Probability and Statistics for Engineering and the Sciences, 8e Devore TB TestBank

Probability and Statistics for Engineering and the Sciences, 8e Devore TB Probability and Statistics for Engineering and the Sciences, 8e Devore TBChapter 1 – Overview and Descriptive Statistics SHORT ANSWER 1. Give one possible sample of size 4 from each of the following populations: a. All daily newspapers published in the United States b. All companies listed on the New York Stock Exchange c. All students at your college or university d. All grade point averages of students at your college or university ANS: a. Houston Chronicle, Des Moines Register, Chicago Tribune, Washington Post b. Capital One, Campbell Soup, Merrill Lynch, Pulitzer c. John Anderson, Emily Black, Bill Carter, Kay Davis d. 2.58. 2.96, 3.51, 3.69 PTS: 1 2. A Southern State University system consists of 23 campuses. An administrator wishes to make an inference about the average distance between the hometowns of students and their campuses. Describe and discuss several different sampling methods that might be employed. Would this be an enumerative or an analytic study? Explain your reasoning. ANS: One could take a simple random sample of students from all students in the California State University system and ask each student in the sample to report the distance from their hometown to campus. Alternatively, the sample could be generated by taking a stratified random sample by taking a simple random sample from each of the 23 campuses and again asking each student in the sample to report the distance from their hometown to campus. Certain problems might arise with self reporting of distances, such as recording error or poor recall. This study is enumerative because there exists a finite, identifiable population of objects from which to sample. PTS: 1 3. A Michigan city divides naturally into ten district neighborhoods. How might a real estate appraiser select a sample of single-family homes that could be used as a basis for developing an equation to predict appraised value from characteristics such as age, size, number of bathrooms, and distance to the nearest school, and so on? Is the study enumerative or analytic? ANS: One could generate a simple random sample of all single family homes in the city or a stratified random sample by taking a simple random sample from each of the 10 district neighborhoods. From each of the homes in the sample the necessary variables would be collected. This would be an enumerative study because there exists a finite, identifiable population of objects from which to sample. PTS: 1 4. An experiment was carried out to study how flow rate through a solenoid valve in an automobile’s pollution-control system depended on three factors: armature lengths, spring load, and bobbin depth. Two different levels (low and high) of each factor were chosen, and a single observation on flow was made for each combination of levels. a. The resulting data set consisted of how many observations? b. Is this an enumerative or analytic study? Explain your reasoning.ANS: a. Number observations equal 2 2 2=8 b. This could be called an analytic study because the data would be collected on an existing process. There is no sampling frame. PTS: 1 5. The accompanying data specific gravity values for various wood types used in construction . .41 .41 .42 .42. .42 .42 .42 .43 .44 .54 .55 .58 .62 .66 .66 .67 .68 .75 .31 .35 .36 .36 .37 .38 .40 .40 .40 .45 .46 .46 .47 .48 .48 .48 .51 .54 Construct a stem-and-leaf display using repeated stems and comment on any interesting features of the display. ANS: One method of denoting the pairs of stems having equal values is to denote the stem by L, for ‘low’ and the second stem by H, for ‘high’. Using this notation, the stem-and-leaf display would appear as follows: 3L 1 stem: tenths 3H 56678 leaf: hundredths 4L 234 5L 144 5H 58 6L 2 6H 6678 7L 7H 5 The stem-and-leaf display on the previous page shows that .45 is a good representative value for the data. In addition, the display is not symmetric and appears to be positively skewed. The spread of the data is .75 - .31 = .44, which is .44/.45 = .978 or about 98% of the typical value of .45. This constitutes a reasonably large amount of variation in the data. The data value .75 is a possible outlier. PTS: 1 6. Temperature transducers of a certain type are shipped in batches of 50. A sample of 60 batches was selected, and the number of transducers in each batch not conforming to design specifications was determined, resulting in the following data: 0 4 2 1 3 1 1 3 4 1 2 3 2 2 8 4 5 1 3 1 2 1 2 4 0 1 3 2 0 5 3 3 1 3 2 4 7 0 2 3 5 0 2 3 2 1 0 6 4 2 1 6 0 3 3 3 6 1 2 3 a. Determine frequencies and relative frequencies for the observed values of x = number of nonconforming transducers in a batch. b. What proportion of batches in the sample has at most four nonconforming transducers? What proportion has fewer than four? What proportion has at least four nonconforming units? ANS: a. Number Nonconforming Frequency Relative Frequency 0 7 0.117 1 12 0.2002 13 0.217 3 14 0.233 4 6 0.100 5 3 0.050 6 3 0.050 7 1 0.017 8 1 0.017 1.001 The relative frequencies don’t add up exactly to 1because they have been rounded b. The number of batches with at most 4 nonconforming items is 7+12+13+14+6=52, which is a proportion of 52/60=.867. The proportion of batches with (strictly) fewer than 4 nonconforming items is 46/60=.767. PTS: 1 7. The number of contaminating particles on a silicon wafer prior to a certain rinsing process was determined for each wafer in a sample size 100, resulting in the following frequencies: Number of particles Frequency Number of particles Frequency 0 1 8 12 1 2 9 4 2 3 10 5 3 12 11 3 4 11 12 1 5 15 13 2 6 18 14 1 7 10 a. What proportion of the sampled wafers had at least two particles? At least six particles? b. What proportion of the sampled wafers had between four and nine particles, inclusive? Strictly between four and nine particles? ANS: a. From this frequency distribution, the proportion of wafers that contained at least two particles is (100-1-2)/100 = .97, or 97%. In a similar fashion, the proportion containing at least 6 particles is (100 – 1-2-3-12-11-15)/100 = 56/100 = .56, or 56%. b. The proportion containing between 4 and 9 particles inclusive is (11+15+18+10+12+4)/100 = 70/100 = .70, or 70%. The proportion that contain strictly between 4 and 9 (meaning strictly more than 4 and strictly less than 9) is (15+ 18+10+12)/100= 55/100 = .55, or 55%. PTS: 1 8. The cumulative frequency and cumulative relative frequency for a particular class interval are the sum of frequencies and relative frequencies, respectively, for that interval and all intervals lying below it. Compute the cumulative frequencies and cumulative relative frequencies for the following data: ANS: Class Frequency Relative Cumulative CumulativeFrequency Frequency Relative Frequency 60 – under 65 3 .075 3 .075 65 – under 70 6 .15 9 .225 70 – under 75 7 .175 16 .40 75 – under 80 1 .025 17 .425 80 – under 85 12 .30 29 .725 85 – under 90 7 .175 36 .90 90 – under 95 2 .05 38 .95 95 – under 100 2 .05 40 1.0 PTS: 1 9. Consider the following observations on shear strength of a joint bonded in a particular manner: 30.0 4.4 33.1 66.7 81.5 22.2 40.4 16.4 73.7 36.6 109.9 a. Determine the value of the sample mean. b. Determine the value of the sample median. Why is it so different from the mean? c. Calculate a trimmed mean by deleting the smallest and largest observations. What is the corresponding trimming percentage? How does the value of this compare to the mean and median? ANS: a. The sum of the n = 11 data points is 514.90, so = 514.90/11 = 46.81. b. The sample size (n = 11) is odd, so there will be a middle value. Sorting from smallest to largest: 4.4 16.4 22.2 30.0 33.1 36.6 40.4 66.7 73.7 81.5 109.9. The sixth value, 36.6 is the middle, or median, value. The mean differs from the median because the largest sample observations are much further from the median than are the smallest values. c. Deleting the smallest (x = 4.4) and largest (x = 109.9) values, the sum of the remaining 9 observations is 400.6. The trimmed mean is 400.6/9 = 44.51. The trimming percentage is 100(1/11) = 9.1%. lies between the mean and median. PTS: 1 10. A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the accompanying data on time (sec) to complete the escape: a. Construct a stem-and-leaf display of the data. How does it suggest that the sample mean and median will compare? b. Calculate the values of the sample mean and median. c. By how much could the largest time, currently 424, be increased without affecting the value of the sample median? By how much could this value be decreased without affecting the value of the sample mean? d. What are the values of and when the observations are re-expressed in minutes? ANS: a. A. stem-and-leaf display of this data appears below: 32 55 stem: ones 33 49 leaf: tenths 34 35 34469 37 03345 38 9 39 2347 40 23 41 42 4 The display is reasonably symmetric, so the mean and median will be close. b. The sample mean is = 9638/26 = 370.7. The sample median is = (369+370)/2=369.50. c. The largest value (currently 424) could be increased by any amount. Doing so will not change the fact that the middle two observations are 369 and 170, and hence, the median will not change. However, the value x = 424 can not be changed to a number less than 370 ( a change of 424-370 = 54) since that will lower the values(s) of the two middle observations. d. Expressed in minutes, the mean is (370.7 sec)/(60 sec) = 6.18 min; the median 6.16 min. PTS: 1 11. A sample of n = 10 automobiles was selected, and each was subjected to a 5-mph crash test. Denoting a car with no visible damage by S (for success) and a car with such damage by F, results were as follows: S S S F F S S F S S a. What is the value of the sample proportion of successes x/n? b. Replace each S with a 1 and each F with a 0. Then calculate for this numerically coded sample. How does compare to x/n? c. Suppose it is decided to include 15 more cars in the experiment. How many of these would have to be S’s to x/n = .80 for the entire sample of 25 cars? ANS: a. 7/10 = .70 b. = .70 = proportion of successes, and x/n have the same value c. = .80 so s = (0.80)(25) = 20 total of 20 successes 20 – 7 = 13 of the new cars would have to be successes. PTS: 1 12. Answer the following two questions: a. If a constant c is added to each in a sample, yielding = + c, how do the sample mean and median of the ’s relate to the mean and median of the ’s? Verify your conjectures. b. If each is multiplied by a constant c, yielding = , answer the question of part (a). Again, verify your conjectures. ANS: a. = the median of = median of b. = PTS: 113. Calculate and interpret the values of the sample mean and sample standard deviation for the following observations on fracture strength. ANS: The sample mean, The sample standard deviation, On average, we would expect a fracture strength of 116.2. In general, the size of a typical deviation from the sample mean (116.2) is about 25.75. Some observations may deviate from 116.2 by more than this and some by less. PTS: 1 14. The first four deviations from the mean in a sample of n = 5 reaction times were .6, .9, 1.0, and 1.5. What is the fifth deviation from the mean? Give a sample for which these are the five deviations from the mean. ANS: Let d denote the fifth deviation. Then .6+.9+1.0+1.5+d = 0 or 4.0 + d =0, so d =-4.0. One sample for which these are the deviations is = 4.6, = 4.9, = 5.0, = 5.5, = 0. (Obtained by adding 4.0 to each deviation; adding any other number will produce a different sample with the desired property). PTS: 1 15. A sample of 20 glass bottles of a particular type was selected, and the internal pressure strength of each bottle was determined. Consider the following partial sample information: Median = 202.2 lower fourth = 196.0 Upper fourth = 216.8 Three smallest observations 125.8 188.1 193.7 Three largest observations 221.3 230.5 250.2 Are there any outliers in the sample? Any extreme outliers? ANS: 1.5(IQR) = 1.5(216.8-196.0) = 31.2 and 3(IQR) = 3(216.8-196.0) = 62.4. Mild outliers: observations below 196-31.2 = 164.6 or above 216.8+31.2=248. Extreme outliers: observations below 196-62.4 = 133.6 or above 216.8+62.4 = 279.2. Of the observations given, 125.8 is an extreme outlier and 250.2 is a mild outlier. PTS: 1CHAPTER 2 - Probability SHORT ANSWER 1. Suppose that vehicles taking a particular freeway exit can turn right (R), turn left (L), or go straight (S). Consider observing the direction for each of three successive vehicles. a. List all outcomes in the event A that all three vehicles go in the same direction. b. List all outcomes in the event B that all three vehicles take different directions. c. List all outcomes in the event C that exactly two of the three vehicles turn right. d. List all outcomes in the even D that exactly tow vehicles go in the same direction. e. List outcomes in , C D, and C D. ANS: a. Event A = {RRR, LLL, SSS} b. Event B = {RLS, RSL, LRS, LSR, SRL, SLR} c. Event C = {RRL, RRS, RLR, RSE, LRR, SRR} d. Event D = {RRL, RRS, RLR, RSE, LRR, SRR, LLR, LLS, LRL, LSL, RLL, SLL, SSR, SSL, SRS, SLS, RSS, LSS} e. Event contains outcomes where all cars go the same direction, or they all go different directions: = {RRR, LLL, SSS, RLS, RSL, LRS, LSR, SRL, SLR} Because Event D totally encloses Event C, the compound event C D = D. Therefore, C D = {RRL, RRS, RLR, RSR, LRR, SRR, LLR, LLS, LRL, LSL, RLL, SLL, SSR, SSL, SRS, SLS, RSS, LSS} Using similar reasoning, we see that the compound event C D = C. Therefore, C D = {RRL, RRS, RLR, RSR, LRR, SRR} PTS: 1 2. Each of a sample of four home mortgages is classified as fixed rate (F) or variable rate (V). a. What are the 16 outcomes in S? b. Which outcomes are in the event that exactly two of the selected mortgages are fixed rate? c. Which outcomes are in the event that all four mortgages are of the same type? d. Which outcomes are in the event that at most one of the four is a variable-rate mortgage? e. What is the union of the events in parts (c) and (d), and what is the intersection of these two events? f. What are the union and intersection of the two events in parts (b) and (c)? ANS: a. Home Mortgage Number Outcome 1 2 3 4 1 F F F F 2 F F F V 3 F F V F 4 F F V V 5 F V F F 6 F V F V 7 F V V F 8 F V V V 9 V F F F 10 V F F V11 V F V F 12 V F V V 13 V V F F 14 V V F V 15 V V V F 16 V V V V b. Outcome numbers 4, 6, 7, 10, 11,13 c. Outcome numbers 1, 16 d. Outcome numbers 1, 2, 3, 5, 9 e. The Union: outcomes 1, 2, 3, 5, 9, 16. The Intersection: outcome 1. f. The Union: outcomes 1, 4, 6, 7, 10, 11, 13, 16. The Intersection: this cannot happen. (There are no outcomes in common) :b c = . PTS: 1 3. A college library has five copies of a certain text on reserve. Two copies (1 and 2) are first printings, and the other three (3, 4, and 5) are second printings. A student examines these books in random order, stopping only when a second printing has been selected. One possible outcome is 4, and another is 125. a. List the S. b. Let A denote the event that exactly one book must be examined. What outcomes are in A? c. Let B be the event that book 4 is the one selected. What outcomes are in B? d. Let C be the event that book 2 is not examined. What outcomes are in C? ANS: a. Outcome Number Outcome 1 123 2 124 3 125 4 213 5 214 6 215 7 13 8 14 9 15 10 23 11 24 12 25 13 3 14 4 15 5 b. Outcome numbers 13, 14, 15, so A={3, 4, 5} c. Outcome numbers 2, 5, 8, 11, 14 so B={124, 214, 14, 24, 4} d. Outcome numbers 7, 8, 9, 13, 14, 15 so C={13, 14, 15, 3, 4, 5} PTS: 1 4. The Department of Statistics at a state university in California has just completed voting by secret ballot for a department head. The ballot box contains four slips with votes for candidate A and three slips with votes for candidate B. Suppose these slips are removed from the box one by one. a. List all possible outcomes.b. Suppose a running tally is kept as slips are removed. For what outcomes does A remain ahead of B throughout the tally? ANS: a. S = {BBBAAAA, BBABAAA, BBAABAA, BBAAABA, BBAAAAB, BABBAAA, BABABAA, BABAABA, BABAAAB, BAABBAA, BAABABA, BAABAAB, BAAABBA, BAAABAB, BAAAABB, ABBBAAA, ABBABAA, ABBAABA, ABBAAAB, ABABBAA, ABABABA, ABABAAB, ABAABBA, ABAABAB, ABAAABB, AABBBAA, AABBABA, AABBAAB, AABABBA, AABABAB, AABAABB, AAABBBA, AAABBAB, AAABABB, AAAABBB} b. {AAAABBB, AAABABB, AAABBAB, AABAABB, AABABAB} PTS: 1 5. Let A denote the event that the next item checked out at a college library is a math book, and let B be the event that the next item checked out is a history book. Suppose that P(A) = .40 and P(B) = .50. a. Why is it not the case that P(A) + P(B) = 1? b. Calculate P( ) c. Calculate P(A B). d. Calculate P( ). ANS: a. The probabilities do not add to 1 because there are other items besides math and history books to be checked out from the library. b. P( ) = 1 – P(A) = 1 - .40 = .60 c. P(A B) = P(A) + P(B) = .40 + .50 = .90 (since A and B are mutually exclusive events) d. P( ) = P[ ] (De Morgan’s law) = 1 – P(A B) = 1 - .90 = .10 PTS: 1 6. A large company offers its employees two different health insurance plans and two different dental insurance plans. Plan 1 of each type is relatively inexpensive, but restricts the choice of providers, whereas plan 2 is more expensive but more flexible. The accompanying table gives the percentages of employees who have chosen the various plans: Dental Plan Health Plan 1 2 1 27% 14% 2 24% 35% Suppose that an employee is randomly selected and both the health plan and dental plan chosen by the selected employee are determined. a. What are the four simple events? b. What is the probability that the selected employee has chosen the more restrictive plan of each type? c. What is the probability that the employee has chosen the more flexible dental plan? ANS: Let H1 and H2 represent the two health plans. Let D1 and D2 represent the two dental plans. a. The simple events are {H1,D1}, {H1, D2}, {H2,D1}, {H2,D2}. b. P({H1,D1}) = .27 c. P({D2}) = P({H1,D2}, {H2,D2}) = .14 + .35 = .49 PTS: 17. An Economic Department at a state university with five faculty members-Anderson, Box, Cox, Carter, and Davis-must select two of its members to serve on a program review committee. Because the work will be time-consuming, no one is anxious to serve, so it is decided that the representative will be selected by putting five slips of paper in a box, mixing them, and selecting two. a. What is the probability that both Anderson and Box will be selected? (Hint: List the equally likely outcomes.) b. What is the probability that at least one of the two members whose name begins with C is selected? c. If the five faculty members have taught for 3, 6, 7, 10, and 14 years, respectively, at the university, what is the probability that the two chosen representatives have at least 15 years’ teaching experience at the university? ANS: Outcomes: (A, B), (A, ), (A, ), (A,D), (B,A), (B, ), (B, ), (B,D), ( ,A), ( ,B), ( ), ( ,D), ( ,A), ( ,B), ( ), ( ,D), (D,A), (D,B), (D, ), (D, ) a. P[(A,B) or (B,A)] = 2 / 20 = .10 b. P(at least one C) = 14 / 20 = .70 c. P(at least 15 years) = 1 – P(at most 14 years) = 1 – P[(3,6) or (6,3) or (3,7) or (7,3) or (3,10) or (10,3) or (6,7) or (7,6)] = 1 – 8 / 20 = .60 PTS: 1 8. Student Engineers Council at an Indiana college has one student representative from each of the five engineering majors (civil, electrical, industrial, materials, and mechanical). In how many ways can a. Both a council president and a vice president be selected? b. A president, a vice president, and a secretary be selected? c. Two members be selected for the President’s Council? ANS: a. (5)(4) = 20 (5 choices for president, 4 remain for vice president) b. (5)(4)(3) = 60 c. = (No ordering is implied in the choice) PTS: 1 9. A real estate agent is showing homes to a prospective buyer. There are ten homes in the desired price range listed in the area. The buyer has time to visit only four of them. a. In how many ways could the four homes be chosen if the order of visiting is considered? b. In how many ways could the four homes be chosen if the order is disregarded? c. If four of the homes are new and six have previously been occupied and if the four homes to visit are randomly chosen, what is the probability that all four are new? (The same answer results regardless of whether order is considered.) ANS: a. (10)(9)(8)(7) = 5040 b. =c. P(all 4 are new) = (# of ways of visiting all new)/(# of ways of visiting) = = PTS: 1 10. An experimenter is studying the effects of temperature, pressure, and type of catalyst on yield from a certain chemical reaction. Three different temperatures, four different pressures, and five different catalysts are under consideration. a. If any particular experimental run involves the use of a single temperature, pressure, and catalyst, how many experimental runs are possible? b. How many experimental runs are there that involve use of the lowest temperature and two lowest pressures? ANS: a. b. PTS: 1 11. A certain sports car comes equipped with either an automatic or a manual transmission, and the car is available in one of four colors. Relevant probabilities for various combinations of transmission type and color are given in the accompanying table. Color Transmission Type White Blue Black Red A .13 .10 .11 .11 M .15 .07 .15 .18 Let A = (automatic transmission), B = {black}, and C = {white}. a. Calculate P(A), P(B), and P(A B). b. Calculate both P(A|B) and P(B|A), and explain in context what each of these probabilities represents. ANS: a. P(A) = .13 + .10 + .11 + .11 =.45, P(B) = .11 + .15 = .26 P(A B) = .11 b. P(A|B) = Knowing that the car is black, the probability that it has an automatic transmission is .4231. P(B|A) = Knowing that the car has an automatic transmission, the probability that it is black is .2444. PTS: 1 12. Consider the following information: where A = {Visa Card}, B = {MasterCard}, P(A) = .5, P(B) = .4, and P(A B) = .25. Calculate each of the following probabilities. a. P(B|A)b. P( |A) c. P(A|B) d. P( |B) e. Given that an individual is selected at random and that he or she has at least one card, what is the probability that he or she has a Visa card? ANS: a. P(B|A) = b. P( |A) = c. P(A|B) = d. P( |B) = e. P(A|A B) = PTS: 1 13. A certain shop repairs both audio and video components. Let A denote the event that the next component brought in for repair is an audio component, and let B be the event that the next component is a compact disc player (so the event B is contained in A). Suppose that P(A) = .625 and P(B) = .05. What is P(B/A)? ANS: Since B is contained in A, A B = B, then P(B|A) = = PTS: 1 14. At a certain gas station, 40% of the customers use regular unleaded gas ( ), 35% use extra unleaded gas ( ), and 25% use premium unleaded gas ( ). Of those customers using regular gas, only 30% fill their tanks (event B). Of those customers using extra gas, 60% fill their tanks, whereas of those using premium, 50% fill their tanks. a. What is the probability that the next customer will request extra unleaded gas and fill the tank? b. What is the probability that the next customer fills the tank? c. If the next customer fills the tank, what is the probability that regular gas is requested? Extra gas? Premium gas? ANS: Therefore,a. P = .21 b. P(B) = c. P( PTS: 1 15. Suppose that the proportions of blood phenotypes in a particular population are as follows: A B AB O .42 .10 .04 .44 Assuming that the phenotypes of two randomly selected individuals are independent of one another, what is the probability that both phenotypes are O? What is the probability that the phenotypes of two randomly selected individuals match? ANS: Using subscripts to differentiate between the selected individuals, P( ) = P( ) P( ) = (.44)(.44) = .1936P(two individuals match) = P( ) + P( ) + P( ) + P( ) = . PTS: 1 16. Two pumps connected in parallel fail independently of one another on any given day. The probability that only the older pump will fail is .15, and the probability that only the newer pump will fail is .05. What is the probability that the pumping system will fail on any given day (which happens if both pumps fail)? ANS: Let = older pump fails, = newer pump fails, and x = P( ). The P( ) = .15 + x, P( ) = .05 + x, and x = P( ) = P( The resulting quadratic equation, has roots x = .0095 and x = .7905. Hopefully the smaller root is the actual probability of system failure. PTS: 1 17. Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the following probabilities. a. P(all of the next three vehicles inspected pass) b. P(at least one of the next three inspected fail) c. P(exactly one of the next three inspected passes) d. P(at most one of the next three vehicles inspected passes) e. Given that at least one of the next three vehicles passes inspection, what is the probability that all three pass? ANS: P(pass) = .70 a. P(three pass) = (.70)(.70)(.70) = .343b. P(at least one fails) = 1 – P(all pass) = 1 - .343 = .657 c. P(exactly one passes) = (.70)(.30)(.30) + (.30)(.70)(.30) + (.30)(.30)(.70) = .189 d. P (at most one passes) = P(0 passes) + P(one passes) = + .189 = .216 e. P(3 pass | 1 or more pass) = PTS: 1CHAPTER 3 - Discrete Random Variables and Probability Distributions SHORT ANSWER 1. Three automobiles are selected at random, and each is categorized as having a diesel (S) or nondiesel (F) engine (so outcomes are SSS, SSF, etc.). If X = the number of cars among the three with diesel engines, list each outcome in S and its associated X value. ANS: Outcome FFF SFF FSF FFS FSS SFS SSF SSS X 3 2 2 2 1 1 1 0 PTS: 1 2. Let X = the number of nonzero digits in a randomly selected zip code. What are the possible values of X? Give three possible outcomes and their associated X values. ANS: In my perusal of a zip code directory, I found no 00000, nor did I find any zip codes with four zeros, a fact which was not obvious. Thus possible X values are 2, 3, 4, 5 (and not 0 or 1). X = % for the outcome 15213, X=4 for the outcome 44074, and X=3 for 94322. PTS: 1 3. If the sample space is an infinite set, does this necessarily imply that any random variable X defined from S will have an infinite set of possible values? If yes, say why. If no, give an example. ANS: No. In the experiment in which a coin is tossed repeatedly until an H results, let Y = 1 if the experiment terminates with at most 5 tosses and Y = 0 otherwise. The sample space is infinite, yet Y has only two possible values. PTS: 1 4. An automobile service facility specializing in engine tune-ups knows that 50% of all tune-ups are done on four-cylinder automobiles, 40% on six-cylinder automobiles, and 10% on eight-cylinder automobiles. Let X = the number of cylinders on the next car to be tuned. What is the pmf of X? ANS: x 4 6 8 P(x) .50 .40 .10 PTS: 1 5. The n candidates for a store manager have been ranked 1,2,3,…,n. Let X = the rank of a randomly selected candidate, so that X has pmf(this is called the discrete uniform distribution). Compute E(X) and V(X) using the shortcut formula. [Hint: The sum of the first n positive integers is n(n + 1)/2, whereas the sum of their squares is n(n + 1)(2n + 1)/6.] ANS: PTS: 1 6. A chemical supply company currently has in stock 100lb of a certain chemical, which it sells to customers in 5-lb lots. Let X = the number of lots ordered by a randomly chosen customer, and suppose that X has pmf x 1 2 3 4 P(x) .2 .3 .3 .2 Compute E(X) and V(X). Then compute the expected number of pounds left after the next customer’s order is shipped, and the variance of the number of pounds left. (Hint: The number of pounds left is a linear function of X.) ANS: Each lot weighs 5 lbs, so weight left = 100 – 5x. Thus the expected weight left is 100 – 5E(X) = 87.5, and the variance of the weight left is V(100 – 5X) = V(-5X) = 25V(x) = 26.25. PTS: 1 7. Twenty-five percent of all telephones of a certain type are submitted for service while under warranty. Of these, 60% can be repaired whereas the other 40% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty? ANS: Let S represent a telephone that is submitted for service while under warranty and must be replaced. Then p = P(S) = P(replaced | submitted) P(submitted) = (.40)(.25) = .10. Thus, X, the number among the company’s 10 phones that must be replaced, has a binomial distribution with PTS: 1 8. A geologist has collected 10 specimens of basaltic rock and 10 specimens of granite. The geologist instructs a laboratory assistant to randomly select 15 of the specimens for analysis. a. What is the pmf of the number of granite specimens selected for analysis? b. What is the probability that all specimens of one of the two types of rock are selected for analysis? c. What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value?ANS: a. Possible values of X are 5,6,7,8,9,10. (In order to have less than 5 of the granite, there would have to be more than 10 of the basaltic). Following the same pattern for the other values, we arrive at the pmf, in table form below. x 5 6 7 8 9 10 P(x) .0163 .1354 .3483 .3483 .1354 .0163 b. P(all 10 of one kind or the other) = P(X = 5) + P(X = 10) = .0163 + .0163 + .0326 c. E(X) = n . ( M / N) = 15 (10 / 20) = 7.5; V(X) = (5/19) (7.5) [1 – (10/20)] = .9869; = .9934 7.5 .9934 = (6.5066, 8.4934), so we want P( X = 7) + P( X = 8) = .3483 + .3483 = .6966 PTS: 1 9. A family decides to have children until it has three children of the same gender. Assuming P(B) = P(G) = .5, what is the pmf of X = the number of children in the family? ANS: The only possible values of X are 3,4, and 5. P(3) = P(X = 3) = P(first 3 are B’s or first 3 are G’s) = P(4) = P(two among the 1st three are B’s and the 4th is a B) + P(two among the 1st three are G’s and the 4th is a G) = P(5) = 1 – p(3) – p(4) = .375 PTS: 1 10. Three brothers and their wives decide to have children until each family has two female children. What is the pmf of X = the total number of male children born to the brothers? What is E(X), and how does it compare to the expected number of male children born to each brother? ANS: This is identical to an experiment in which a single family has children until exactly 6 females have been born(since p = .5 for each of the three families), so p(x) = nb(x;6,5) and E(X) = 6 (=2+2+2, the sum of the expected number of males born to each one.) PTS: 1 11. Suppose the number X of tornadoes observed in Kansas during a 1-year period has a Poisson distribution with a. Compute b. Compute c. Compute ANS:a. b. c. PTS: 1 12. Assume that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene? Use this distribution to calculate the approximate probability that a. Between 6 and 9 (inclusive) carry the gene. b. At least 10 carry the gene. ANS: a. b. PTS: 1 13. The number of tickets issued by a meter reader for parking-meter violations can be modeled by a Poisson process with a rate parameter of five per hour. a. What is the probability that exactly three tickets are given out during a particular hour? b. What is the probability that at least three tickets are given out during a particular hour? c. How many tickets do you expect to be given during a 45-min period? ANS: a. b. c. Tickets are given at the rate of 5 per hour, so for a 45minute period the rate is which is also the expected number of tickets in a 45 minute period. PTS: 1 14. Automobiles arrive at a vehicle equipment inspection station according to a Poisson process with rate = 10 per hour. Suppose that with probability .5 an arriving vehicle will have no equipment violations. a. What is the probability that exactly ten arrive during the hour and all ten have no violations? b. For any fixed what is the probability that y arrive during the hour, of which ten have no violations? c. What is the probability that ten “no-violation” cars arrive during the next hour? [Hint: Sum the probabilities in part (b) from y = 10 to ] ANS: a. = (.)(.125) =. b. P(y arrive and exactly 10 have no violations) = P(exactly 10 have no violations / y arrive). P(y arrive)c. P(exactly 10 without a violation) = In fact, generalizing this argument shows that the number of “no-violation” arrivals within the hour has a Poisson distribution with parameter 5; the 5 results from PTS: 1 15. A mail-order computer business has five telephone lines. Let X denote the number of lines in use at a specified time. Suppose the pmf of X is as given in the accompanying table. x 0 1 2 3 4 5 P(x) .10 .15 .20 .25 .22 .08 Calculate the probability of each of the following events. a. {at most 3 lines are in use} b. {fewer than 3 lines are in use} c. {at least 3 lines are in use} d. {between 2 and 5 lines, inclusive, are in use} e. {between 2 and 4 lines, inclusive, are not in use} f. {at least 4 lines are not in use} ANS: a. b. c. d. e. f. PTS: 1 16. Suppose that in one area in California, 40% of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random; let X denote the number among the four who have earthquake insurance. a. Find the probability distribution of X. [Hint: Let S denote a homeowner who has insurance and F one who does not. Then one possible outcome is SFSS, with probability (.3)(.7)(.3)(.3) and associated X value 3. There are 15 other outcomes.] b. What is the most likely value for X? c. What is the probability that at least two of the four selected have earthquake insurance? ANS: a. x Outcomes P(x)0 FFFF =.1296 1 FFFS, FFSF, FSFF, SFFF =.3456 2 FFSS, FSFS, SFFS, FSSF, SFSF, SSFF =.3456 3 FSSS, SFSS, SSFS, SSSF =.1536 4 SSSS =.0256 b. P(x) is largest for x = 1 and 2 c. P( ) = P(2) + P(3) + P(4) =.5248 PTS: 1 17. An insurance company offers its policyholders a number of different payment options. For a randomly selected policyholder, let X = the number of months between successive payments. The cdf of X is as follows: a. What is the pmf of X? b. Using just the cdf, compute c. Using just the pmf, compute P(X>6). ANS: a. Possible X values are those values at which F(x) jumps, and the probability of any particular value is the size of the jump at that value. Thus we have: x 1 3 4 6 12 P(x) .30 .10 .05 .15 .40 b. , c. PTS: 1 18. The pmf for X = the number of major defects on a randomly selected gas stove of a certain type is x 0 1 2 3 4 P(x) .10 .15 .45 .25 .05 Compute the following: a. E(X) b. V(X) directly from the definition c The standard deviation of X d. V(X) using the shortcut formula ANS: a.b. = .40 +.15+0.0 +.25 +.20 = 1.0 c. d. PTS: 1 19. An appliance dealer sells three different models of upright freezers having 13.5, 15.9, and 19.1 cubic feet of storage space, respectively. Let X = the amount of storage space purchased by the next customer to buy a freezer. Suppose that X has pmf x 13.5 15.9 19.1 P(x) .2 .4 .4 a. Compute b. If the price of a freezer having capacity X cubic feet is 25X – 8.5, what is the expected price paid by the next customer to buy a freezer? c. What is the variance of the price 25X – 8.5 paid by the next customer? d. Suppose that although the rated capacity of a freezer is X, the actual capacity is What is the expected actual capacity of the freezer purchased by the next customer? ANS: a. b. c. d. PTS: 1 20. Compute the following binomial probabilities directly from the formula for b(x;n,p). a. b(3; 8, .7) b. b(5; 8, .7) c. d. ANS: a. b. c. d. PTS: 121. Use the cumulative binomial probabilities table available in your text to obtain the following probabilities: a. B(4; 10, .4) b. b(4; 10, .4) c. b(6; 10, .6) d. when Bin(10,.4) e. when Bin(10,.3) f. when Bin(10,.7) g. when Bin(10,.3) ANS: a. B(4;10,.4) = .633 b. b(4;10,.4) = B(4;10,.4) – B(3;10,.4) = .251 c. b(6;10,.6) = B(6;10,.6) – B(5;10,.6) = .251 d. e. f. g. PTS: 1 22. Suppose that only 25% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible. What is the probability that, of 20 randomly chosen drivers coming to an intersection under these conditions, a. At most 6 will come to a complete stop? b. Exactly 6 will come to a complete stop? c. At least 6 will come to a complete stop? d. How many of the next 20 drivers do you expect to come to a complete stop? ANS: Let S = comes to a complete stop, so p = .2, n = 20 a. b. c. d. We expect 5 of the next 20 to stop. PTS: 1 23. Each of 12 refrigerators of a certain type has been returned to a distributor because of the presence of a high-pitched oscillating noise when the refrigerator is running. Suppose that 5 of these 12 have defective compressors and the other 7 have less serious problems. If they are examined in random order, let X = the number among the first 6 examined that have a defective compressor. Compute the following: a. (X = 1) b. ANS: a.b. PTS: 1 24. The number of pumps in use at both a six-pump station and a four-pump station will be determined. Give the possible values for each of the following random variables. a. T = the total number of pumps in use b. X = the difference between the numbers in use at stations 1 and 2 c. U = the maximum number of pumps in use at either station d. Z = the number of stations having exactly two pumps in use ANS: a. T = total number of pumps in use at both stations. Possible values: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 b. X: -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 c. U: 0, 1, 2, 3, 4, 5, 6 d. Z: 0, 1, 2 PTS: 1Chapter 4 - Continuous Random Variables and Probability Distributions SHORT ANSWER 1. Let X denote the amount of time for which a book on 2-hour reserve at a college-library is checked out by a randomly selected student and suppose that X Calculate the following probabilities: a. b. ANS: a. b. PTS: 1 2. A college professor always finishes his lectures within 2 minutes after the bell rings to end the period and the end of the lecture. Let X = the time that elapses between the bell and the end of the lecture and suppose the pdf of X is a. Find the value of k. [Hint: Total area under the graph of f(x) is 1.] b. What is the probability that the lecture ends within 1minutes of the bell ringing? c. What is the probability that the lecture continues beyond the bell for between 60 and 90 seconds? d. What is the probability that the lecture continues for at least 90 seconds beyond the bell? ANS: a. b. c. d. PTS: 1 3. The time X (minutes) for a lab assistant to prepare the equipment for a certain experiment is believed to have a uniform distribution with A = 20 and B = 30. a. Write the pdf of X and sketch its graph. b. What is the probability that preparation time exceeds 27 minutes? c. Find the preparation mean time, then calculate the probability that preparation is within 2 minutes of the mean time?d. For any a such that 20 < a < a + 2 < 30, what is the probability that preparation time is between a and a + 2 minutes? ANS: a. b. c. 25 2 is from 23 to 27 minutes: d. since the interval has length 2. PTS: 1 4. “Time headway” in traffic flow is the elapsed time between the time that one car finishes passing a fixed point and the instant that the next car begins to pass that point. Let X = the time headway for two randomly chosen consecutive cars on a freeway during a period of heavy flow. The following pdf of X is What is the probability that the time headway is a. At most 6 seconds? b. At least 6 seconds? c. At most 5 seconds? d. Between 5 and 6 seconds? ANS: a. b. c. d. PTS: 1 5. The cdf of checkout duration X for a book on a 2-hour reserve at a college library is given by:Use this cdf to compute the following: a. b. c. d. The median checkout duration e. ANS: a. b. c. d. e. PTS: 1 6. Let X denote the amount of space occupied by an article placed in a packing container. The pdf of X is a. Obtain the cdf of X.. b. What is c. What is ? d. What is the 75th percentile of the distribution? e. Compute f. What is the probability that X is within 1 standard deviation of its mean value? ANS: a. Therefore,b. c. d. The 75th percentile is the value of x for which F(x) = .75 e. f. =.8465 - .1602 = .6863 PTS: 1 7. Let X have a uniform distribution on the interval [a, b]. a. Obtain an expression for the (100p) th percentile. b. Compute E(X), V(X), and . c. For n a positive integer, compute . ANS: a. b. c. PTS: 1 8. Let X be the temperature in at which a certain chemical reaction takes place, and let Y be the temperature in (so Y = 1.8X + 32). a. If the median of the X distribution is , show that 1.8 + 32 is the median of the Y distribution. b. How is the 90th percentile of the Y distribution related to the 90th percentile of the X distribution? Verify your conjecture. c. More generally, if Y = aX + b, how is any particular percentile of the Y distribution related to the corresponding percentile of the X distribution? ANS: a. b.c. The (100p)th percentile for Y is , verified by substituting p for .9 in the argument of b. When Y = aX + b, (i.e. a linear transformation of X), and the (100p)th percentile of the X distribution is , then the corresponding (100p)th percentile of the Y distribution is (same linear transformation applied to X’s percentile) PTS: 1 9. Let Z be a standard normal random variable and calculate the following probabilities: a. b. c. d. e. f. g. h. i. j. ANS: a. b. = c. = d. = e. = f. = g. = h. = i. = j. PTS: 1 10. In each case, determine the value of the constant c that makes the probability statement correct. a. b. c. d. e. ANS: a. b. c.d. e. PTS: 1 11. If X is a normal random variable with mean 85 and standard deviation 10, compute the following probabilities by standardizing. a. b. c. d. e. f. ANS: a. b. c. d. e. f. = = .6915 - .0668 = .6247 PTS: 1 12. The distribution of resistance for resistors of a certain type is known to be normal, with 10% of all resistors having a resistance exceeding 10.634 ohms, and 5% having a resistance smaller than 9.7565 ohms. What are the mean value and standard deviation of the resistance distribution? ANS: Since 1.28 is the 90th z percentile , the given information implies that , from which PTS: 1 13. Let X have a binomial distribution with parameters n = 25 and p. Calculate each of the following probabilities using the normal approximation (with the continuity correction) for the cases p = .5 and .6, and compare to the exact probabilities calculated from the cumulative binomial probabilities table available in your text. a. b. c. ANS: P: .5 .612.5 15 2.50 2.45 a. Binomial Normal Approximation P .5 12.5 2.5 .770 = .7689 .6 15 2.45 .562 = .5671 b. Binomial Normal Approximation P .5 12.5 2.5 .212 = .2119 .60 15 2.45 .034 = .0329 c. Binomial Normal Approximation P .5 12.5 2.5 .212 = ..2119 .6 15 2.45 .586 = .5793 PTS: 1 14. Suppose only 40% of all drivers in Florida regularly wear a seatbelt. A random sample of 500 drivers is selected. What is the probability that a. Between 170 and 220 (inclusive) of the drivers in the sample regularly wear a seatbelt? b. Fewer than 175 of those in the sample regularly wear a seatbelt? ANS: a. b. PTS: 1 15. Let X = the time between two successive arrivals at the drive-up window of a local bank. If X has an exponential distribution with = 1 (which is identical to a standard gamma distribution with =1), compute the following: a. The expected time between two successive arrivals. b. The standard deviation of the time between two successive arrivals. c. . d. . ANS:a. =1 b. =1 c. = .9933 d. = .043 PTS: 1 16. Let X have a standard gamma distribution with Evaluate the following: a. b. c. d. e. f. ANS: a. b. c. d. e. f. PTS: 1 17. Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime X (in weeks) has a gamma distribution with mean of 40 and variance of 320. a. What is the probability that a transistor will last between 1 and 40 weeks? b. What is the probability that a transistor will last at most 40 weeks? Is the median of the lifetime distribution less than 40? Why or why not? ANS: a. b. so while the mean is 40, the median is less than 40. This is the result of the positive skew of the gamma distribution. PTS: 1 18. The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters Compute the following: a. E(X) and V(X) b. c. ANS:a. , and b. c. PTS: 1 19. Let X = hourly median power (in decibels) of received radio signals transmitted between two cities. It is believed that the lognormal distribution provides a reasonable probability model for X. If the parameter values are calculate the following: a. The mean value and standard deviation of received power b. The probability that received power is between 50 and 250 dB c. The probability that X is less than its mean value. Why is this probability not .5? ANS: a. and b. c. The lognormal distribution is not a symmetric distribution. PTS: 1 20. Suppose the proportion X of surface area in a randomly selected quadrat that is covered by a certain plant has a standard beta distribution with a. Compute E(X) and V(X). b. Compute c. Compute d. What is the expected proportion of the sampling region not covered by the plant? ANS: a. b. so c. d.PTS: 1 21. Stress is applied to a 20-in. steel bar that is clamped in a fixed position at each end. Let X = distance from the left end at which the bar snaps. Suppose Y/20 has a standard beta distribution with E(Y) = 10 and V(Y) = a. What are the parameters of the relevant standard beta distribution? b. Compute c. Compute the probability that the bar snaps more than 2 in. from where you expect it to. ANS: a. after some algebra. b. c. We expect it to snap at 10, so PTS: 1 22. Consider the following ten observations on bearing lifetime (in hours): 152.7 172.0 172.5 173.3 193.0 204.7 216.5 234.9 262.6 422.6 Construct a normal probability plot and comment on the plausibility of the normal distribution as a model for bearing lifetime.ANS: The z percentiles and observations are shown on the normal probability plot below. The accompanying plot is quite straight except for the point corresponding to the largest observation.This observation is clearly much larger than what would be expected in a normal random sample. Because of this outlier, it would be inadvisable to analyze the data using any inferential method that depended on assuming a normal population distribution. PTS: 1 23. Construct a normal probability plot for the following sample of observations on coating thickness for low-viscosity paint. Would you feel comfortable estimating population mean thickness using a method that assumed a normal population distribution? .83 .98 1.06 1.14 1.20 1.25 1.29 1.40 1.48 1.49 1.51 1.62 1.65 1.71 1.76 1.83 ANS: The z percentile values and observations are shown on the normal probability plot below. The accompanying probability plot is reasonably straight, and thus it would be reasonable to use estimating methods that assume a normal population distribution. PTS: 1 24. The propagation of fatigue cracks in various aircraft parts has been the subject of extensive study in recent years. The accompanying data consists of propagation lives to reach crack size in fastener holes intended for use in military aircraft..736 .863 .865 .913 .915 .937 .983 1.007 1.011 1.064 1.109 1.132 1.140 1.153 1.253 1.394 Construct a normal probability plot for the data. Does it appear plausible that propagation life has a normal distribution? Explain. ANS: The z percentile and observations are shown on the normal probability plot below. The accompanying probability plot is very straight, suggesting that an assumption of population normality is extremely plausible. PTS: 1Chapter 5 - Joint Probability Distributions and Random Samples SHORT ANSWER 1. Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressure in each tire is a random variable—X for the right tire and Y for the left tire, with joint pdf a. What is the value of K? b. What is the probability that both tires are underfilled? c. What is the probability that the difference in air pressure between the two tires is at most 2 psi? d. Determine the (marginal) distribution of air pressure in the right tire alone. e. Are X and Y independent random variables? ANS: a. b. c. = = = .3593 (after much algebra) d. e. is obtained by substituting y for x in (d); clearly are not independent. PTS: 1 2. Let X denote the number of brand X VCRs sold during a particular week by a certain store. The pmf of X is x 0 1 2 3 4 .1 .2 .3 .25 .15 Seventy percent of all customers who purchase brand X VCRs also buy an extended warranty. Let Y denote the number of purchasers during this week who buy an extended warranty.a. What is P(X = 4, Y = 2)? [Hint: This probability equals P(Y = 2/X = 4) P(X = 4); now think of the four purchases as four trials of a binomial experiment, with success on a trial corresponding to buying an extended warranty.] b. Calculate P(X =Y). c. Determine the joint pmf of X and Y and then the marginal pmf of Y. ANS: a. b. c. For any such pair, PTS: 1 3. Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y: a. What is the probability that the lifetime X of the first component exceeds 3? b. What are the marginal pdf‖s of X and Y? Are the two lifetimes independent? Explain. c. What is the probability that the lifetime of at least one component exceeds 3? ANS: a. b. The marginal pdf of X is It is now clear that f(x,y) is not the product of the marginal pdf‖s, so the two random variables are not independent. c.PTS: 1 4. The joint pdf of pressures for right (X) and left (Y) front tires is given by . a. Determine the conditional pdf of Y given that X = x and the conditional pdf of X given that Y = y if you are given b. If the pressure in the right tire is found to be 22 psi, what is the probability that the left tire has a pressure of at least 25 psi? Compare this to c. If the pressure in the right tire is found to be 22 psi, what is the expected pressure in the left tire, and what is the standard deviation of pressure in this tire? ANS: a. b. c. =25. PTS: 1 5. An instructor has given a short test consisting of two parts. For a randomly selected student, let X = the number of points earned on the first part and Y = the number of points earned on the second part. Suppose that the joint pmf of X and Y is given in the accompanying table. p(x,y) 0 5 10 15 0 .02 .06 .02 .10 5 .04 .15 .20 .10 10 .01 .15 .14 .01 a. If the score recorded in the grade book is the total number of points earned on the two parts, what is the expected recorded score E(X + Y)? b. If the maximum of the two scores is recorded, what is the expected recorded score?ANS: a. + (0 + 5)(.06) + … + (10 + 15)(.01) = 14.10 b. = (0)(.02) + (5)(.06) +…+(15)(.01) = 9.60 PTS: 1 6. Abby and Bianca have agreed to meet for lunch between noon and 1:00 P.M. Denote Abby’s arrival time by X, Bianca’s by Y, and suppose X and Y are independent with pdf’s. What is the expected amount of time that the one who arrives first must wait for the other person? [Hint: h(X, Y ) = |X – Y|.] ANS: = PTS: 1 7. Show that if X and Y are independent random variables, then ANS: PTS: 1 8. Show that if Under what conditions will ANS: therefore Corr PTS: 1 9. A particular brand of dishwasher soap is sold in three sizes: 25oz, 40oz, and 65 oz. Twenty percent of all purchasers select a 25 oz box, fifty percent select a 40 oz box, and the remaining thirty percent choose a 65 oz box. Let denote the package sizes selected by two independently selected purchasers.a. Determine the sampling distribution of , calculate , and compare to b. Determine the sampling distribution of the sample variance ANS: 20 .50 .30 25 40 65 .20 25 .04 .10 .06 .50 40 .10 .25 .15 .30 65 .06 .15 .09 a. 25 32.5 40 45 52.5 65 .04 .20 .25 .12 .30 .09 b. 0 112.5 312.5 800 .38 .20 .30 .12 PTS: 1 10. It is known that 80% of all brand A zip drives work in a satisfactory manner throughout the warranty period (are ―success‖). Suppose that n = 10 drives are randomly selected. Let X = the number of successes in the sample. The statistic X/n is the sample proportion (fraction) of successes. Obtain the sampling distribution of this statistic. [Hint: One possible value of X/n is .3, corresponding to X = 3. What is the probability of this value (what kind of random variable is X)?] ANS: x 0 1 2 3 4 5 6 7 8 9 10 x/n 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 p(x/n) .000 .000 .000 .001 .005 .027 .088 .201 .302 .269 .107 X is a binomial random variable with p = .8. PTS: 1 11. Let X be the number of packages being mailed by a randomly selected customer at a certain shipping facility. Suppose the distribution of X is as follows: x 1 2 3 4 p(x) .4 .3 .2 .1 a. Consider a random sample of size n = 2 (two customers), and let be the sample mean number of packages shipped. Obtain the probability distribution of . b. Refer to part (a) and calculatec. Again consider a random sample of size n = 2, but now focus on the statistic R = the sample range (difference between the largest and smallest values in the sample). Obtain the distribution of R. [Hint: Calculate the value of R for each outcome and use the probabilities from part (a).] d. If a random sample of size n = 4 is selected, what is ? (Hint: You should not have to list all possible outcomes, only those for which ANS: Outcome 1,1 1,2 1,3 1,4 2,1 2,2 2,3 2,4 Probability .16 .12 .08 .04 .12 .09 .06 .03 1 1.5 2 2.5 1.5 2 2.5 3 0 1 2 3 1 0 1 2 Outcome 1,1 3,2 3,3 3,4 4,1 4,2 4,3 4,4 Probability .08 .06 .04 .02 .04 .03 .02 .01 2 2.5 3 3.5 2.5 3 3.5 4 2 1 0 1 3 2 1 2 a. 1 1.5 2 2.5 3 3.5 4 .16 .24 .25 .20 .10 .04 .01 b. c. 0 1 2 3 .30 .40 .22 .08 d. P(2,2,1,1) + P (3,1,1,1) +…+P(1,1,1,3) PTS: 1 12. A company maintains three offices in a certain region, each staffed by two employees. Information concerning yearly salaries (1000’s of dollars) is as follows: Office 1 1 2 2 3 3 Employee 1 2 3 4 5 6 Salary 19.7 23.6 20.2 23.6 15.8 19.7 a. Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary b. Suppose one of the three offices is randomly selected. Let denote the salaries of the two employees. Determine the sampling distribution of c. How does from parts (a) and (b) compare to the population mean salaryANS: a. 17.75 18.0 19.7 19.95 21.65 21.9 23.6 b. 17.75 21.65 21.9 1/3 1/3 1/3 c. all three values are the same: 20.4333 d. PTS: 1 13. The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of 500 psi. a. What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9950 and 10,250? b. If the sample size had been 15 rather than 40, could the probability requested in part (a) be calculated from the given information? ANS: a. n = 40 =.9992 - .2643 =.7349 b. According to the Rule of Thumb given in your text, n should be greater than 30 in order to apply the C.L.T., thus using the same procedure for n = 15 as was used for n = 40 would not appropriate. PTS: 1 14. The lifetime of a certain type of battery is normally distributed with mean value 12 hours and standard deviation 1 hour. There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages? ANS: X is normally distributed with Sample size n = 4. We desire the 95th percentile: 48 + (1.645)(2) = 51.29 PTS: 1 15. The number of parking tickets issued in Grand Rapids on any given weekday has a Poisson distribution with parameter What is the approximate probability thata. Between 40 and 70 tickets are given out on a particular day? (Hint: When is large, a Poisson random variable has approximately a normal distribution.) b. The total number of tickets given out during a 5-day week is between 215 and 265? ANS: a. With Y = # of tickets, Y has approximately a normal distribution with b. Here PTS: 1 16. Let represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent normal random variables with expected values respectively. a. If Calculate What is b. Using the given in part (a), calculate c. Using the given in part (a), calculate d. If calculate ANS: a. b. c. d. We wantPTS: 1 17. Suppose your waiting time for a bus in the morning is uniformly distributed on [0,5], whereas waiting time in the evening is uniformly distributed on [0,10] independent of morning waiting time. a. If you take the bus each morning and evening for a week, what is your total expected waiting time? [Hint: Define random variables and use a rule of expected value.) b. What is the variance of your total waiting time? c. What are the expected value and variance of the difference between morning and evening waiting times on a given day? d. What are the expected value and variance of the difference between morning waiting time and total evening waiting time for a particular week? ANS: Let denote morning times and denote evening times. a. b. c. d. PTS: 1 18. Three different roads feed into a particular freeway entrance. Suppose that during a fixed time period, the number of cars coming from each road onto the freeway is a random variable, with expected value and standard deviation as given in the table. Road 1 Road 2 Road 3 Expected value Standard deviation 16 24 18 a. What is the expected total number of cars entering the freeway at this point during the period? (Hint: Let b. What is the variance of the total number of entering cars? Have you made any assumptions about the relationship between the numbers of cars on the different roads? c. With denoting the number of cars entering from road I during the period, suppose that (so that the three streams of traffic are not independent). Compute the expected total number of entering cars and the standard deviation of the total.ANS: a. = 750 + 1000 + 550 = 2300. b. Assuming independence of c. =1696, with standard deviation = 41.1825 PTS: 1 19. In an area having sandy soil, 50 small trees of a certain type were planted, and another 50 trees were planted in an area having clay soil. Let X = the number of trees planted in sandy soil that survive 1 year and Y = the number of trees planted in clay soil that survive 1 year. If the probability that a tree planted in sandy soil will survive 1 year is .7 and the probability of 1-year survival in clay soil is .6, compute an approximation to (do not bother with the continuity correction). ANS: X is approximately normal with as is Y with PTS: 1Chapter 6 – Point Estimation COMPLETION 1. The objective of __________ is to select a single number such as , based on sample data, that represents a sensible value (good guess) for the true value of the population parameter, such as . ANS: point estimation PTS: 1 2. Given four observed values: would result in a point estimate for that is equal to __________. ANS: 5 PTS: 1 3. An estimator that has the properties of __________ and __________ will often be regarded as an accurate estimator. ANS: unbiasedness, minimum variance PTS: 1 4. A point estimator is said to be an __________ estimator of if for every possible value of . ANS: unbiased PTS: 1 5. The sample median and any trimmed mean are unbiased estimators of the population mean if the random sample from a population that is __________ and __________. ANS: continuous, symmetric PTS: 1 6. Among all estimators of parameter that are unbiased, choose the one that has minimum variance. The resulting is called the __________ of . ANS: minimum variance unbiased estimator (MVUE) PTS: 1 7. The standard error of an estimator is the __________ of . ANS: standard deviation PTS: 18. In your text, two important methods were discussed for obtaining point estimates: the method of __________ and the method of __________. ANS: moments, maximum likelihood PTS: 1 9. Let be a random sample from a probability mass function or probability density function f(x). For k = 1,2,3,……, the kth population moment is denoted by __________, while the kth sample moment is __________. ANS: PTS: 1 10. Let be a random sample of size n from an exponential distribution with parameter . The moment estimator of = __________. ANS: PTS: 1 11. Let be the maximum likelihood estimates (mle’s) of the parameters . Then the mle of any function h( ) of these parameters is the function of the mle’s. This result is known as the __________ principle. ANS: invariance PTS: 1 MULTIPLE CHOICE 1. Which of the following statements are true? a. A point estimate of a population parameter is a single number that can be regarded as a sensible value of . b. A point estimate of a population parameter is obtained by selecting a suitable statistic and computing its value from the given sample data. The selected statistic is called the point estimator of . c. The sample mean is a point estimator of the population mean . d. The sample variance is a point estimator of the population variance . e. All of the above statements are true. ANS: E PTS: 1 2. Which of the following statements are not true? a. The symbol is customarily used to denote the estimator of parameter and the point estimate resulting from a given sample. b. The equality is read as “the point estimator of c. The difference between and the parameter is referred to as error of estimation. d. None of the above statements is true.ANS: B PTS: 1 3. Which of the following statements are not always true? a. A point estimator is said to be an unbiased estimator of parameter if for every possible value of . b. If the estimator is not unbiased of parameter , the difference is called the bias of . c. A point estimator is unbiased if its probability sampling distribution is always “centered” at the true value of the parameter , where “centered” here means that the median of the distribution of . d. All of the above statements are true. ANS: C PTS: 1 4. Which of the following statements are not always true? a. It is necessary to know the true value of the parameter to determine whether the estimator is unbiased. b. When X is a binomial random variable with parameters n and p, the sample proportion is an unbiased estimator of p. c. When choosing among several different estimators of parameter , select one that is unbiased. d. All of the above statements are not always true. ANS: A PTS: 1 5. Which of the following statements are true if is a random sample from a distribution with mean ? a. b. c. d. e. All of the above statements are true provided that the sample size n > 30. ANS: D PTS: 1 6. Which of the following statements are true if is a random sample from a distribution with mean ? a. The sample mean is always an unbiased estimator of . b. The sample mean is an unbiased estimator of if the distribution is continuous and symmetric. c. Any trimmed mean is an unbiased estimator of if the distribution is continuous and symmetric. d. None of the above statements are true. e. All of the above statements are true. ANS: E PTS: 1 7. Which of the following statements are not true? a. Maximum likelihood estimators are generally preferable to moment estimators because of certain efficiency properties. b. Maximum likelihood estimators often require significantly mor