Thurs Final Exam Solutions - Sh/ofimg g Pr“ ‘p’w STT...

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Unformatted text preview: Sh/ofimg g: Pr“ ‘p’w STT 315 (Secs 1 - 11) Final Exam - FORM A Thurs, Dec 16, 2010 Gilliland (Closed book - 120 minutes) fit S‘Wfi 7:45 .— 9:45 am Name PID TA 0. You may use a calculator. All electronic devices capable of communication and/or web access must be turned off 2. Be sure to code your name, section number and PID number correctly on the answer sheet that has been provided. Sign the answer sheet. 3. On our answer sheet on must also code the letter A as the FORM 0 our examination in the s ace rovia'ed. i i I. In the places above, please put your name, your PH), and the name of your TA. 4. After the exam, return our answer sheet as on leave ran: the rant 0 the room. You may keep your final exam paper. Solutions will be placed on the class pages. E 5. There are 48 questions, each worth 1.459 course points. (Total 70. 032 course points possible.) Tables are found on the last sheet. 6. Your score is determined from your answer sheet and not from what you have written on your exam paper. 1 - 2. An experiment has sample space S. A and B are independent events with P(A) = .50 and P(B) = .40. 1. P(A|B)= (a) .10 (c) .60 (d) .20 (e) .12 2. P(AUB)= (a) .90 (c) .60 (d) .80 (e) ,68 Home) -- set—Homo Paws) main—.20 g. Here is a Venn diagram of a sample space with nine outcomes. Beside each outcome is its probability. Two events A, B are shown. Poms): (b) .60 (o) .65 (d) .75 (e) .40 4. Ten percent of a company‘s accounts are in error. An auditor selects 10 of the accounts at random from this large set of accounts. What is the probability that at least two of the 4 ~ ._ : cunts will be in error? (a) .070 (b) .651 _ (d) .736 (6) none of these WHOM) m 32.3 31-» PM = -\-— .734 .244 T550101 5 - 6. Suppose it is known that 40% of the people who inquire about investment opportunities at a brokerage house end up purchasing stock, and 50% end up purchasing bonds. It is also known that 30% of the inquirers end up getting a portfolio with both stocks and bonds. A person is selected at random from those who make inquiries. 5. What is the probability that she or he will get stocks or, (a) .20 (b) .90 . 9.145 or both (i.e., open any portfolio)? (d) .80 (e) none of these OH’0*3%@W0‘%§ 6. What is the probability that the selected person will get bonds given _t ,. . (a) .20 (b) .90 (c) .70 easier :3” ,. .30 We} W _7_. The positions of President, Vice President, Treasurer, and Secretary are to be filled by choosing 4 persons from the 10 persons in the Club. Ho . o possibilities are there for filling the positions? a) ,i (b) 104 (c) 210 (d) 4! (e) none of these 10.41917 :3? D .. gets stocks? (6) none of these §__-__9,. At a certain university, 30% of the students who take basic statistics are first-year students, 40% are sophomores, 20% are juniors, and 10% are seniors. From records of the statistics department it is found that out of the first-year students who take the basic statistic course 10% get A‘s; out of the sophomores who take the course 20% get A's; out of the juniors 30% get A‘s; and out of the seniors who take the course 40% get A‘s. 8. What percentage of all students who take._ I .‘. . (a) 25% * ' .urse get A's? ' ' (c) 16% (d) 29.25% (c) none of these : or he is a senior? (e) .35 9. Given that a student received an A in the course, what is the probabili ___ - (a) .14 (b) .90 (c) .25 ' 10 - 12. Here is the probability mass function of a discrete random variable X. n ulative distribution function at x = 2? (b) .60 (c) .80 (d) .90 (e) 1.00 10. What is the value oftn .=- ... 11. What is the expected value of X? Recall that E (X) = Z X1906). (a) 1.40 (b) 2.00 (c) 1.60 (d) 1.35 6. gm «1» 1.6.2.) '6 2.6.1") 166.2.) 12. P(0<X<3)= (a) .50 (b) .70 (c) .40 (e) .80 E90) +?(7.\ =' (1041.10 1_3. A random variable X has a mean of 50 and a standard n .- ---:»-..nn of 10. By Chebyshev, P(35 < X < 65) is at least (a) 8/9 (b) 997 _ g . (d) 15/16 (e) 3/4 5=Bo+i.6£ao) |-.L., ... tanning-1'65? '35 = 59 4.5003 h.“ 1.5 14 — 15. A large population of parts is 30% defective. A simple random-sample of 20 parts is to be drawn from the population. Let X denote the number of defective parts in the sample. 14. P(3SX<6)= (a) .573 (b) .309 (c) .524 (d) .608 )(N'BCZDJ303 P(3§nghx. P[3¢X¢-5j :P6X‘gflw POCfi'ZE I Wr‘iléwflag” Tabifli 15. Determine E00 and SD(X). - (d) 6, 1.78 (e) .3, .214 (a) 6, 1.56 ‘ (b) 4, 3.24 ECX‘) =. 109391 : 5 SW” - "1}. E. In applications, selecting a simple random sample from a population may not be a simple task. " (b) False 12. A carton of 12 eggs has 4 rotten eggs and 8 good eggs. Three eggs are chosen at random from the carton to make a three egg omelet. Let X = the number of rotten eggs chosen. What is the probability that the sample will consist of two rotten eggs and one good egg, that is, what is P ”X —h_ _ . (a) 81/220 (c) 112/220 ((1) 56/220 (e) none of these i2. 3 3.2.0» Q. A large population of parts is 8% defective and 92% nondefective. Parts are drawn at random one afier the other until a defective part is drawn for-£77,.-= ime. LetX denote the draw when a defective is observed for the first time. P(X >10) = - (b) .743 (c) .257 (d) .599 (e) none of these PM >193$F01§ 1e aw: are ass-1.8a minis 3.532? 2:834 g. Model the return X from an investment as uniformly distributed on the intervai from -8% to 12%. (X is taken to be a continuous random variable.) Find the 75th percentile return, that is, a po ' ' ' ~- . h that P(X S x) = .75. (a) .75 (b) 6.5% (c) 8% m (e) 75% y ;- -3 4. ..75( 11—53)) . ' H i? a: a? 6i” {”8557 (10) “‘3 at ‘ y ' -: -» s +~ is a: "? 20 - 21. The yield X at a given coal mine in tons of ore on a working day is normally distributed with mean 800 tons and standard deviation 40 tons. 20. Find the probability that between 75 tons are mined on a working day. (a) .400 (b) .789 (c) .832 (d) .886 (e) .904 . . i. x" f _ . 750 gee 350 an" 0 ms 3 X '3 LN 21. Find the 90th percentile of this normal distribution, that is a position x such that P(X , - (a) 878 tons (b) 895 tons (c) .90 (d) 866 tons - ' 0 1:1st '390 sea-i- 1.1311290) " 3&3 Q. A manufacturer of a 100W light bulb has data that suggest that the lifetime of the light bulb follows an exponential distribution with mean 2000 hours. Use this exponential model to answer the following question about a randomly selected light bulb from the manufacturer. What is the probability that the lifeti ,n we“ '9 t bulb exceeds 2000 hours? (a) .12 (b) .50 (c) .45 (e) .25 Lilw ..... —p\><' '— PtXea) = Ive )tK 2000 P6X > 7;) :: e .1": e i .: e = . 57 2;. A computer system contains 50 identical microchips. The probability that any microchip is in working order at any given time is .80. A certain operation requires that at least 35 chips be in working order. What is the probability that the operation will be carried out successfully? (Use the Binomial model. If you have a calculator that has the Binomial cumulative distribution on it, you may use that function to calculate the proba-Q'. ., : nerwise use Normal approximation.) Pick the closest. (a) .999 (b) .88 (d) .90 (e) .84 X” 8650430). F (X23, “l ‘PCXfEflLD Em) 2' 40 5°“): 232$ vse 3'4»? " 4i a-—-érmpmm-#—r+wmmmmmmwmnwmmm 23. Suppose that x in units of feet is normally distributed across a large population of interest with unknown mean y and known standard deviation 0': 8 feet. Suppose that a random sample of size n = 64 is to be selected from a population. Use a normal distribution for t_-. z. .._._- n e average a? to determine the probability that 3? will fall within 2 feet of y? 1 (b) .38 (c) .87 (d) .68 (e) .12 .. ‘ (4+ 2- E. A survey by an electric company contains questions on the following: Age of household head, Gender of household ‘ head, and Use of electric heating (yes or no). What type of variable is ”Gender ofho __, _ .- d"? (b) numerical _2_§. Below is a boxplot of the seal strengths of n z 100 sample potato chip bags. o 1 o 20 30 4o 50 so 70 so 90 1 00 Seal Strength The distribution of seal strengths is (a) skewed to the right .(b) skewed to the left (c) symmetric 2. Below is a stem—and—leaf display of the ages of 49 children ranging in ages from 2 months to 38 months. 0 [2234444 0 l5677778333999 1 l00012244 1 l5566899 2 |012344 2 I556? 3 I134 3 l8 H to Hwhox-qoom-qa Find the 15th percentile age using the book’s method. (a) .15 (b) 4_ months (c) 5 months (gt—Dias: 508165335? w Li ma 5W“ thbmgm I ‘ (d) 4.5 months e) 7.5 28. A sample of 100 persons have weights with Q]: 150 lbs Q2.— — 160 lbs and Q3— — 166 lbs. The three smallest weights 1n the sample are 124 lbs, 131 Ebs and 132 lbs. Thelo .3... _ - (a) 124 lbs ' (c) 132 lbs (d) 126 lbs (e) none of these .21? z: 1519 -11‘..6[1e) ===Ilé a 7:] . 131 1%! 2. Below is a dotplot of the weights of a sample of n = 43 deer weighed at a MDNR station on the last day of hunting season. Each dot represents one sample weight. 1,” (”a . : . . . : : : : . : . . . . . : . . : : : : . . /"'/./.—::‘“\ +———i.;_—.:.—- :11. ————————— + ————————— + --------- + ————————— +—Mweight (lbs) 150 170 180 190 200 210 Assuming that the sample mean is 190 and the sample standard deviation 1s 10, how many of the deer had weights within two standard deviations of the mean? - (a) 30 (b) 34 w (d) 36 (e) 95% 0M3 (“aid sari-mm 1701115 and 9.10 lbs 3 3’3 Cases) 30, A random sample of n = 100 Medicaid payments to a clinic from a large p0puiation of payments is audited. It is found to contain 30 of the payments which were in error. The 95% confidence interval estimate to - nulation proportion of payments in error is Pick the closest. (a) [.250, .350] (b) {.194, .406] (c) [.147, .453] 31. A confidence interval estimate for the mean p of a large, highly skewed population is needed. A random sample of size n= 3 1s drawn and the t interval estimate x. + Ems/J3 1s calculated. The t- interval estimate may not deliver the designed confidence level 95%' m this case. -‘ " (b) False 3;. A random sample of n = 19 from a normal population has me. - confidence interval estimate of the population mean is 40.6 i ' (a) 1.99 (b) 4.74 .31 ~40.6 and standard deviation 5 = 5.6. The 98% (d) 2.99 (e) 5.60 33 — 34. A company that conducts surveys of current jobs for low level executives wants to estimate the average salary of all low level executives to within $2,000 with 95% confidence with the z-interval. From previous surveys it is known that the standard deviation of low level executive salaries is about $24,000. Use the figure $24,000 in determining sample size. 33. What is the minimum required sample size based on the book’s formula for sample size? (a)1,231 (b) 944 (c)712 (d) 868 l: (e) 554 3 l 34. What is the minimum required sample size ' M . . . .Ition of is size 3 000? (a) 951 (c) 662 (d) 784 (e) 596 N g. An interest group in Michigan submits a large number signatures in its petition. The Michigan Bureau of Elections wishes to obtain a lower estimate for the proportion p of valid signatures in this population of signatures The Bureau selects a simple random sample of n— — 100 signatures to investigate. It finds that the sample contains 80 valid signatures and 20 invalid signatures. The 95% confidence lower estimate for p is _ ' (a) .704 (b) .760 (c) .722 (e) .787 1% .30 ' Fall-55:32) 933w LMS‘ (#3231 3,754 widest interval? I (a) 95% confidence, n= 289 (b) 95% confidence, n= 324 (c) 99% confidence, n= 400 “3% “iii .115 w W §_7. In hypothesis testing, the significance level 01 is chosen to be small toIcIu, .. Q. A Type 11 error occurs when (a ,_ Hp is rejected when it is true. -' d) H1 1s1eta11eEl \1l1en iisnote .9' 3_. Which of these statements is correct? H . ‘- (a) The p- -value of a test statistic' [S the probability that the null hypothesis H; is true. b pI-Ivalue of a test statistic is the probability that the alternative hypothesis H1 15 true. r;. m- (c) The —value'ol’ Ia teslf si’Tist’iWi‘s 11 1111111 oT‘W’edIB‘hT‘y iat111g ’T’Wlil’fin ligT it’l‘im eev1 e11.ce E (d) A large p—value leads to Rejection of 1““ Q. A t-test of H0: 11 = Ho versus H0: p 3:6 110 is conducted. The population is normal, Flo is a specified value, the sample size is n = 7. Suppose that tob, = l. 943 u e-ms- e 3 or your calculator to determine the p—value of tabs. .. 1 49w (b) .05 (e) .056 (d) .028 (e) .025 5(63 /wad+41/%€5+‘ P‘leug I 2X.0€5’:l0 41 — 44. The ACME survey of schools of business indicates that 10% of the faculty positions in schools of business are currently vacant. A placement service working for a renowned university wants to test whether the claim is true and collects information on a random sample of 200 business faculty positions chosen from universities around the country. The results indicate that 30 out of the 200 positions surveyed are vacant. Is there evidence to conclude that the ACME survey result is wrong at the 0.01 level of significance? (Use a two—tail test.) 41. The value of the test statistic for testing H0: p = .10 based on th- .1. - I data is zobs = (a) 2.24 (b) 2.04 (d) 1.54 (e) 2.58 30 — “5" ‘D :7 2.36 A P’% —- 2&5' M 2,9?) 42. What is the approximate p-val 5-4; e test statistic? (a) 0 (b) .01 (c) .05 (d) .09 (e) less than .01 v 0:“! a that. .o—lee :: 1K.ooase.015?— a ass &‘ 43. With the critical value approach to hypothesis t ' what are the critical value(s) of the z-test? (a) :l:1.960 (b) 3:12.576 (0) $2.326 (d) $2.750 (e) $1.645 Mile 3 44. What decision is reached at level a = 0.01? I (a) Reject Ho (Decide that the population proportion is different from .10) (b) Retain Ho 45 - 48. A study was undertaken to determine customer satisfaction in Canadian automobile markets following certain changes in customer service. Suppose that before the changes, the average customer satisfaction rating, on a scale of 0 to 100, was 70. A survey questionnaire was sent to a random sample of 121 residents who bought new cars after the changes in customer service were instituted. The data were to be used to decide whether the population average satisfaction was raised above 70 or remained at 70. It was decided that the probability of reaching the conclusion that the changes in customer semisfaction when it had not was to be controlled to 0. 10 or less. The sample data showed x =73 0; the sample standard deviation was found to be 5— — 19.9. due. in watt—+0 .1 +e5'l' 45. The value of the test statistic based on the given data is tubs = (a) 1.96 (b) 2.04 (c) 2.34 £72m :' 73,0 ’70 3 [-55 is a/ifiiT 46. What is the approximate p-v . “-:‘-'.\l e test statistic? w (b) .10 (c) .20 (d) .025 (e) less than .01 no a? gist... use. a w fiéia 2. 4 fig. '01” (e) 1.30 a L66 47. With the critical value approach to hypothesis testing, what are t ' ical value(s) of the t-test? (a) 1.960 (b) 2.576 ' (c) l 289 (d) 1.645 (e) 1.484 Twila :5 0L :7 3 f, C9 48. What decision is r 11651. at level (110310? W. W .2... WW ‘ RejeCt Ho (Decide that the population aveiage is greater than 70) (b) Retain H0 "WM“ ...
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