MATH 3090 Spring 2014 Test 2 Version A

Transcription

1 Multiple Choice: (Questions 1 24) Answer the following questions on the scantron provided. Give the response that best answers the question. Each multiple choice correct response is worth 2.5 points. Please use a #2 pencil. For your record, also circle your choice on your exam since the scantron will not be returned to you. Only the responses recorded on your scantron sheet will be graded. 1. The probability distribution of the random variable X is given in the table below: Which of the following is false? A. P(X = 3) = B. P(X > 4) = C. P(X 4) = D. P(X 2) = X P(x) A large tax preparation company would like to estimate the proportion of its prepared tax returns that resulted in an IRS refund for its clients last year. In a random sample of 30 tax returns the company prepared last year, 21 resulted in the client receiving a refund from the IRS. What is a 99% confidence interval for the true proportion of this company s prepared tax returns that resulted in the client receiving an IRS refund last year? A. B. C. D. E. (.70)(.30) 0.70 ± (.70)(.30) 0.70 ± (.68)(.28) 0.68 ± (.68)(.32) 0.68 ± (.68)(.32) 0.68 ± If you want to be 99% confident of estimating the population proportion to within 0.04, what sample size is needed? A B. 601 C. 151 D

2 4. Suppose the average height of a U.S. Navy Seaman is 71 inches tall. Further assume that the heights of all Navy Seamen are normally distributed with a variance of What is the proportion of U.S. Navy Seamen are shorter than 67.5 inches. A B C D Suppose the number of raccoons that raid Woody s hot dog cart has a Poisson probability distribution with an average of 3.2 per day. Find the probability that exactly 2 raccoons raid Woody s hot dog cart today. A B. 3.2 C D To help determine how much to charge for a membership, a local gym is interested in estimating the average number of visits per week for its members. The mean number of days a random sample of 35 of its members visited the gym last week was 2.12, resulting in a 95% confidence interval of (1.78, 2.46). What is the target parameter in this study? A. The target parameter is the proportion of all the gym s members that visited the gym last week. B. The target parameter is the mean number of visits last week for all of the gym s members. C. The target parameter is 2.12 days. D. The target parameter is the interval 1.78 days to 2.46 days. 7. Suppose that the distance a new rail gun can be shot is uniformly distributed between 24 and 52 miles. Which of the following represents the proportion of shots that go beyond 45 miles? A B C. 0.8 D Suppose you re trying to find a 97% confidence interval for a population proportion. What is z α/2? A. z α/2 = 0.04 B. z α/2 = 2.17 C. z α/2 = 1.96 D. z α/2 =

3 9. You are given a data set from which you calculate: Count 17 Mean 1775 Median 1928 Std. Dev Variance Range 1184 Min 1172 Max 2356 IQR th% th% From this summary information can one of the methods of assessing normality be found to suggest a normal distribution for this data? A. Yes B. No 10. At ABC Car Rental, the distribution of daily demand of rental cars is as shown in the table: X P(x) What is the expected daily demand for rental cars at ABC? A. 2 cars B. 2.5 cars C. 2.9 cars D. 3 cars 11. In an experiment of rolling a die three times, which of the following random variables is continuous? A. X is the number of times 2 turns up. B. X is the time required to roll a die three times. C. X is the sum of the numbers turn up in the three rolls. D. X is 1 if three identical numbers turn up, otherwise X is The average number of cars sold by the Ford dealership in Anderson, SC, is 4 cars per week and a standard deviation of 2. With this info, what probability model could you use to calculate the probability that exactly 2 cars will be sold in a week? A. binomial B. exponential C. Poisson D. none of the above 3

4 13. Find the probability that the standard normal random variable z falls between -2 and 2. A B..2 C D A marketing research company is estimating which of two soft drinks college students prefer. A random sample of 100 college students produced the following 90% confidence interval for the proportion of college students who prefer drink A: (.255,.575). Identify the point estimate for estimating the true proportion of college students who prefer that drink. A. pp =.415 B. pp =.83 C. XX =.415 D. XX =.83 E. Cannot be determined 15. The average time between fighter takeoffs while in combat from the deck of the USS Nimitz is 6 minutes. Assuming the times between takeoffs is exponentially distributed, what is the probability the time between takeoffs is greater than 9 minutes? A B C D randomly selected students were asked if they were interested in studying abroad in Paris, France and 35 responded with yes. The margin of error for a 90% confidence interval for the proportion of all students who would like to study abroad in Paris, would be: A (.7)(.3) 50 B (.7)(.3) 50 C (.685)(.315) 54 D (.685)(.315) 50 4

5 17. Suppose the number of raccoons that raid Woody s hot dog cart has a Poisson probability distribution with an average of 3.2 per day. Find the standard deviation of the number of raccoon who raid Woody s cart in a day. A. 3.2 raccoons 2 B raccoons C raccoons 2 D raccoons 18. According to government data, 25% of employed women have never been married. What is the expected number of women who have been married in a sample of 80 employed women? A. 20 employed women B. 15 employed women C. 75 employed women D. 60 employed women 19. You are the manager of a cereal company that would like to estimate the population mean weight of the boxes of cereal produced by your assembly line (in grams). You would like to estimate this population mean to within.01 grams with 95% confidence. An estimate of the standard deviation of the boxes is found to be 0.2 grams. What sample size is the minimum needed to make this estimate? A. 30 B. 40 C D E A service firm identifies and records their returned satisfaction surveys in two ways satisfied or notsatisfied. They have determined that the responses are independent. Thus, no relevant response or selection bias exists. How should we expect these data to be distributed? A. normally distributed B. exponentially distributed C. binomially distributed D. the uniformly distributed 21. According to government data, 22% of American children under the age of 6 live in households with incomes less than the official poverty level. In a survey of 300 children, what is the standard deviation of the number of children who come from below poverty level households? The number of children out of 30 who come from below poverty level households is a binomial random variable. A children B children C. 66 children D children 5

6 22. When creating a 95% confidence interval for a population mean, what does the value 95% represent? A. There is a 95% chance the population mean falls within the interval created. B. 95% of the time the sample mean falls within the interval created. C. 95 out of 100 intervals created using the same method will contain the population mean. D. There is a 95% probability that the interval created is a correct estimate for population mean. 23. A research group has accurately determined that 13% of the tadpoles in a pond have a certain genetic defect. Assuming this is true, what s the probability that out of a thousand sampled tadpoles 180 had the defect? The number of tadpoles out of 1000 sampled that had the defect is a binomial random variable. A (0.87) 820 B (0.87) 820 C (0.13) 820 D (0.87) If a random sample of 45 observations is selected from an exponentially distributed population μ=σ=.1, which of the following best describes the sampling distribution of xx? A. XX follows an exponential distribution with θθ = 0.1 B. XX follows a normal distribution with μμ xx = and σσ XX = 0.1 C. XX follows a normal distribution with μμ xx = 0.1 and σσ XX = D. XX follows a normal distribution with μμ xx = 0.1 and σσ XX = 0.1 6

7 Free Response: The Free Response questions will count 40% of your total grade. Read each question carefully. In order to receive full credit you much show legible and logical (relevant) justification which supports your final answers. You MUST show your work. Answers with no justification will receive no credit. Show work by drawing a relevant picture and showing calculations (calculator work will not be counted as work shown). 1. Television viewing reached a new high when the Nielsen Company reported a mean daily viewing time of 7.4 hours per household. Use a normal probability distribution with a standard deviation of 1.75 hours to answer the following questions about daily television viewing per household. How many hours of television viewing per day must a household have in order to be in the top 10% of viewing times for all television viewing households? (5pts) 1.28 = xx xx 0 = hoooooooo 1.5 pt Drawing a picture - shows mean - indicates areas - approximately normally distributed 1 pt Correct z-value of pt Correctly identifies as a positive value 1 pt Uses correct algebra to solve for x 0.5 pts Units 7

8 Give the appropriate probability statement, show correct work (calculator work will not be counted as work shown), and calculations to 4 decimal places. 2. According to the Mars/M&M corporation M&M plain candies have a mean weight of g and a standard deviation of g. Assume the weights of M&M plain candies are normally distributed. a. If 1 M&M plain candy is randomly selected, find the probability that it weighs more than g. (5pts) Let X = weight of one plain M&M candy PP(XX > 0.854) = PP(zz > 0.06) = = zz = = pt Correct probability statement (do not have to define X) 1 pt Correct z-score value 1 pt Correct Work shown for z-score calculation and/or picture drawn and labeled 1 pt Correctly looked up probability in table for z- score calculated 1 pt Correctly added 0.5 to table value Note: Do not take off points for additionally showing calculator work b. If 465 M&M plain candies are randomly selected, find the probability that their mean weight is at least g. (5pts) PP(XX > 0.854) = PP(zz > 1.24) = = zz = 0.052/ 465 = pt Correct probability statement (do not have to define XX ) but must use XX 1 pt Correct z-score value 1 pt Correct Work shown for z-score calculation and/or picture drawn and labeled 1 pt If use SD - Correctly looked up probability in table 1 pt If use SD - Correctly added 0.5 to table value Note: Do not take off points for additionally showing calculator work Note: If the student does not use sampling distribution they will not earn credit for problem except may earn 1 pt for correct probability statement 8

9 Show relevant work, give calculations to 2 decimal places. 3. In a test of a weight loss program, a random sample of 25 adults used the Atkins weight loss program. After a year, their mean weight loss was found to be 2.1 lb, with a standard deviation of 4.8 lb. Assume weight loss in the Atkins program is normally distributed. a. What conditions are needed to construct a 90% confidence interval estimate of the mean weight loss for all such subjects? Are these conditions met and why? (4 pts) 1. Random Sample selected from the Population stated in problem 2. Distribution of XX is normal stated in problem Population Normally Distributed 1 pt Random Sample condition 1 pt stated in problem 1 pt Population Normally Distributed 1 pt Stated in problem Note: Any additional incorrect conditions listed should be docked a point for each b. Construct a 90% confidence interval estimate of the mean weight loss for all adults using the Atkins weight loss program. (5 pts) xx ± tt αα 2 ss nn 2.1 ± (1.711) ± ( , ) c. Interpret the interval found in part (b). (5 pts) 1 pt Using correct formula 1 pt Uses correct d.f. 1 pt Finds correct critical value for their d.f. from table 1 pt Plugs values into formula correctly 1 pt Shows relevant work Note: if student only gives calculator work then they lose points for (finding correct critical value from table and showing relevant work) We are 90% confident that the mean weight loss for all adults using the Atkins weight loss program lies between 0.46 and 3.74 lbs. d. Does the Atkins program appear to be effective? Explain. (3 pts) 1 pt Confident not probability/chance 1 pt All or population or true 1 pt Mean not proportion 1 pt Context 1 pt Units (lbs.) Yes because 0 is not in the interval (entire interval is positive) but it is possible weight loss may not be by much. Answers will vary make sure correct and gives reasonable explanation. 9

10 Give the appropriate probability statement, show correct work (calculator work will not be counted as work shown), and calculations to 4 decimal places. 4. Twenty-five percent of high school students plan to choose a career in information technology. A random sample of 24 high school students were selected. a. Define a random variable for this experiment. (2 pts) X = # of high school students out of 24 that choose IT as a career Right or Wrong can give partial credit if not full description (ie, # of high school students) b. Assuming your random variable in part (a) follows a binomial distribution, find the probability that among 24 randomly selected high school students, at least 3 plan to choose a career in information technology. (5pts) PP(XX 3) = 1 [PP(XX = 0) + PP(XX = 1) + PP(XX = 2)] = = 1 [(1) (24) (276) ] = = pt Correct probability statement 1pt Correctly uses binomial formula by showing work (not calculator syntax) 1 pt Finding probability of P(0), P(1), and P(2) 1 pt Knows to find complement of 0,1, and 2 1 pt Calculations to 4 decimal places Did you correctly fill in your scantron? Are your bubbles dark enough? Did you use your XID with the C bubbled as a 0? Is your section number listed? Is your instructor filled out? Is your test Version bubbled in correctly? 10