Stat 104: Quantitative Methods for Economists. Study Guide Solutions, part 2

Transcription

1 Stat 104: Quantitative Methods for Economists Study Guide Solutions, part 2

2 1) The table below shows, for credit card holders with one to three cards, the joint probabilities for number of cards owned (X) and number of credit purchases made in a week (Y). I have already calculated part of the marginal distribution for X. Y (3) 0.4 X (2) (1) a) For this to be a legal probability table, the sum of all the joint probabilities has to be what value? 1 b) Fill in the empty cells in the above joint probability table. (1) = 0.02 (2) = 0.09 (3) = 0.04 c) What is the probability that X equals Y? P(X=1 and Y=1)+P(X=2 and Y=2) +P(X=3 and Y=3)= = 0.29 d) Are X and Y are independent? Why? P(X=1 Y=1) = 0.13/0.23 = which does not equal P(X=1)=0.40 So Not Independent. 2) Suppose the sample mean score on a national test is 500 with a standard deviation of 100. If each score is increased by 25%, what are the new mean and standard deviation? (a) 500, 100 (b) 525, 100 (c) 525, 125 (d) 625, 100 (e) 625, 125

3 3) A random variable Y has the following probability distribution: y p(y) 3k 7k The value of the constant k is: (a) 0.05 (b) 0.10 (c) 0.15 (d) 0.20 (e) ) If X and Y are independent random variables, then (a) knowledge of the outcome of X would not change your beliefs about the outcome of Y. (b) p(x y) = p(x). (c) p(x, y) = p(x) p(y). (d) p(x, y) = p(y) p(x y). (e) All of the above. The following is used for the next 3 questions. Let X be a discrete random variable with the following probability function 5) What is P(0 < X < 3)? a).3 b) 1 c).8 d).5 e).7 x P(X=x) ) The mean µ (expected value) of X is a) 4 b) 1.5 c) -1 d) 2.7 e) 1.1

4 7) The variance of X is a) 2.50 b) 1.21 c) 1.29 d) 2 e) 1.73 Questions 8-10 were just repeats of the above three questions. 11) Which of the following is a legal probability table? 12) A television game show has three payoffs with the following probabilities: Payoff ($) : ,000 Probability : What are the expected value and standard deviation for the payoff variable?

5 13) Alex, Alicia, and Juan fill orders in a fast-food restaurant. Alex incorrectly fills 20% of the orders he takes. Alicia incorrectly fills 12% of the orders she takes. Juan incorrectly fills 5% of the orders he takes. Alex fills 30% of all orders, Alicia fills 45% of all orders, and Juan fills 25% of all orders. An order has just been filled. We know P(Wrong Alex) =.2, P(Wrong Alicia)=.12 and P(Wrong Juan)=.05. We also know that P(Alex)=.3, P(Alicia)=.45 and P(Juan)=.25. Using P(A B) = P(A and B)/P(B) we can build the following table: Correct Wrong Alex Alicia Juan a. What is the probability that Alicia filled the order? 0.45 b. If the order was filled by Juan, what is the probability that it was filled correctly? P(correct Juan) = P(correct and Juan)/P(Juan) =.2375/.25 = 0.95 (or P(wrong Juan) = 1-P(correct Juan) = ) c. Who filled the order is unknown, but the order was filled incorrectly. What are the revised probabilities that Alex, Alicia, or Juan filled the order? P(Alex wrong) =.06/.1265=0.474 P(Alicia wrong) =.054/.1265=0.427 P(Juan wrong) =.0125/.1265 = d. Who filled the order is unknown, but the order was filled correctly. What are the revised probabilities that Alex, Alicia, or Juan filled the order? P(Alex correct) =.24/.8735=0.275 P(Alicia correct) =.396/.8735=0.453 P(Juan correct) =.2375/.8735 = 0.272

6 14) A researcher calls a random sample of 10 households in a community on the night that 80% of the households have someone at home. What would be an expected number of successful calls? a) 4 b) 8 c).2 d) 10 e) 12 15) Suppose that 30% of travelers at the Atlanta airport carry laptop computers. Determine the expected value and standard deviation of the total number of laptop computers in a random sample of 200 travelers at the Atlanta airport. (a) expected value= 60 laptops (b) expected value= 60 laptops (c) expected value= 30 laptops (d) expected value= 30 laptops standard deviation = 6.5 laptops standard deviation = 42 laptops standard deviation = 6.5 laptops standard deviation = 42 laptops E(X) = np = 200(.3) = 60 Var(X) = np(1-p) = 60(.7) = 42 Std Dev = ) Jessica Simpson is not a professional bowler, and 40% of her bowling swings are gutter balls. She is planning to take 90 bowling swings. What are the expected value and standard deviation of the number of gutter balls? a) 36, 3.65 b) 54, 4.65 c) 54, 21.6 d) 36, 21.6 e) None of the above E(X) = np = 90(.4) = 36 Var(X) = np(1-p) = 36(.6) = 21.6 Std Dev = ) Andrew is not a professional photographer, and 10% of the pictures he takes of the sorority pledges are of unsatisfactory quality. He is planning to take 36 pictures. What are the expected value and variance of the number of satisfactory pictures? a) 3.0,3.24 b) 3.6,0.36 c) 3.6,3.24 d) 3.0,0.36 e) None of the above E(X) = np = 36(.9) = 32.4 Var(X) = np(1-p) = 32.4(.1) = 3.24 Std Dev = 1.8

7 18) Mortimer s steak house advertises that it is the home of the 16 ounce steak. By this they mean that the weight of their steaks is normally distributed with mean 16 and standard deviation 2. a) If their advertising is true, what is the median weight of the steaks they sell? The mean and the median are both 16 because of symmetry b) If their advertising is true, what is the probability that you will get a steak that weighs between 12 and 14 ounces? P( 12 < X < 14) = P ((12-16)/2 < Z < (14-16)/2) = P(-2 < Z < -1) (be able to use the table in the book to find this area!) P( -2 < Z < -1) = c) You are out to dinner with three friends and all four of your steaks are less than 12 ounces. What is the probability that this will happen if Mortimer s claim is true? P(X < 12) = P(Z < (12-16)/2) = P(Z < -2) = P(all 4 below 12 ounces) = P(X1<12 and X2<12 and X3<12 and X4<12) = P(X1<12)*P(X2<12*P(X3<12)*P(X4<12) = ^4 = ) The distribution of weights in a large group is approximately normally distributed. The mean is 80 kg. and approximately 68% of the weights are between 70 and 90 kg. The standard deviation of the distribution of weights is equal to: a. 20 d. 50 b. 5 e. 40 c. 10 f. none of the above 20) Among subscribers, suppose the amount of time spent reading BusinessWeek magazine is approximately normally distributed with a mean of 50 minutes and a standard deviation of 15 minutes. Roughly what percentage of subscribers spend more than an hour reading BusinessWeek? (a) 10% (b) 25% (c) 50% (d) 75% P(X>60) = P(Z > ((60-50)/15) = P(Z > 2/3) =

8 21) The time needed to complete an exam is normally distributed with mean 40 minutes and Std. 4 minutes. a) What is the proportion of students who complete the exam before 44 minutes? P(X < 44) = P(Z < (44-40)/4) = P(Z < 1) = b) If 3 friends took the exam, what is the probability that at least one of them will complete the exam before 44 minutes? P(at least 1) = 1-P(no one completes before 44 minutes) P(takes more than 44 minuutes) = = P(at least 1) = 1 (0.1587) 3 22) Which of the following statements about the Central Limit Theorem (CLT) are correct? a) The CLT states that for large samples, the sample mean X is equal to µ b) The CLT states that for large samples, the distribution of the population mean is approximately normal. c) The CLT states that the sample mean X is always equal to µ d) The CLT states that for large samples, the distribution of the sample mean is approximately normal. e) Both (b) and (c) are correct f) None of the above 23) The lifetime for a WonderWaffle waffle iron is approximately normally distributed, with a mean of 8 years and a standard deviation of 2 years. One potential buyer (Finicky Frank) wants to have high confidence that the waffle iron will last 5 years; he will only buy it if there's a 90% or better chance that it will last at least 5 years. Another potential buyer (Lenient Linda) plans on making waffles for a longer period of time (7 years), but needs less certainty that the waffle iron will last that long; Linda will only buy the waffle iron if there's a 75% or better chance that the machine will last at least 7 years. Based on the information above, will Frank and/or Linda buy the WonderWaffle waffle iron? (Assume that both Frank and Linda are aware of the distribution of WonderWaffle lifetimes.) Your answer will be one of the following but you need to show your work: Neither Frank nor Linda will buy. Frank will buy, but Linda will not buy. Linda will buy, but Frank will not buy. Both Frank and Linda will buy.

9 Let X represent the lifetime of a wonderwaffle. P(X>5) = P((X-8)/2 > (5-8)/2) = P(Z > -1.5) = 93.3% so Frank will buy P(X>7) = P((X-8)/2 > (7-8)/2) = P(Z > -.5) = 69% so Linda will not buy 24) On the isle of Pacifica, the length of tree snakes is a normal random variable with mean 18 inches and standard deviation 3 inches. a) What is the probability that a snake will be more than 20 inches long? P(X>20) = P((X-18)/3) > (20-18)/3) = P(Z>2/3) =.2525 b) What is the probability that a randomly selected tree snake is exactly 19 inches long? P(X=19)=0 c) If length is measured to the nearest half-inch, then what proportion of tree snakes would have their length reported as 19 inches? P(18.75 < X < 19.25) = d) Select five tree snakes at random. What is the probability that at least one has a length above twenty inches long? 1-(.75^5)= ) Which of the following is NOT CORRECT about a standard normal distribution? a. P( 0 < Z < 1.50) =.4332 b. P( Z < -1.0 ) =.1587 c. P( Z > 2.0 ) =.0228 d. P( Z < 1.5 ) =.9332 e. P( Z > -2.5) = ) Homer Simpson likes to go fishing in Lake Springfield. Unfortunately, the nuclear power plant dumps mercury into water, which eventually contaminates the fish. The concentration in fish will vary among individual fish because of differences in eating patterns, movements around the lake, etc. Suppose that the concentrations of mercury in individual fish follows an approximate normal distribution with a mean of 0.25 ppm and a standard deviation of 0.08 ppm. Fish are safe to eat if the mercury level is below 0.30 ppm. What proportion of fish are safe to eat? a..632 d..278 b..231 e..375 c..734 f. none of the above

10 We have X~N(.25,.08 2 ). Then P(X<.3) = P((X-.25)/.08 < (.3-.25)/.08) or P(Z<.625) = ) Suppose that warranty records show the probability that a new car needs a warranty repair in the first 90 days is If a sample of three new cars is selected, what is the probability that a) none needs a warranty repair? X = # that need repair X ~ Bin(n=3,p=.05) P(X=0) =.95^3 = (or.86) b) at least one needs a warranty repair? P(X>=1) = 1-P(X=0) = =.14 c) more than one needs a warranty repair? Need a computer for this part:

11 28) Below are binomial probabilities for a certain n and p. a) What is n? Y = # successes in n trials so since Y=0,1,2,,5, so n=5 b) What is p? There are many ways to find p. One way is you can realize that P(Y=5)=p 5. Solving for p we find that p =.02 c) What is E(Y)? Since n=5 and p=.02 we have that E(Y)=np=1

12 29) A university claims that 80% of its basketball players get degrees. An investigation examines the fate of all 20 players entered the program over a period of several years that ended six years ago. Of these players, 11 graduated and the remaining 9 are no longer in school. If the university s claim is true, the number of players among the 20 who graduate should have the binomial distribution with n = 20 and p = 0.8. a) What is the probability that exactly 11 out of 20 players graduate? Need a computer for this: b) Find the mean number of graduates out of 20 players if the university s claim is true. The mean is np = 20(.8) = 16

13 c) What is the most probable number of graduates out of 20 players? Looking at the computer generated graph, we see the most probable number graduates is 16 (the highest probability is at 16). d) Find the standard deviation of the number of graduates out of 20 players. The std deviation is sqrt(n*p*(1-p)) = sqrt(20*.8*.2) = 1.78 e) Based on the answers in (a) (d) (especially (a)), Do you believe the university s claim that 80% of its basketball players get degrees? Why? No we do not believe the claim. If p=.8 for the graduation rate you would expect around 16 students to graduate; not 11. In fact, if p=.8 the probability is only.007 what 11 would graduate; this is an unlikely event if p=.8.

14 30) Let the random variable Z follow a standard normal distribution. You can answer this question using the computer or by hand-it is good to do by hand for the practice. a) Find P(Z< 1.20) b) Find P(Z > 1.33) c) Find P(Z < -1.70)

15 d) Find P(Z > -1.00) e) Find P(l.20 < Z< 1.33) f) Find P(-1.70 < Z < 1.20)

16 31) It is known that amounts of money spent on textbooks in a year by students on a particular campus follow a normal distribution with mean $380 and standard deviation $50. a) What is the probability that a randomly chosen student will spend less than $400 on textbooks in a year? P(X < 400) = P((X-380)/50 < ( )/50) = P(Z <.4) = b) What is the probability that a randomly chosen student will spend more than $360 on textbooks in a year? P(X>360) = P((X-380)/50 > ( )/50) = P(Z > -.4) = c) Draw a graph to illustrate why the answers to parts (a) and (b) are the same. d) What is the probability that a randomly chosen student will spend between $300 and $400 on textbooks in a year? P(300 < X < 400) = P(( )/50 < Z < ( )/50) = P(-1.6 < Z <.4) = F(.4) F(-1.6) = =

17 32) An automobile dealer mounts a new promotional campaign, in which it is promised that purchasers of new automobiles may, if dissatisfied for any reason, return them within two days of purchase and receive a full refund. It is estimated that the cost to the dealer of such a refund is $250. The dealer estimates that 15% of all purchasers will indeed return automobiles and obtain refunds. Suppose that fifty automobiles are purchased during the campaign period. a) Find the mean and standard deviation of the number of these automobiles that will be returned for refunds. X = # returned X~Bin(50,.15) E(X) = np = 50(.15) = 7.5 Var(X) = np(1-p) = 50(.15)(.85) = b) Find the mean and standard deviation of the total refund costs that will accrue as a result of these fifty purchases (hint: a+bx rule). Refund = $250X E(Refund) = 250E(X) = 250(7.5) = $1,875 Var(Refund) = 250*250*Var(X) = 250*250* = $ ) Suppose that on a business calculus test, the probability is that Brittany would get 80% of the items right. a) For a 10-item quiz, calculate the probability that Brittany will get at least 7 items right.

18 b) fewer than 6 items right (and therefore fail the quiz ). c) 9 or 10 items right (and get an A on the quiz - ).

19 d) What is the expected number of items that Brittany will get right? What proportion of the time will she get that number right? On average she will get np = 10(.8) = 8 items correct. The probability that she gets 8 items correct is given by the computer: e) What is the standard deviation of the number of items that Brittany will get right? Sqrt(np(1-p)) = sqrrt(10*.8*.2) = ) Ronald McDonald is conducting a binomial experiment. He heard a rumor that his fast food restaurant is becoming unpopular for children because his Happy Meal toys are not as cool as his competitors toys. In fact, a psychological study showed that only five percent of kids even like going to McDonald s. However, Ronald thinks this study is biased because the psychologist who conducted it was the wife of Burger King s CEO. Ronald wants to find out for himself how many kids truly love McDonald s. He conducts an observational study one day on 60 children who pass by and counts how many enter the McDonald s versus how many did not enter. a) If the psychologist s research was absolutely correct, in theory, what is the expected number of kids who will enter that McDonald s? The np rule implies 60(.05) = 3 kids b) Unfortunately, Ronald McDonald must not know much about statistics. There is something very wrong with this study. What is it? In other words, why can t this experiment fall under a binomial distribution? (Hint: Look at the requirements for the number of successes of an experiment to have a binomial distribution) The kids are probably in groups (families) so one isn t observing 60 independent trials of an experiment.

20 35) You are taking a multiple-choice final exam with 150 questions. You have about three hours to complete the exam, so you aren t worried about time at all. However, you just looked up at the clock and realized you have only 10 minutes left and exactly 40 questions to go! Since you are filling in a bubble sheet, you also realize that you don t have any time left to actually read the questions if you want to finish the entire exam. You decide on the strategy that you are going to randomly guess and fill in the bubble sheet for the rest of the 40 questions. Assuming that each question is independent, there are 5 answer choices for each question, and there is only one correct answer, find the following: a. Expected value of random variable correct answers (from guessed portion only): X = # correct X = Binomial(n=40,p=.2) E(X) = 40(.2) = 8 b. Variance of random variable correct answers ( from guessed portion only): Var(X)=n*p*(1-p) = 40*.2*.8 36) We-Mobile Cellular, a phone company, is working on new pricing plans. Jane has reviewed the company files and found that the number of minutes that typical business customers use the phone each month is approximately normal, with a mean of 810 and a standard deviation of 154. a) What percentage of these customers use the phone more than 1000 minutes? b) What is the maximum number of minutes used by the bottom 25% of customers?

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