BCTC STATS FINAL PRACTICE EXAM QUESTIONS WITH CORRECT DETAILED ANSWERS

BCTC STATS FINAL PRACTICE EXAM QUESTIONS WITH CORRECT DETAILED ANSWERS a) 75.9-61 = 14.9 b) 12.42 c) z= (61-75.9)/1.2 = -12.42 d) Nelson's height is unusual - Answer-With a height of 61in, Nelson was the shortest president of a particular club in the past century. The club presidents of the past century have a mean height of 75.9in. and a standard deviation of 1.2in. a.What is the positive difference between Nelson's height and the mean? b.How many standard deviations is that [the difference found in part (a)]? c.Convert Nelson's height to a z score. d.If we consider "usual" heights to be those that convert to z scores between -2 and 2, is Nelson's height usual or unusual? unusual - Answer-A data value is considered _______ if its z-score is less than -2 or greater than 2. a) 1/8 b) 1/8 c) 3/8 - Answer-To the right are the outcomes that are possible when a couple has three children. Refer to that list, and find the probability of each event. a. Among three children, there are exactly 0 girls b. Among three children, there are exactly 3 boys. c. Among three children, there are exactly 2 girls probability: 0.773 is it close to 3/4? = yes, it is reasonably close - Answer-In a genetics experiment on peas, one sample of offspring contained 364 green peas and 107 yellow peas. Based on those results, estimate the probability of getting an offspring pea that is green. Is the result reasonably close to the value of 3/4 that was expected? a) 1-0.73= 0.27 b) 1-0.0084 = 0.9916 - Answer-a. If P(A)=0.73, find the probability of the complement of A, P(A-overbar). b. A certain group of women has a 0.84% rate of red/green color blindness. If a woman is randomly selected, what is the probability that she does not have red/green color blindness? (350+797-283)/927 = 0.932 - Answer-The following data summarizes results from 927 pedestrian deaths that were caused by accidents. If one of the pedestrian deaths is randomly selected, find the probability that the pedestrian was intoxicated or the driver was not intoxicated. a) total subjects: 159 + 142 = 301 b) did not use drug: 157 + 30 = 187 c) What is the probability that a randomly selected subject did not use marijuana? 187/301 = 0.621 - Answer-Use the following results from a test for marijuana use, which is provided by a certain drug testing company. Among 142 subjects with positive test results, there are 30 false positive results. Among 159 negative results, there are 2 false negative results. Complete parts (a) through (c). (Hint: Construct a table.) a) (91/223)*(91/223) = 0.1665 b) (91.223)*(90/222) = 0.1654 - Answer-Refer to the table below. Given that 2 of the 223 subjects are randomly selected, complete parts (a) and (b). a. Assume that the selections are made with replacement. What is the probability that the 2 selected subjects are both group Upper O and type Rh+ a) 4 tires= (1-240/6300)^4 b) 100 tires = (1-240/6300)^100 c) yes, because there is a very small chance that all 100 tires are good. - Answer-A tire company produced a batch of 6,300 tires that includes exactly 240 that are defective. a.) If 4 tires are randomly selected for installation on a car, what is the probability that they are all good? b). If 100 tires are randomly selected for shipment to an outlet, what is the probability that they are all good? c) Should this outlet plan to deal with defective tires returned by consumers? a) p= 1-(1-0.0497)^4 = 0.1845 b) Yes, the four cars are not randomly selected. - Answer-The probability of a randomly selected car crashing during a year in a certain country is 0.0497. a) If a family has four cars, find the probability that at least one of them has a car crash during a year. b)Is there any reason why the probability might be wrong? (11!) / (11-2)!2! = 55 - Answer-When testing for current in a cable with eleven color-coded wires, the author used a meter to test two wires at a time. How many different tests are required for every possible pairing of two wires? a) 42*41*40*39 = 2,686,320 b) 42C4 = 111,930 - Answer-There are 42 members on the board of directors for a certain non-profit institution. a.If they must elect a chairperson, first vice chairperson, second vice chairperson, and secretary, how many different slates of candidates are possible? b.If they must form an ethics subcommittee of four members, how many different subcommittees are possible? a) (from the table given) 7 = 0.305 b) (from table add 7 + 8) = 0.392 c) the result from part b d) No, since the appropriate probability is greater than 0.05, it is not an unusually high number. - Answer-The accompanying table describes results from eight offspring peas. The random variable x represents the number of offspring peas with green pods. a) Find the probability of getting exactly 7 peas with green pods. b) Find the probability of getting 7 or more peas with green pods. c) Which probability is relevant for determining whether 7 is an unusually high number of peas with green pods, the result from part (a) or part (b)? d)Is 7 an unusually high number of peas with green pods? Why or why not? Use 0.05 as the threshold for an unusual event. (7!) / 2!*(7-2)! * (0.45)^2(1-0.45)^(7-2) = 0.214 - Answer-Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. n = 7 , x = 2 , p = 0.45 a) mean= 20*(0.5) = 10 s.d = 20*(0.5)(0.5) = 5 ... sqrt(5) = 2.2 b) yes because 18 is greater than the maximum usual value - Answer-Several psychology students are unprepared for a surprise true/false test with 20 questions, and all of their answers are guesses. a. Find the mean and standard deviation for the number of correct answers for such students. b. Would it be unusual for a student to pass by guessing (which requires getting at least 18 correct answers)? f(x)= 1 / (57-47) = 0.1 p = 0.1 * (51.5-51.25) = 0.025 - Answer-A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 47.0 and 57.0 minutes. Find the probability that a given class period runs between 51.25 and 51.5 minutes. a) shaded region is less than -1.26 b) p = 0.1038 (figure out how to calc p-score) - Answer-Assume the readings on thermometers are normally distributed with a mean of 0C and a standard deviation of 1.00C. Find the probability that a randomly selected thermometer reads less than negative 1.26 and draw a sketch of the region. z= (85-100)/15 = -1.00 find that on table = 0.1587 - Answer-Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.