. The values varied from 0 to 7, and a distribution of relative frequencies for the values was created. Let the random variable X represent the number of days per week that a member visits. The mean of X is 3.12. Which of the following statements is the best interpretation of the mean? A Each member visits the fitness center 3 or 4 days per week. B The average number of days that each member visits the fitness center is 3.12 days per week. C Half the members visit the fitness center 3 days per week or less, and the other half visit 4 days per week or more. D The long-run average resulting from repeated sampling of members of the fitness center will approach 3.12 days per week. E For a random sample of members selected from the population, the average number of visits for the sample will be 3.12 days per week.
Answer D Correct. The mean of a random variable is the long-run average based on repeated sampling from the population.
The quality control manager at a factory records the number of equipment breakdowns each day. Let the random variable Y represent the number of breakdowns in one day. The standard deviation of Y is 0.28. Which of the following is the best interpretation of the standard deviation? A The number of breakdowns on a randomly selected day is expected to be 0.28. B The number of breakdowns on a randomly selected day will be 0.28 away from the mean. C The average number of breakdowns per day for a random sample of days is expected to be 0.28. D On average, the number of breakdowns per day varies from the mean by about 0.28. E The number of breakdowns per day for a random sample of days is expected to be 0.28 away from the mean.
Answer D Correct. The standard deviation of a random variable is, roughly speaking, the average or typical amount of variability from the mean for the population.
A company that ships crystal bowls claims that bowls arrive undamaged in 95 percent of the shipments. Let the random variable GG represent the number of shipments with undamaged bowls in 25 randomly selected shipments. Random variable GG follows a binomial distribution with a mean of 23.75 shipments and a standard deviation of approximately 1.09 shipments. Which of the following is the best interpretation of the mean? A Every shipment of 25 bowls will have 23.75 undamaged bowls. B Every shipment of 25 bowls will have 23.75 damaged bowls. C On average, the company receives 23.75 shipments before receiving the first shipment with a damaged bowl. D For all possible shipments of size 25, the average number of damaged shipments is equal to 23.75. E For all possible shipments of size 25, the average number of undamaged shipments is equal to 23.75.
Answer E Correct. For the random variable GG, the mean number of undamaged bowls is equal to 23.75 for all possible shipments of size 25.
Students at a local elementary school were shown a painting and asked which emotion—joy, happiness, love, or anger—they felt by looking at the painting. The students were classified by their age. The following table summarizes the responses of the students by age-group. JoyHappinessLoveAngerTotal6 to 8 years282040181069 to 11 years61258060226Total894512078332 One student from the school will be selected at random. What is the probability that the student is in the age-group of 6 to 8 years given that the selected student responded joy? A 8933289332 B 2833228332 C 2810628106 D 106332106332 E 2889
Answer E Correct. Of the 89 students who responded joy, 28 were in the age-group of 6 to 8 years, so 28892889 is the probability that a student will be in the age-group of 6 to 8 years given that the student responded joy.
The probability that a randomly selected visitor to a certain website will be asked to participate in an online survey is 0.40. Avery claims that for the next 5 visitors to the site, 2 will be asked to participate in the survey. Is Avery interpreting the probability correctly? A Yes, because 2 out of 5 is equal to 40%. B Yes, because participants in the survey are selected at random. C No, because there could be voluntary response bias. D No, because only 40% of all people will visit the site. E No, because 0.40 represents probability in the long run over many visits to the site
Answer E Correct. Over many visits, the long-run relative frequency of being asked to participate is 0.40, or 2 out of 5. Such a ratio does not apply to a single observation of 5 visits.
Let random variable R represent the the number of visitors to a certain museum during a given day. The following table shows the probability distribution of the random variable. Which of the following claims about the distribution of random variable R is best supported by the histogram? A The most likely number of visitors to the museum on a given day is between 450 and 500. B The mean number of visitors to the museum is much less than the median number of visitors. C The mean number of visitors to the museum is much greater than the median number of visitors. D On a given day, the number of visitors from 100 through 500 will occur with equal probabilities. E On a given day, it is equally likely for the museum to have less than 300 visitors as it is to have more than 300 visitors.
Answer E Correct. The distribution displayed in the histogram is symmetric, which indicates values below the center and above the center are equally likely.
Mateo plays on his school basketball team. From past history, he knows that his probability of making a basket on a free throw is 0.8. Suppose he wants to create a simulation using random numbers to estimate the probability of making at least 3 baskets on his next 5 free throw attempts. Which of the following assignments of the digits 0 to 9 could be used for the simulation? A Let the even digits represent making a basket and the odd digits represent not making a basket. B Let the digits 0 and 1 represent making a basket and the digits from 2 to 9 represent not making a basket. C Let the digits from 0 to 3 represent making a basket and the digits from 4 to 9 represent not making a basket. D Let the digits from 0 to 6 represent making a basket and the digits from 7 to 9 represent not making a basket. E Let the digits from 0 to 7 represent making a basket and the digits 8 and 9 represent not making a basket.
Answer E Correct. The eight digits 0 to 7 represent 80% of the 10 digits. This matches Mateo's probability of making a basket.
An online customer service department estimates that about 15 percent of callers have to wait more than 8 minutes to have their calls answered by a person. The department conducted a simulation of 1,000 trials to estimate the probabilities that a certain number of callers out of the next 10 callers will have to wait more than 8 minutes to have their calls answered. The simulation is shown in the following histogram. Based on the simulation, what is the probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to have their calls answered? A 0.150 B 0.190 C 0.474 D 0.526 E 0.810
Answer E Correct. The phrase "at most 2" means 0, 1, or 2. The sum of the relative frequencies for the bars corresponding to 0, 1, and 2 callers is 0.181+0.345+0.284=0.8100.181+0.345+0.284=0.810.
The seniors at three high schools were surveyed about their plans after graduation. The following table shows the responses, classified by high school. WorkMilitaryCollegeUndecidedTotalHigh School A994913863349High School B622615654298High School C833112471309Total244106418188956 One senior from the high schools will be selected at random. What is the probability that the senior selected will not be from High School B given that the senior responded with a choice other than college? A 156418156418 B 538956538956 C 262418262418 D 658956658956 E 396538
Answer E Correct. There were 956−418=538956−418=538 seniors who gave a response other than college. Of those, 99+49+63=21199+49+63=211 were from High School A and 83+31+71=18583+31+71=185 were from High School C. Thus 211+185538=396538211+185538=396538 is the probability that a senior selected will not be from High School B given that the senior gave a response other than college.
