Elementary Statistics Picturing the World 7th Edition by Ron Larson - Test Bank

Elementary Statistics Picturing the World 7th Edition by Ron Larson – Test Bank

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ExamName___________________________________MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.1)Find the area of the indicated region under the standard normal curve.A)0.0968B)0.4032C)0.9032D)0.0823Answer:A2)Find the area of the indicated region under the standard normal curve.A)0.0968B)0.9032C)0.0823D)0.9177Answer:B3)Find the area of the indicated region under the standard normal curve.A)0.3438B)0.309C)1.309D)0.6562Answer:D4)Find the area of the indicated region under the standard normal curve.A)0.1504B)0.0212C)0.1292D)0.8489Answer:A1
5)Find the area under the standard normal curve to the left of z=1.5.A)0.7612B)0.0668C)0.5199D)0.9332Answer:D6)Find the area under the standard normal curve to the left of z=1.25.A)0.8944B)0.2318C)0.7682D)0.1056Answer:A7)Find the area under the standard normal curve to the right of z=1.A)0.1587B)0.5398C)0.1397D)0.8413Answer:A8)Find the area under the standard normal curve to the right of z=-1.25.A)0.6978B)0.7193C)0.8944D)0.5843Answer:C9)Find the area under the standard normal curve between z=0 and z=3.A)0.0010B)0.4987C)0.9987D)0.4641Answer:B10)Find the area under the standard normal curve between z=1 and z=2.A)0.2139B)0.1359C)0.8413D)0.5398Answer:B11)Find the area under the standard normal curve between z=-1.5 and z=2.5.A)0.7182B)0.6312C)0.9270D)0.9831Answer:C12)Find the area under the standard normal curve between z=1.5 and z=2.5.A)0.9816B)0.9938C)0.9332D)0.0606Answer:D13)Find the area under the standard normal curve between z=-1.25 and z=1.25.A)0.8817B)0.2112C)0.6412D)0.7888Answer:D14)Find the sum of the areas under the standard normal curve to the left of z=-1.25 and to the right ofz=1.25.A)0.3944B)0.1056C)0.2112D)0.7888Answer:C2
15)Determine whether the graph could represent a variable with a normal distribution. Explain your reasoning.A)No, the graph is skewed left.B)No, the graph is not symmetric.C)Yes, the graph fulfills the properties of the normal distributionD)No, the graph crosses the x-axis.Answer:C16)Determine whether the graph could represent a variable with a normal distribution. Explain your reasoning.A)No, the graph crosses the x-axis.B)Yes, the graph fulfills the properties of the normal distribution.C)No, the graph is skewed right.D)No, the graph is not symmetric.Answer:A3
17)Determine whether the graph could represent a variable with a normal distribution. Explain your reasoning.A)No, the graph crosses the x-axis.B)No, the graph is skewed right.C)No, the graph is skewed left.D)Yes, the graph fulfills the properties of the normal distributionAnswer:B4
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.18)You are performing a study about the weight of preschoolers. A previous study found the weights to be normallydistributed with a mean of 30 and a standard deviation of 4. You randomly sample 30 preschool children and find theirweights to be as follows.25252626.5272727.5282828.52929303030.531313232.532.533333434.5353537373838a) Draw a histogram to display these data. Use seven classes. Do the weights appear to be normallydistributed? Explain.b) Find the mean and standard deviation of your sample.c) Is there a high probability that the mean and standard deviation of your sample are consistentwith those found in previous studies? Explain your reasoning.Answer:(a)It is not reasonable to assume that the heights are normally distributed since the histogram is skewed.(b)μ=31,Η=3.86(c) Yes. The mean and standard deviation are close.19)The SAT is an exam used by colleges and universities to evaluate undergraduate applicants. The test scores arenormally distributed. In a recent year, the mean test score was1481and the standard deviation was292. The testscores of four students selected at random are 1930, 1340, 2150, and 1450.a) Find the z-scores that correspond to each valueb) Determine whether any of the values are unusual.Answer:a) x=1930zY1.54x=1340zY-0.48x=2150zY2.29x=1450zY-0.11b) x=2150 is unusual because its corresponding z-score (2.29) lies more than 2 standard deviations fromthe mean.5
20)The ACT is an exam used by colleges and universities to evaluate undergraduate applicants. The test scores arenormally distributed. In a recent year, the mean test score was22.4and the standard deviation was5.2. The testscores of four students selected at random are 16, 24, 9, and 35.a) Find the z-scores that correspond to each value.b) Determine whether any of the values are unusual.Answer:a) x=16zY-1.23x=24zY0.31x=9zY-2.58x=35zY2.42b) x=9 is unusual because its corresponding z-score (-2.58) lies more than 2 standard deviations from themean.x=35 is unusual because its corresponding z-score (2.42) lies more than 2 standard deviations from themean.MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Find the probability of z occurring in the indicated region.21)0 1.82A)0.4656B)0.9656C)0.9772D)0.0344Answer:B22)-0.590A)0.2776B)0.7224C)0.1894D)0.2224Answer:A6
23)-1.330A)0.9082B)0.0668C)0.9332D)0.0918Answer:A24)01.75A)0.0401B)0.0228C)0.0668D)0.9599Answer:A7
25)-2 03A)0.0456B)0.0228C)0.9772D)0.9544Answer:C26)01.50A)0.4332B)0.9332C)0.5668D)0.0668Answer:AProvide an appropriate response.27)Use the standard normal distribution to find P(0<z<2.25).A)0.4878B)0.7888C)0.5122D)0.8817Answer:A28)Use the standard normal distribution to find P(-2.25<z<0).A)0.0122B)0.5122C)0.6831D)0.4878Answer:D29)Use the standard normal distribution to find P(-2.25<z<1.25).A)0.8822B)0.4878C)0.0122D)0.8944Answer:A8
30)Use the standard normal distribution to find P(-2.50<z<1.50).A)0.8822B)0.6167C)0.5496D)0.9270Answer:D31)Use the standard normal distribution to find P(z<-2.33 or z>2.33).A)0.0198B)0.7888C)0.0606D)0.9802Answer:A32)For the standard normal curve, find the z-score that corresponds to the third quartile.A)-0.23B)-0.67C)0.67D)0.77Answer:C33)For the standard normal curve, find the z-score that corresponds to the first quartile.A)0.67B)-0.23C)-0.67D)0.77Answer:C34)For the standard normal curve, find the z-score that corresponds to the first decile.A)-1.28B)0.16C)-2.33D)1.28Answer:A35)Assume that the random variable X is normally distributed, with meanμ=70and standard deviationΗ=16.Compute the probability P(X<90).A)0.8849B)0.8944C)0.1056D)0.9015Answer:B36)Assume that the random variable X is normally distributed, with meanμ=100and standard deviationΗ=15.Compute the probability P(X>112).A)0.2420B)0.1977C)0.2119D)0.7881Answer:C37)Assume that the random variable X is normally distributed, with meanμ=90and standard deviationΗ=12.Compute the probability P(57<X<105).A)0.8914B)0.7888C)0.8819D)0.8944Answer:AProvide an appropriate response. Use the Standard Normal Table to find the probability.38)IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. An individual’s IQscore is found to be 110. Find the z-score corresponding to this value.A)-0.67B)1.33C)-1.33D)0.67Answer:D39)IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. An individual’s IQscore is found to be 90. Find the z-score corresponding to this value.A)-1.33B)-0.67C)1.33D)0.67Answer:B40)IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. An individual’s IQscore is found to be 120. Find the z-score corresponding to this value.A)0.67B)-1.33C)-0.67D)1.33Answer:D9
41)IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the IQ score thatcorresponds to a z-score of 1.96.A)115.6B)122.4C)132.1D)129.4Answer:D42)IQ test scores are normally distributed with a mean of105and a standard deviation of13. An individual’s IQscore is found to be126. Find the z-score corresponding to this value.A)0.62B)-1.62C)-0.62D)1.62Answer:D43)IQ test scores are normally distributed with a mean of105and a standard deviation of17. An individual’s IQscore is found to be112. Find the z-score corresponding to this value.A)-0.41B)2.43C)-2.43D)0.41Answer:D44)The lengths of pregnancies of humans are normally distributed with a mean of 268 days and a standarddeviation of 15 days. Find the probability of a pregnancy lasting more than 300 days.A)0.9834B)0.2375C)0.3189D)0.0166Answer:D45)The lengths of pregnancies of humans are normally distributed with a mean of 268 days and a standarddeviation of 15 days. Find the probability of a pregnancy lasting less than 250 days.A)0.1151B)0.0606C)0.1591D)0.0066Answer:A46)The distribution of cholesterol levels in teenage boys is approximately normal withμ=170andΗ=30(Source:U.S. National Center for Health Statistics). Levels above 200 warrant attention. Find the probability that ateenage boy has a cholesterol level greater than 200.A)0.3419B)0.8413C)0.1587D)0.2138Answer:C47)The distribution of cholesterol levels in teenage boys is approximately normal withμ=170andΗ=30(Source:U.S. National Center for Health Statistics). Levels above 200 warrant attention. Find the probability that ateenage boy has a cholesterol level greater than 225.A)0.0606B)0.0718C)0.0336D)0.0012Answer:C48)An airline knows from experience that the distribution of the number of suitcases that get lost each week on acertain route is approximately normal withμ=15.5 andΗ=3.6. What is the probability that during a givenweek the airline will lose less than 20 suitcases?A)0.8944B)0.3944C)0.4040D)0.1056Answer:A49)An airline knows from experience that the distribution of the number of suitcases that get lost each week on acertain route is approximately normal withμ=15.5 andΗ=3.6. What is the probability that during a givenweek the airline will lose more than 20 suitcases?A)0.1056B)0.4040C)0.3944D)0.8944Answer:A10
50)An airline knows from experience that the distribution of the number of suitcases that get lost each week on acertain route is approximately normal withμ=15.5 andΗ=3.6. What is the probability that during a givenweek the airline will lose between 10 and 20 suitcases?A)0.1056B)0.3944C)0.8314D)0.4040Answer:C51)Assume that the salaries of elementary school teachers in the United States are normally distributed with amean of $32,000 and a standard deviation of $3000. If a teacher is selected at random, find the probability thathe or she makes more than $36,000.A)0.4040B)0.9082C)0.0918D)0.1056Answer:C52)Assume that the salaries of elementary school teachers in the United States are normally distributed with amean of $32,000 and a standard deviation of $3000. If a teacher is selected at random, find the probability thathe or she makes less than $28,000.A)0.9981B)0.2113C)0.9827D)0.0918Answer:D53)Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standarddeviation of 2.5 inches. The cheerleaders for a local professional basketball team must be between 65.5 and 68.0inches. If a woman is randomly selected, what is the probability that her height is between 65.5 and 68.0 inches?A)0.9608B)0.7881C)0.1844D)0.3112Answer:C54)The lengths of pregnancies of humans are normally distributed with a mean of 268 days and a standarddeviation of 15 days. A baby is premature if it is born three weeks early. What percent of babies are bornprematurely?A)6.81%B)9.21%C)10.31%D)8.08%Answer:D55)The distribution of cholesterol levels in teenage boys is approximately normal withμ=170andΗ=30(Source:U.S. National Center for Health Statistics). Levels above 200 warrant attention. What percent of teenage boyshave levels between 170 and 225?A)56.13%B)46.64%C)3.36%D)6.06%Answer:C56)Assume that blood pressure readings are normally distributed withμ=120andΗ=8.A blood pressurereading of 145 or more may require medical attention. What percent of people have a blood pressure readinggreater than 145?A)99.91%B)11.09%C)6.06%D)0.09%Answer:D57)Assume that the heights of American men are normally distributed with a mean of 69.0 inches and a standarddeviation of 2.8 inches. The U.S. Marine Corps requires that men have heights between 64 and 78 inches. Findthe percent of men meeting these height requirements.A)3.67%B)31.12%C)96.26%D)99.93%Answer:C11
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.58)Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standarddeviation of 2.5 inches. The U.S. Army requires that the heights of women be between 58 and 80 inches. If awoman is randomly selected, what is the probability that her height is between 58 and 80 inches?Answer:If x=58, then z=-2.24 and P(x)=0.0125. If x=80, then z=6.56 and P(x)=0.9999.P(58<x<80)=0.9999-0.0125=0.9874.MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.59)The graph depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and astandard deviation of 15. Find the probability that a member selected at random is from the shaded area of thegraph.A)0.8051B)0.6293C)0.7486D)0.4400Answer:C60)The graph depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and astandard deviation of 15. Find the probability that a member selected at random is from the shaded area of thegraph.A)0.7745B)0.7303C)0.7938D)0.7619Answer:CSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.Provide an appropriate response. Use the Standard Normal Table to find the probability.61)The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15days. Out of 50 pregnancies, how many would you expect to last less than 250 days?Answer:About 6 pregnancies62)The distribution of cholesterol levels in teenage boys is approximately normal withμ=170andΗ=30.Levelsabove 200 warrant attention. If 95 teenage boys are examined, how many would you expect to have cholesterollevels greater than 225?Answer:About 3 teenage boys12
63)An airline knows from experience that the distribution of the number of suitcases that get lost each week on acertain route is approximately normal withμ=15.5 andΗ=3.6. In one year, how many weeks would youexpect the airline to lose between 10 and 20 suitcases?Answer:About 43 weeks64)Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standarddeviation of 2.5 inches. The U.S. Army requires that the heights of women be between 58 and 80 inches. If 200women want to enlist in the U.S. Army, how many would you expect to meet the height requirements?Answer:About 197 women65)Assume that the heights of men are normally distributed with a mean of 69.0 inches and a standard deviation of2.8 inches. The U.S. Marine Corps requires that the heights of men be between 64 and 78 inches. If 500 menwant to enlist in the U.S. Marine Corps, how many would you not expect to meet the height requirements?Answer:About 19 menProvide an appropriate response.66)Find the z-score that corresponds to the given area under the standard normal curve.Answer:z=-0.5867)Find the z-score that corresponds to the given area under the standard normal curve.Answer:z=-1.7168)Find the z-score that corresponds to the given area under the standard normal curve.Answer:z=0.4213
69)Find the z-score that corresponds to the given area under the standard normal curve.Answer:z=3.07MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.70)Find the z-scores for which 90% of the distribution’s area lies between-z and z.A)(-1.96, 1.96)B)(-0.99, 0.99)C)(-2.33, 2.33)D)(-1.645, 1.645)Answer:D71)Find the z-scores for which 98% of the distribution’s area lies between-z and z.A)(-1.96, 1.96)B)(-0.99, 0.99)C)(-2.33, 2.33)D)(-1.645, 1.645)Answer:C72)Find the z-score for which 70% of the distribution’s area lies to its right.A)-0.53B)-0.81C)-0.47D)-0.98Answer:A73)Find the z-score that is greater than the mean and for which 70% of the distribution’s area lies to its left.A)0.47B)0.81C)0.98D)0.53Answer:D74)Use a standard normal table to find the z-score that corresponds to the cumulative area of0.0188.A)2.08B)2.1C)-2.1D)-2.08Answer:D75)Find the z-score that has84.85% of the distribution’s area to itsright.A)-1.03B)0.39C)1.03D)-0.39Answer:A76)Find the z-score for which99% of the distribution’s area lies between-z and z.A)(-2.33,2.33)B)(-1.96,1.96)C)(-2.575,2.575)D)(-1.645,1.645)Answer:C77)For the standard normal curve, find the z-score that corresponds to the90thpercentile.A)2.81B)1.52C)1.28D)0.28Answer:C78)For the standard normal curve, find the z-score that corresponds to the30thpercentile.A)-0.12B)-0.47C)-0.53D)-0.98Answer:C14
79)For the standard normal curve, find the z-score that corresponds to the7thdecile.A)0.47B)0.53C)0.12D)0.98Answer:B80)Use a standard normal table to find the z-score that corresponds to the80thpercentile.A)1.405B)2.41C)0.84D)0.2Answer:C81)IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the x-score thatcorresponds to a z-score of 2.33.A)142.35B)125.95C)139.55D)134.95Answer:D82)IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the x-score thatcorresponds to a z-score of-1.645.A)75.3B)91.0C)79.1D)82.3Answer:A83)The scores on a mathematics exam have a mean of68and a standard deviation of6. Find the x-value thatcorresponds to the z-score2.33.A)74.0B)70.3C)54.0D)82.0Answer:DSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.84)A mathematics professor gives two different tests to two sections of his college algebra courses. The first classhas a mean of 56 with a standard deviation of 9 while the second class has a mean of 75 with a standarddeviation of 15. A student from the first class scores a 62 on the test while a student from the second class scoresan 83 on the test. Compare the scores.Answer:z=(62-56)/9=0.667; z=(83-75)/15=0.533. The student with the score of 62 has the better score.MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.85)Compare the scores: a score of 75 on a test with a mean of 65 and a standard deviation of 8 and a score of 75 on atest with a mean of 70 and a standard deviation of 4.A)A score of 75 with a mean of 65 and a standard deviation of 8 is better.B)You cannot determine which score is better from the given information.C)The two scores are statistically the same.D)A score of 75 with a mean of 70 and a standard deviation of 4 is better.Answer:C86)Compare the scores: a score of 88 on a test with a mean of 79 and a score of 78 on a test with a mean of 70.A)A score of 75 with a mean of 65 and a standard deviation of 8 is better.B)The two scores are statistically the same.C)You cannot determine which score is better from the given information.D)A score of 75 with a mean of 70 and a standard deviation of 4 is better.Answer:C15
87)Compare the scores: a score of 220 on a test with a mean of 200 and a standard deviation of 21 and a score of 90on a test with a mean of 80 and a standard deviation of 8.A)A score of 220 with a mean of 200 and a standard deviation of 21 is better.B)The two scores are statistically the same.C)You cannot determine which score is better from the given information.D)A score of 90 with a mean of 80 and a standard deviation of 8 is better.Answer:D88)Two high school students took equivalent language tests, one in German and one in French. The student takingthe German test, for which the mean was 66 and the standard deviation was 8, scored an 82, while the studenttaking the French test, for which the mean was 27 and the standard deviation was 5, scored a 35. Compare thescores.A)The two scores are statistically the same.B)A score of 82 with a mean of 66 and a standard deviation of 8 is better.C)A score of 35 with a mean of 27 and a standard deviation of 5 is better.D)You cannot determine which score is better from the given information.Answer:B89)SAT scores have a mean of 1026 and a standard deviation of 209. ACT scores have a mean of 20.8 and astandard deviation of 4.8. A student takes both tests while a junior and scores 1130 on the SAT and 25 on theACT. Compare the scores.A)A score of 25 on the ACT test was better.B)You cannot determine which score is better from the given information.C)A score of 1130 on the SAT test was better.D)The two scores are statistically the same.Answer:A90)SAT scores have a mean of 1026 and a standard deviation of 209. ACT scores have a mean of 20.8 and astandard deviation of 4.8. A student takes both tests while a junior and scores 860 on the SAT and 16 on theACT. Compare the scores.A)You cannot determine which score is better from the given information.B)The two scores are statistically the same.C)A score of 16 on the ACT test was better.D)A score of 860 on the SAT test was better.Answer:DSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.91)Assume that blood pressure readings are normally distributed withμ=122andΗ=6. A researcher wishes toselect people for a study but wants to exclude the top and bottom 10 percent. What would be the upper andlower readings to qualify people to participate in the study?Answer:(114.3,129.7)92)Assume that the salaries of elementary school teachers in the United States are normally distributed with amean of$33,000and a standard deviation of$3000. What is the cutoff salary for teachers in the top 10%?Answer:x=μ+zΗ=33,000+(1.28)(3000)=$36,84016
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.93)Assume that the salaries of elementary school teachers in the United States are normally distributed with amean of$39,000and a standard deviation of$3000. What is the cutoff salary for teachers in the bottom 10%?A)$34,065B)$42,840C)$43,935D)$35,160Answer:DSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.94)The times for completing one circuit of a bicycle course are normally distributed with a mean of68.0minutesand a standard deviation of4.9minutes. An association wants to sponsor a race but will cut the bottom 25% ofriders. In a trial run, what should be the cutoff time?Answer:x=μ+zΗ=68.0+(0.675)(4.9)=71.3195)Assume that the heights of men are normally distributed with a mean of70.7inches and a standard deviation of1.1inches. If the top 5 percent and bottom 5 percent are excluded for an experiment, what are the cutoff heightsto be eligible for this experiment? Round your answers to one decimal place.Answer:68.9inches,72.5inchesMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.96)Assume that the heights of women are normally distributed with a mean of65.0inches and a standarddeviation of1.3inches. FindQ3, the third quartile that separates the bottom 75% from the top 25%.A)64.1B)65.9C)66.5D)66.7Answer:B97)The body temperatures of adults are normally distributed with a mean of 98.6°F and a standard deviation of0.81°F. What temperature represents the 95th percentile?A)99.64°FB)97.27°FC)99.93°FD)100.19°FAnswer:C98)In a certain normal distribution, find the standard deviationΗwhenμ=50 and 10.56% of the area lies to theright of 55.A)5B)3C)4D)2Answer:C99)In a certain normal distribution, find the meanμwhenΗ=5 and 5.48% of the area lies to the left of 78.A)62B)94C)70D)86Answer:DSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.100)In a certain normal distribution, 6.3% of the area lies to the left of 36 and 6.3% of the area lies to the right of 42.Find the meanμand the standard deviationΗ.Answer:μ=39,Η=1.9617
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.101)A tire company finds the lifespan for one brand of its tires is normally distributed with a mean of48,100milesand a standard deviation of2000miles. If the manufacturer is willing to replace no more than 10% of the tires,what should be the approximate number of miles for a warranty?A)51,390B)44,810C)45,540D)50,660Answer:C102)The distribution of room and board expenses per year at a four-year college is normally distributed with amean of $5850 and standard deviation of $1125. Random samples of size 20 are drawn from this populationand the mean of each sample is determined. Which of the following mean expenses would be consideredunusual?A)$6350B)$6180C)$5180D)none of theseAnswer:C103)The lengths of pregnancies are normally distributed with a mean of271days and a standard deviation of20days. If64women are randomly selected, find the probability that they have a mean pregnancy between271days and273days.A)0.2881B)0.7881C)0.5517D)0.2119Answer:A104)The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15days. If 64 women are randomly selected, find the probability that they have a mean pregnancy between 266days and 268 days.A)0.3577B)0.2881C)0.5517D)0.7881Answer:A105)Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standarddeviation of 2.5 inches. If 100 women are randomly selected, find the probability that they have a mean heightgreater than 63.0 inches.A)0.2881B)0.9918C)0.0082D)0.8989Answer:B106)Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standarddeviation of 2.5 inches. If 75 women are randomly selected, find the probability that they have a mean heightbetween 63 and 65 inches.A)0.2119B)0.0188C)0.9811D)0.3071Answer:C107)Assume that the heights of men are normally distributed with a mean of68.1inches and a standard deviation of2.8inches. If64men are randomly selected, find the probability that they have a mean height greater than69.1inches.A)0.9005B)9.9671C)0.8188D)0.0021Answer:D108)Assume that the heights of men are normally distributed with a mean of 69.0 inches and a standard deviation of2.8 inches. If 64 men are randomly selected, find the probability that they have a mean height between 68 and 70inches.A)0.9979B)0.0021C)0.9958D)0.9015Answer:C18
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.109)The body temperatures of adults are normally distributed with a mean of 98.6°F and a standard deviation of0.60°F. If 25 adults are randomly selected, find the probability that their mean body temperature is less than 99°F.Answer:0.9996MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.110)The body temperatures of adults are normally distributed with a mean of 98.6°F and a standard deviation of0.60°F. If 36 adults are randomly selected, find the probability that their mean body temperature is greater than98.4°F.A)0.8188B)0.9772C)0.9360D)0.0228Answer:BSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.111)Assume that the salaries of elementary school teachers in the United States are normally distributed with amean of $32,000 and a standard deviation of $3000. If 100 teachers are randomly selected, find the probabilitythat their mean salary is less than $32,500.Answer:0.9525MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.112)Assume that the salaries of elementary school teachers in the United States are normally distributed with amean of $32,000 and a standard deviation of $3000. If 100 teachers are randomly selected, find the probabilitythat their mean salary is greater than $32,500.A)0.1312B)0.0475C)0.3312D)0.9525Answer:B113)Assume that blood pressure readings are normally distributed with a mean of117and a standard deviation of8. If100people are randomly selected, find the probability that their mean blood pressure will be less than119.A)0.9938B)0.8615C)0.8819D)0.0062Answer:A114)Assume that blood pressure readings are normally distributed with a mean of 120 and a standard deviation of8. If 100 people are randomly selected, find the probability that their mean blood pressure will be greater than122.A)0.9938B)0.0062C)0.8615D)0.8819Answer:B115)The average number of pounds of red meat a person consumes each year is 196 with a standard deviation of 22pounds (Source: American Dietetic Association). If a sample of 50 individuals is randomly selected, find theprobability that the mean of the sample will be less than 200 pounds.A)0.9015B)0.7613C)0.8815D)0.0985Answer:A116)The average number of pounds of red meat a person consumes each year is 196 with a standard deviation of 22pounds (Source: American Dietetic Association). If a sample of 50 individuals is randomly selected, find theprobability that the mean of the sample will be greater than 200 pounds.A)0.9015B)0.7613C)0.0985D)0.8815Answer:C19
117)A coffee machine dispenses normally distributed amounts of coffee with a mean of 12 ounces and a standarddeviation of 0.2 ounce. If a sample of 9 cups is selected, find the probability that the mean of the sample will beless than 12.1 ounces. Find the probability if the sample is just 1 cup.A)0.9332; 0.6915B)0.4332; 0.6915C)0.4332; 0.1915D)0.9332; 0.1915Answer:A118)A coffee machine dispenses normally distributed amounts of coffee with a mean of 12 ounces and a standarddeviation of 0.2 ounce. If a sample of 9 cups is selected, find the probability that the mean of the sample will begreater than 12.1 ounces. Find the probability if the sample is just 1 cup.A)0.0668; 0.8085B)0.5668; 0.8085C)0.5668; 0.3085D)0.0668; 0.3085Answer:DUse the Central Limit Theorem to find the mean and standard error of the mean of the indicated sampling distribution.119)The amounts of time employees of a telecommunications company have worked for the company are normallydistributed with a mean of5.6years and a standard deviation of1.9years. Random samples of size18aredrawn from the population and the mean of each sample is determined.A)5.6years,0.11yearsB)5.6years,0.45yearsC)1.32years,0.45yearsD)1.32years,1.9yearsAnswer:B120)The monthly rents for studio apartments in a certain city have a mean of$960and a standard deviation of$200.Random samples of size30are drawn from the population and the mean of each sample is determined.A)$960,$6.67B)$960,$36.51C)$175.27,$36.51D)$175.27,$200Answer:BProvide an appropriate response.121)Scores on a test have a mean of 72 and a standard deviation of 12. Random samples of size 64 are drawn fromthis population and the mean of each sample is determined. Use the Central Limit Theorem to find the meanand standard error of the mean of the sampling distribution. Then sketch a graph of the sampling distribution.A)72, 945546372819099B)9, 1.54.567.5910.51213.5C)72, 1.567.56970.57273.57576.5D)72, 12364860728496108Answer:C20
122)The weights of people in a certain population are normally distributed with a mean of158lband a standarddeviation of24lb.Find the mean and standard error of the mean for this sampling distribution when usingrandom samples of size5.A)158,10.73B)158,5C)158,24D)158,4.8Answer:A123)The mean annual income for adult women in one city is $28,520 and the standard deviation of the incomes is$5000. The distribution of incomes is skewed to the right. Find the mean and standard error of the mean for thissampling distribution when using random samples of size77. Round your answers to the nearest dollar.A)$28,520, $5000B)$28,520, $65C)$28,520, $77D)$28,520, $570Answer:D124)What happens to the mean and standard deviation of the distribution of sample means as the size of the sampledecreases?A)The mean of the sample means stays constant and the standard error decreases.B)The mean of the sample means increases and the standard error stays constant.C)The mean of the sample means decreases and the standard error increases.D)The mean of the sample means stays constant and the standard error increases.Answer:D125)If the sample size is multiplied by 4, what happens to the standard deviation of the distribution of samplemeans?A)The standard error is increased by a factor of 4.B)The standard error is doubled.C)The standard error is decreased by a factor of 4.D)The standard error is halved.Answer:DSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.126)A study of 800 homeowners in a certain area showed that the average value of the homes is $182,000 and thestandard deviation is $15,000. If 50 homes are for sale, find the probability that the mean value of these homes isless than $185,000. Remember: check to see if the finite correction factor applies.Answer:50/800=0.0625=6.25%, hence the finite correction factor applies; P(x<185,000)=0.9279127)A study of 800 homeowners in a certain area showed that the average value of the homes is $182,000 and thestandard deviation is $15,000. If 50 homes are for sale, find the probability that the mean value of these homes isgreater than $185,000. Remember: check to see if the finite correction factor applies.Answer:50/800=0.0625=6.25%, hence the finite correction factor applies;P(x>185,000)=1-0.9279=0.0721128)A soda machine dispenses normally distributed amounts of soda with a mean of 20 ounces and a standarddeviation of 0.2 ounce. Are you more likely to randomly select one bottle with more than 20.3 ounces or are youmore likely to select a sample of eight bottles with a mean amount of more than 20.3 ounces? Explain.Answer:It is more likely to select one bottle with more than 20.3 ounces because a large percentage of the data isnow closer to the mean.21
129)A soda machine dispenses normally distributed amounts of soda with a mean of 20 ounces and a standarddeviation of 0.2 ounce. Are you more likely to randomly select one bottle with an amount between 19.8 ouncesand 20.2 ounces or are you more likely to select a sample of eight bottles with a mean amount between 19.8ounces and 20.2 ounces? Explain.Answer:It is more likely to select a sample of eight bottles with an amount between 19.8 ounces and 20.2 ouncesbecause a large percentage of the data is now closer to the mean.130)In a recent survey,83% of the community favored building a police substation in their neighborhood. Yourandomly select18citizens and ask each if he or she thinks the community needs a police substation.Decide whether you can use the normal distribution to approximate the binomial distribution. If so, find the mean andstandard deviation. If not, explain why.Answer:cannot use normal distribution, nq=(18)(0.17)=3.06<5131)A recent survey found that73% of all adults over 50 wear glasses for driving. You randomly select32adultsover 50, and ask if he or she wears glasses. Decide whether you can use the normal distribution to approximatethe binomial distribution. If so, find the mean and standard deviation, If not, explain why.Answer:use normal distribution,μ=23.36andΗ=2.51.132)According to government data, the probability than an adult was never married is20%. You randomly select48adults and ask if he or she was ever married. Decide whether you can use the normal distribution toapproximate the binomial distribution, If so, find the mean and standard deviation, If not, explain why.Answer:use normal distribution,μ=9.6andΗ=2.77.133)Decide if it is appropriate to use the normal distribution to approximate the random variable x for a binomialexperiment with sample size of n=7and probability of success p=0.1.Answer:cannot use normal distribution134)Decide if it is appropriate to use the normal distribution to approximate the random variable x for a binomialexperiment with sample size of n=29and probability of success p=0.4.Answer:can use normal distribution135)Decide if it is appropriate to use the normal distribution to approximate the random variable x for a binomialexperiment with sample size of n=25 and probability of success p=0.2Answer:can use normal distributionMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.136)Ten percent of the population is left-handed. A class of 100 students is selected. Convert the binomialprobability P(x>12) to a normal probability by using the correction for continuity.A)P(xL11.5)B)P(x>12.5)C)P(xK12.5)D)P(x<11.5)Answer:B137)Ten percent of the population is left-handed. A class of6150students is selected. Convert the binomialprobability P(xL25) to a normal probability by using the correction for continuity.A)P(xL24.5)B)P(xL25.5)C)P(xK25.5)D)P(x<24.5)Answer:A22
138)Ten percent of the population is left-handed. A class of 100 students is selected. Convert the binomialprobability P(x<12) to a normal probability by using the correction for continuity.A)P(xK12.5)B)P(x>12.5)C)P(xL11.5)D)P(x<11.5)Answer:D139)Ten percent of the population is left-handed. A class of7650students is selected. Convert the binomialprobability P(x>20) to a normal probability by using the correction for continuity.A)P(xK20.5)B)P(x<19.5)C)P(x>20.5)D)P(xL19.5)Answer:CSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.140)The failure rate in a statistics class is20%. In a class of30students, find the probability that exactly five studentswill fail. Use the normal distribution to approximate the binomial distribution.Answer:P(4.5<X<5.5)=P(-0.68<z<-0.23)=0.409-0.2483=0.1607141)A local motel has50rooms. The occupancy rate for the winter months is 60%. Find the probability that in agiven winter month at least35rooms will be rented. Use the normal distribution to approximate the binomialdistribution.Answer:P(xL34.5)=0.0968142)A local motel has100rooms. The occupancy rate for the winter months is 60%. Find the probability that in agiven winter month fewer than70rooms will be rented. Use the normal distribution to approximate thebinomial distribution.Answer:P(x<69.5)=0.9738MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.143)A student answers all 48 questions on a multiple-choice test by guessing. Each question has four possibleanswers, only one of which is correct. Find the probability that the student gets exactly 15 correct answers. Usethe normal distribution to approximate the binomial distribution.A)0.7967B)0.0606C)0.0823D)0.8577Answer:C144)If the probability of a newborn child being female is 0.5, find the probability that in 100 births, 55 or more willbe female. Use the normal distribution to approximate the binomial distribution.A)0.7967B)0.8159C)0.1841D)0.0606Answer:C145)An airline reports that it has been experiencing a 15% rate of no-shows on advanced reservations. Among 150advanced reservations, find the probability that there will be fewer than 20 no-shows.A)0.7549B)0.3187C)0.7967D)0.2451Answer:D146)Find the probability that in 200 tosses of a fair six-sided die, a five will be obtained at least 40 times.A)0.3875B)0.8810C)0.0871D)0.1210Answer:D147)Find the probability that in 200 tosses of a fair six-sided die, a five will be obtained at most 40 times.A)0.0853B)0.1190C)0.9131D)0.8810Answer:C23
148)A telemarketer found that there was a 1% chance of a sale from his phone solicitations. Find the probability ofgetting 5 or more sales for 1000 telephone calls.A)0.9599B)0.0401C)0.0871D)0.8810Answer:ASHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.149)The author of a statistics book has trouble deciding whether to use the words “he” or “she” in the book’sexamples. To solve the problem, the author flips a coin each time the problem arises. If a head shows, the authoruses “he” and if a tail shows, the author uses “she”. If this problem occurs 100 times in the book, what is theprobability that “she” will be used 58 times?Answer:P(57.5<x<58.5)=P(1.50<z<1.70)=0.9554-0.9332=0.0222MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.150)Match the binomial probability P(x<23) with the correct statement.A)P(there are at most 23 successes)B)P(there are fewer than 23 successes)C)P(there are at least 23 successes)D)P(there are more than 23 successes)Answer:B151)Match the binomial probability P(x<47) with the correct statement.A)P(there areat most47successes)B)P(there areat least47successes)C)P(there aremore than47successes)D)P(there arefewer than47successes)Answer:D152)Match the binomial probability P(xK23) with the correct statement.A)P(there are at least 23 successes)B)P(there are more than 23 successes)C)P(there are at most 23 successes)D)P(there are fewer than 23 successes)Answer:C153)Match the binomial probability P(x>23) with the correct statement.A)P(there are more than 23 successes)B)P(there are fewer than 23 successes)C)P(there are at least 23 successes)D)P(there are at most 23 successes)Answer:A154)Ten percent of the population is left-handed. In a class of 100 students, write the binomial probability for thestatement “There are at least 12 left-handed students in the class.”A)P(x>12)B)P(xL12)C)P(xK12)D)P(x<12)Answer:B155)Ten percent of the population is left-handed. In a class of 100 students, write the binomial probability for thestatement “There are more than 12 left-handed students in the class.”A)P(x=12)B)P(x>12)C)P(xK12)D)P(x<12)Answer:B156)Ten percent of the population is left-handed. In a class of 100 students, write the binomial probability for thestatement “There are at most 12 left-handed students in the class.”A)P(xL12)B)P(x=12)C)P(xK12)D)P(x<12)Answer:C24
157)Ten percent of the population is left-handed. In a class of145students, write the binomial probability for thestatement “There areat most21left-handed students in the class.”A)P(xL21)B)P(xK21)C)P(x<21)D)P(x>21)Answer:B158)Ten percent of the population is left-handed. In a class of 100 students, write the binomial probability for thestatement “There are exactly 12 left-handed students in the class.”A)P(x=12)B)P(xK12)C)P(xL12)D)P(x>12)Answer:A25