Exam 1 Additional Practice Solutions

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Solutions (Exam 1 Additional Practice) 1. Suppose that the observational units in a study are the 21 purchases that I made on amazon.com in the year 2018. Identify each of the following as a categorical variable, a quantitative variable, or not a variable. a) How much did I spend on the purchase? quantitative b) Was the purchase shipped to me or to someone else? categorical c) Do I tend to spend more on purchases sent to others than on purchases sent to me? not a variable (research question) d) Did the purchase include free shipping? categorical e) What was the average price of these purchases? not a variable (summary statistic) 2. The National Retail Federation conducted a national survey of 8526 consumers on September 1-9, 2009. Among the findings reported were that: • 29.6% of those surveyed said that the state of the U.S. economy would affect their Halloween spending plans; • the average amount that the respondents said they expect to spend on Halloween is $56.31. a) Identify the observational units in this study. each consumer b) Identify a quantitative variable in this study. amount spent on Halloween c) Are the numbers 29.6% and $56.31 parameters or statistics? Statistics (as the numbers are based on the sample) 3. In the following scenario describe the population, sample, observational unit, variable(s), parameter of interest, and the statistic which could be used to estimate the parameter. Is (are) the variable(s) quantitative or categorical? “A study is to be done on the social interaction skills of California second graders. A random sample of 50 second graders is taken. A psychologist watches each of the selected children for a period of 3 days, and records the number of positive interactions the child had with other children in his/her class at school.” population: all California second graders sample: 50 California second graders OU: each CA second grader variable: number of positive interactions with other children in class - quantitative parameter: average number of positive interactions for all CA second graders statistic: average number of positive interactions for the sample of 50 CA second graders

4. In a May 2013 survey of a random sample of 2,076 U.S. cell phone owners, the Pew Research Center found that 52% use their cell phones to send and/or receive email. Describe the following, in the context of this study: (a) Observational unit: each U.S. cell phone owner (b) Variable: whether or not use cell phone to send and/or receive email (categorical/binary) (c) Parameter of interest? Proportion of % of all U.S. cell phone owners to use their cell phones to send and/or receive email. (d) Which of the following is the best description of the population of interest in this study? A. The proportion of all U.S. cell phone owners who use their cell phones to send and/or receive email. B. All U.S. cell phone owners who use their cell phones to send and/or receive email. C. All U.S. cell phone owners. D. All U.S. adults. 5. The following graph summarizes data collected on a sample of 432 domestic wines. (a) Identify the observational unit in this study. each domestic wine (b) What variable(s) is/are graphed? Be sure to note the variable type. amount of lead (in parts per billion) - quantitative (c) For these data, how will the mean and median compare? Is one bigger than the other or both are the same? Explain how you know. mean will be larger than median, because distribution is skewed to the right. (d) TRUE or FALSE: More than 50% of the sampled wines have a lead content of less than 50 parts per billion. TRUE, because 300/432 is more than 50%

| 6. Automotive News tracks and reports monthly inventories for both domestic and foreign auto manufacturers. The table below illustrates inventories for the Big Three U.S. manufacturers in November of 1998. Manufacturer Cars Trucks Daimler Chrysler 178,300 381,900 Ford 321,300 467,300 General Motors 550,500 433,900 (a) What percentage of the Big Three’s inventory is cars? 45% (b) What percentage of the inventory in the table above comes from Ford? 33.8% (c) What percentage of the General Motors inventory is trucks? 44.1% (d) Of the car inventory, what percentage comes from Ford? 30.6% (e) What percentage of the inventory is manufactured by Ford and is a truck? 20% (f) If a randomly selected automotive from these is a truck, what’s the probability it is made by Ford? P(Ford) = 0. 364 (g) If a randomly selected automotive from these is made by Ford, what’s the probability it is a truck? P(Truck|Ford) = 0. 593 7. Based on the following statistics, describe the distribution of the data: 𝑥𝑥̅ = 13 . 7; median = 17.9; sd = 3.3 distribution is skewed to the left, mean < median. 8. What is the standard deviation of the following data set: 6, 6, 6, 6, 6, 6, 6? 0 (no variability) 9. Under what circumstances will we have a negative valued standard deviation? never 10. Suppose that among the pre-owned cars available in a large dealership, 80% have air conditioning, 70% have a CD player, and 65% have both. CD no CD total AC 0.65 0.15 0.80 no AC 0.05 0.15 0.20 Total 0.70 0.30 1.0 (a) Determine the proportion of these cars that have neither air conditioning nor a CD player. (As always, show your method of solution.) P(no AC and no CD) = 0.15, see table above (b) Determine the conditional probability that a randomly selected car has a CD player, given that it has air conditioning. P ( CD|AC ) = 0.65/0.80 = 0.8125 (c) Are the events randomly selected car has air conditioning and randomly selected car has a CD player independent? Justify your answer. no, P ( CD|AC ) = 0 . 8125 = P ( CD ) = 0 . 70. Having AC increases the car’s chance of also having a CD player. (d) Are the events randomly selected car has air conditioning and randomly selected car has a CD player disjoint? Explain briefly. No, can have both. P(CD and AC) = 0.65 = 0

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11. Suppose that Jose has applied for a job with three different companies. He thinks that he has a 40% chance of getting a job offer from company A, a 60% chance of getting a job offer from company B, and a 90% of getting a job offer from company C. Furthermore, he believes that whether or not he receives a job offer from any company is independent of whether or not he receives a job offer from any other company. a) What is the probability that Jose receives a job offer from all three companies? P(all 3 offer) = 0 . 4 ∗ 0 . 6 ∗ 0 . 9 = 0 . 216 b) What is the probability that Jose receives a job offer from at least one of these three companies? P(at least 1 offer) = 1 − (1 − . 4) ∗ (1 − . 6) ∗ (1 − . 9) = 0 . 976 12. A construction company must deal with the possibility of union strikes on many of their contract jobs. By studying their records for the last 5 years, the company management came up with the following results. The probability the company finishes any contract job on time is 0.70. The probability there is a strike on the job is 0.20. The probability that a particular contract job is not finished on time and there is not a strike is 0.15. finish on time not finish on time total strike 0.05 0.15 0.20 not strike 0.65 0.15 0.80 Total 0.70 0.30 1.0 a. Find the probability there is not a strike on a contract job. P(no strike on contract)= 0 . 80 b. What is the probability of finishing a job on time or having a strike? P(on time or strike) = 0 . 05+0 . 15+0 . 65 = 0 . 85; c. What is the probability they finish the job on time and do not have a strike? P(on time and no strike) = 0 . 65 d. What is the probability of finishing the job on time if they have a strike? P(on time|strike) = 0 . 05 /0. 20 = 0 . 25 13. I have 2 students writing different computer programs for me so that the correct analysis of my data depends on at least one of the programs working properly. The probability the first program works is 0.80 and the probability the second program works is 0.70. The students know each other and have spent some time working together on their programs. The probability that both programs work is 0.60. first works first does not work total second works 0.60 0.10 0.70 second does not work 0.20 0.10 0.30 Total 0.80 0.20 1.0 a. What is the probability my data get analyzed correctly? P(correct)= 0.6 + 0 . 1 + 0 . 2 = 0 . 90 b. What is the probability that neither program works to correctly analyze my data? P(neither works) = 0 . 10; c. What is the probability that the first program works and the second does not? P(1 st works and 2 nd doesn’t) = 0.20 d. What is the probability that the second program works and the first does not? P(1 st doesn’t work and 2 nd does) = 0.10 e. What is the probability that exactly one of the programs works? P(1 works) =0 . 1 + 0 . 2 = 0 . 3

14. A local pizza place has determined from last year’s sales that 12,753 orders were for dine-in; 5,893 were for delivery; and 3,122 were take-out orders. a. If we select a customer at random from last year’s sales, what is the probability s/he had a take- out order? P(takeout) = 3122 / (12753 + 5893 + 3122) = 0 . 1434 b. Suppose we select two customers at random, list all of the outcomes in the sample space indicating the type of order for both customers. II, ID, IT, DI, DD, DT, TI, TD, TT c. Suppose we select two customers at random, what is the probability both were take-out? P(2 takeout) = (0 . 1434) 2 = . 0206 d. Suppose we select three customers at random, what is the probability at least one was take-out? P(of 3 at least 1 takeout) = 1 − (1 − 0 . 1434) 3 = 0 . 371 15. Suppose that 15% of spam messages that you receive contain the word “greeting” and 1% of non- spam messages that you receive contain the word “greeting.” Also suppose that 25% of the messages that you receive are spam. Consider selecting one of your email messages at random and define the following events: G = {message contains “greeting”}, M = {message is spam}. spam not spam total greeting (0.15)(0.25) = 0.0375 (0.01)(0.75) = 0.0075 0.045 not greeting 0.2125 0.7425 0.955 Total 0.25 0.75 1.0 (a) Express the given values 0.15, 0.01, and 0.25 as probabilities in terms of these events. P(M) = 0.25, P(G|M ) = 0.15, P(G|notM ) = 0 . 01 (b) Determine the probability that a randomly selected e-mail message contains the word “greeting.” P(G) = 0.045 (c) Given that a randomly selected e-mail message contains the word “greeting,” what is the probability that the message is spam? P(spam|G) = 0 . 0375 / 0 . 045 = 0 . 833 16. Experts project that this year 20% of all taxpayers will file a tax return with mistakes. The IRS itself is not perfect. IRS auditors claim there is an error on the return if it is mistake free about 10% of the time. The audits also indicate no error on a return when in fact it has mistakes about 30% of the time. taxpayer made mistakes taxpayer did NOT make mistakes total IRS claims errors 0.14 (0.10)(0.80) = 0.08 0.22 IRS claims NO errors (0.30)(0.2) = 0.06 0.72 0.78 Total 0.20 0.80 1.0 a. Suppose we select a tax return at random. What is the probability it was filed with no mistakes? P(no mistakes) = 0.80 b. Suppose we select a tax return at random. What is the probability the IRS will say it has an error? P(error) = 0.22 What is the probability the IRS will say it has no errors? P(no errors)= 0.78 c. Suppose we select a tax return at random. What is the probability it has a mistake if the IRS says it has an error? P(mistake|error) = 0.14/0. 22 = 0 . 636

17. Suppose in a certain state a license plate can have four (0-9) numbers followed by three letters. (a) If duplication of numbers and letters is allowed, how many different license plates are possible? (10)(10)(10)(10)(26)(26)(26) = 10 4 ∗ 26 3 = 175,760,000 (b) If duplication of numbers is not allowed, how many different license plates are possible? (10)(9)(8)(7)(26)(26)(26) = 88,583,040 (c) Assuming duplication is not allowed, what is the probability of getting the plate 1234ABC or 9876ZYX? P(1 of the plates listed) = 1 88 , 583 , 040 + 1 88 , 583 , 040 = 0.000000023 (d) Assuming duplication is allowed, what is the probability of getting the plate 1234ABC or 9876ZYX? P(1 of the plates listed) = 1 175 , 760 , 000 + 1 175 , 760 , 000 = 0.000000011 (e) Given your answers for (c) and (d), interpret your result (i.e., why does this make sense)? In either case, the probability of getting one of those particular plates is essentially zero as there are so many possible arrangements available. Your odds are a bit better when duplication of letters is not allowed as it reduces the total number of possible arrangements. 18. Suppose that a photographer wants a picture for an upcoming animal calender. In the picture, the photographer wants 3 cats and 3 dogs but isn’t worried about how they are grouped. How many different photos (assuming no change to background, etc) can be formed when there are 10 cats and 6 dogs to choose from? � 10 3 � × � 6 3 � = 2,400 19. In a process that manufactures aluminum cans, the probability that a can has a flaw on top is 0.03, the probability a can has a flaw on its side is 0.02, and the probability it has a flaw on both the top and side is 0.01. Flaw on top no flaw on top total Flaw on side 0.01 0.01 0.02 no flaw on side 0.02 0.96 0.98 Total 0.03 0.97 1.00 a) What is the probability a randomly chosen can has at least one flaw (top or side or both)? P(at least 1 flaw) = 0.04 b) What is the probability a randomly chosen can has no flaws at all? P(no flaws) = 0.96 c) Let F = # of flaws on a randomly chosen can. Then F can take on the values 0, 1 or 2. Create the table for the probability distribution of F. F 0 1 2 P(F=f) 0.96 0.03 0.01 d) Determine the expected number of flaws on a can. E(F) = 0*0.96 + 1*0.03 + 2*0.01 = 0.05 (Note: This is the number, in the long run, over many such aluminum cans, we expect to see 0.05 flaws per can, on average .)

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20. Oil wells drilled in region A have a probability 0.20 of producing. Oil wells drilled in region B have a probability of only 0.09 producing. Oil wells in region C have probability 0.35 producing. One well is drilled in each region. Assume the wells produce independently. a. Let W be a random variable, such that W = # of wells that produce (out of 3 wells). What are the possible values of W? 0, 1, 2, 3 b. Create the table for the probability distribution of W. (Hint: What must these probabilities add up to? It may be helpful to write down all of the possibilities in the sample space. Keep in mind that these outcomes are NOT equally likely.) P(W = 0) = P(none produce) = P(A doesn’t) . P(B doesn’t) . P(C doesn’t) = (0.80)(0.91)(0.65) = 0.4732 Note: We get to multiply here because we have independent wells! P(W = 3) = P(all 3 produce) = P(A does) . P(B does) . P(C does) = (0.20)(0.09)(0.35) = 0.0063 P(W = 1) → There are 3 ways for this to happen: A does, B doesn’t, C doesn’t; A doesn’t, B does, C doesn’t; A doesn’t, B doesn’t, C does so P(W = 1) = (0.20)(0.91)(0.65) + (0.80)(0.09)(0.65) + (0.80)(0.91)(0.35) = 0.4199 P(W = 2) = 1 - [ P(W = 0) + P(W = 1) + P(W = 3) ] = 1 - 0.8994 = 0.1006 W 0 1 2 3 P(W=w) 0.4732 0.4199 0.1006 0.0063 c. Using the table above, determine the expected number of wells that will produce. E(W ) = 0*0.4732 + 1*0.4199 + 2*0.1006 + 3*0.0063 = 0.64 (Aside: This means that in the long run, over many such sets of oil wells, we expect to see 0.64 of them produce oil, on average.)

Exam 1 Additional Practice Solutions

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