ARME 210 Prelim2 +KEY - ARME210 Name flaswf Kg}; Fall 2000...

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Unformatted text preview: ARME210 Name flaswf Kg}; Fall 2000 ID. Section PRELIM II Please circle the Correct Answer. Each multiple choice question is worth 2 points. 1.) A discrete random variable: a.) cannot take on the values 1/8, 2/8, 3/8, and 5/8. b.) cannot have negative values. 0. cannot have the value zero. d. does not have probabilities greater then 0.01. a) does not give probabilities to intervals. 2.) The reliability of a manufacturing process is defined to be the proportion of nondefective items that are produced. An electronics firm manufacturing transistors claims that their reliability is 0.98. A random sample of 1,000 transistors is tested. Iftheir claim is correct, what is the distribution of the number of defectives in the sample? binomial (n = 1,000, p = 0.02) . binomial (n = 1,000, p = 0.5) c.) binomial (n = 1,000, p = 0.98) (1.) Normal (11 = 1,000, o = 0.98) e.) Normal (ii = 0.98, o = 1/(1,000) ) 3.) A person with a Z—Score of -2.00 has performed below approximately what percent of the students taking the test? a.) 2 percent (1. 84 percent b.) 15 percent a 97 percent c.) 50 percent 4.) All of the following are characteristics of the normal distribution, except: a.) symmetric about mean b.) bell-shaped curve c area under the curve is always one @i it is a discrete distribution e. probability that x is equal to any specific value is zero 5.) An instructor has five sections of a course: A, B, C, D, and E. She wants to randomly select three sections for a special teaching evaluation. She labels the classes as follows: A = 1, B = 2, C = 3, D=4 and E =5. She starts at the beginning of this list of random digits: ~ 1568914227 06565 14374 Which classes did she select? a.) A, E, and A d. B, C, and D b.) AandD A,D,andE c.) A, B, and C 6.) For work with the normal approximation to the binomial distribution, we used a continuity correction because: a.) the sample size was small. b.) the population variance was unknown. 0. the mean was estimated by n? 9 we were approximating a discrete distribution with a continuous one. e.) we were approximating a continuous distribution with a discrete one. 7.) Which is NOT a characteristic of a (random) sampling distribution of means? a Its mean is the same as the mean of the population. @ Its stande deviation is greater than that of the population of scores. c.) It tends to resemble the normal distribution irrespective of the shape of the population of scores with sufficient n. d.) Its standard deviation changes with variation in sample size. 8.) As a rule of thumb, the sampling distribution of the sample proportions can be approximated by a normal probability distribution whenever a.) npz 5 b.) n(1- p) Z 5 0 n2 30 3 Both a and b are true. e. None of the above answers is correct. 9.) 10.) 11.) 12.) When the probability that an estimator is close to the parameter it estimates increases as the sample size, n, increases that estimator is said to be: b.) biased d. c.) efficient The margin of error for a poll is 4%. This means that: a.) unbiased sufficient consistent 4% of those sampled did not answer the question asked we have 95% confidence that the sample statistic is within 4% of the population parameter c.) 4% of those sampled gave the wrong answer to the question asked (1.) 4% of the population were in the sample f.) the confidence we have in the statistic is 4%. Which of the following is a true statement regarding the comparison of t—distributions to the standard normal distribution? a.) The normal distribution is symmetrical whereas the twdistributions are slightly skewed. b.) The proportion of area beyond a specific value of t is less than the proportion of area beyond the corresponding value of z. The greater the (if, the more the t-distributions resemble the standard normal distribution. d.) All of the above. e.) None of the above. In repeated constructions of 95% confidence intervals for a population mean, u, which of the following is most precise: a.) it falls in the interval approximately 95 times out of a 100. ® the interval brackets the unknown u_approximately 95 times out of a 100. c.) 95 out of a 100 populations will have their means in the interval. d.) 3? falls in the interval approximately 95 times out of a 100. e.) The interval brackets Tc' approximately 95 times out of a 100. Please show your work and circle the correct answer. 13.) According to Runzheimer International, the average cost of a domestic trip for business travelers in the financial industry is $1250. Suppose another travel industry research company takes a random sample of 22 business travelers in the financial industry and determines that the sample average cost of a domestic trip is $1192, with a sample standard deviation of $279. Construct a 98% confidence interval for the population mean from these sample data. Assume that the data are normally distributed in the population. Now go back and examine the $1250 figure published by Runzheimer International. Do you think that this value is reasonable? Why or why not @ :l/q3,3:°?79 dFrQ/3 24.01;.025/g 2‘ (5) 07-7 #92 3: 3.5/9 [fifl fl #9:? : N9. 75’ = («t/Ma 2:2, 1‘ /3I//. 78’) 9% c1 / 11 M50 sums reamab/e. 5712a nL falls (Ll/Mm W}, Jen/410 I 14.) You are planning to invest in a new high-tech company, and figure your rate of return over the coming year as in the Table below (where 100% says that you doubled your money, -- 50% says you lost half, etc.): - Rate of , Return Probability £5,615) 742 100% 0.20 ,20 0700 0 50% 0.40 a? 0 /0 0 0 0% 0.25 0 0 —50% 0.15 37 5 a.) Find the mean rate of return. 32' 5 33 7 5 (3) 1).) Find the standard deviation of the rate of return. ' 7 (4) 0‘ : 3375 v (32.52% = 337$?5 : @ @ , c.) Find the probability that you will earn more than 40%, according to the table. (a) P(x740%)== 15.) A USA T oday/IntelliQuest survey of computer users revealed that 23% log onto the Internet and/or an online service more than 20 times per month. Suppose a random sample of 600 computer users is taken. What is the probability that more than 25% of computer users log onto the Internet and/or an online service more than 20 times per month? (6) 72:6”) “'93 @ Z: “JEWJB _} ‘ (‘>. : 7 PP 075) ' \/.a367;¢)@ 600 f(2 7/./é)@ Q = ,5—.37¥ = J23 16.) According to the Internal Revenue Service (IRS), the chances of your tax return being audited are about 6 in 1,000 if your income is less than $50,000; 10 in 1,000 if your income is between $50,000 and $99,999; and 49 in 1,000 if your income is $100,000 or more (Statistical Abstract of the United States; 1995). a.) If five taxpayers with incomes under $50,000 are randomly selected, what is the probability that more than one will be audited? ’7Z= 5/ fawn Q.) @ Mm) = /-[(§)dooe)°6994)5+ (fldwa’mwy (5) = /— (9,104 + , 0,?93) = ,0003 @ b.) What assumptions did you have to make in order to answer part a? ‘ (2) 777“ whim!" -' taxpayem are my, Jeo/ or 007‘ aka await/0 simian—d eve/7&5) a—t’mL 4km? Mgrabaét'fly/ our being auduéas/ remams Cmsfilzé, c.) If two taxpayers with incomes under $50,000 are randomly selected and two with incomes more than $100,000 are randomly selected, what is the probability that none of these taxpayers will be audited by the IRS? 72:2 p:,004 m>=,9a>3 69 (6) 72:3.) [pact/9 13(0): -?0‘i’@ d.) If 1000 taxpayers with incomes more than $100,000 are randomly selected, what is the probability that more than 60 of them will be audited? / (8) 7z=/000i f='05’9 /a= 99,0‘=./%.599=é.91 4 Ha 740)e/’(Xy740.5’) O Q Q Z;M=M? flszzé9):.5—,45s/5 @ [/16 W “8— 5,95 was) 17.) According to Nielsen Media Research, the average number of hours of TV viewing per household per week in the United States is 50.4 hours. Suppose the standard deviation is 11.8 hours and a random sample of 42 US. households is taken. ,a =50,€/) 0':— //,8) 71: 6/02 a.) What is the probability that the sample average is less than 40 hours? If the sample average actually is less than 40 hours, what could this mean in terms of the Nielsen Media Research figures? — — 40 - 5o 4 a PM <40) 2- ' ='-5,7,z :P(z<-5I¥Q)(:~¢O Myflfé} (6) 7V0£0J3ly / I5 noi’ 50, b.) Suppose the population standard deviation is unknown. If71% of all sample means (from samples of size 42) are greater than 49 hours and the population mean is still 50.4 hours, what is the value of the population - standard deviation. “.55 =- 49—50,? ,_ -55: —9.0;1 (9 Wm: 6‘ (4) 8.) According to figures released by the National Agricultural Statistics Service of the US. Department of Agriculture, the US production of wheat over the past 20 years has been approximately uniformly distributed. Suppose the mean production £ over this period was 2.165 billion bushels. If the height of this distribution is .862 £02 #1— billion bushels, what are the values of a and b of this distn'bution? 015%.! you CW" M; +5 back m “u— (6)@ I ' - H6 , a: “if” =7 5‘“ ’ ' W) 0'25 .1 3L @ a=/.585@ .9.) Scores by women on the SAT-I test are normally distributed with a mean of 998 and a standard deviation of 202. Scores by women on the ACT test are normally distributed with a mean of 20.9 and a standard deviation of 4.6. Assume that the two tests use different scales to measure the same aptitude. If a woman gets an SAT score that is in the 67th percentile, find her actual SAT score and her equivalent ACT score. 6) @411: X—‘i’CIS’ :7 Xi/ogéXXQ 203 @ ’A/L/s jive/20.9 :7 Xfizggt9ay 4.9 20.) According to Fortune (April 1994), profits as a percentage of sales in the oil industry during 1993 ranged from —5% to 7%. If we wish to estimate average profits of sales with a 99% confidence, what size sample should be drawn? Assume a margin of error of 2%. $05): 3 ,6) $91 a e ‘ 72 : (a?- 7) (3) z “(875 w 77%) 21.) A one-sided confidence interval for p can be written as p < E +E or p_ > p —E , where the margin of errorE is modified by replacing zm,2 with za . If Air America wants to report an on-time performance of at least x percent with 95% confidence, construct the appropriate one-sided confidence interval and then find the percent in question. Assume that a simple random sample of 750 flights results in 630 that are on time. -—, $30 (7) 725-255) [0‘ —— :ag’7/ 750 @ @ l?2/_/,(01715 .8404, : Igyd'ogl 5 '9/8 750 p 7, 8/? a Bonus 22.) A 95% confidence interval for the lives (in minutes) of Kodak AA batteries is (430,470). Assume that this result is based on a sample of size 100. Construct. the 99% confidence interval. (5) a 450 + £94012?) = #20 0'63?) = /0,.:L @ Q 450 i 31.5? (/0) 450 "f— 24.2] (4.23.74, 4744/) 99 ‘70 c1 \/ @ ...
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ARME 210 Prelim2 +KEY - ARME210 Name flaswf Kg}; Fall 2000...

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