27_20110318fri_stats_86728_p11

27_20110318fri_stats_86728_p11 - 4 i i I More €513:...

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Unformatted text preview: 4 i i I More €513: Condihbnal Hubabwy- P234 (Contmgemc‘j Table) 50600 a 99600 .. (000° Probabih'f o' Beliz- Audifad AUAlW'ed :2 In coMe (Those. ads HA .n" 0-3828 0,3580 0, {354 ,‘ 4 2.6 25—49 50-99 >/ L00 T‘Jal [33 A” I IA come <e .— x=xmmcuem P A11) Audihd ( I) f {K {In 1 m Bey-“s -- '11—:5T RESULTS, (P, 23¢) @ Fa‘se Posfifvet The 712% resulfs mks that the COhdihbn [being teSivct fir) F3 F‘K’ESENB 8g] H' adoally :‘5 NOT! (a Fake Negaftite: The Tesf results erafe 413% {he Comm (‘5 ABSENT; 12%]— ;H: aafuafly PRESENT. [Z] Conérhgenctj Table —‘L|—RIPLE Bum Tsar” Dew-Ml SYMDROME' Down gynelrome. '6’:qu “(35+ s+a+u5 , , ‘ 3 Section 5.3 Conditional Probability:What’s the Probability of A, Given B? I 237 il- The Triple Blood Test screens pregnant women for the genetic disorder, [iii Down syndrome. This syndrome, which occurs in about 1 in 800 live births, "86W arises from an error in cell division that results in a fetus having an extra copy of chromosome 21. It is the most common genetic cause of mental impair- ment. The chance of having a baby with Down syndrome increases after a woman is 35 years...o d. 13(5) = 5 383 A study2 of 5282 men aged 35 or WZed the Triple Blood Test to test its _accura>;Y’I‘fwas reported that of t 5282 omen, “hases of Down syndrome would have been identified using th percent of __‘.J_“__ the gnaffec’ngregnancies ‘would have been identified as being at high risk 5226 for Downfisyndrome (these are false positives).” I ,3 O 4 2:3 2??? m lga? ' ° 5W - “arr?” Questions to Explore “if , 5,3,2 _ a. Construct the contingency table that shows the counts for the possible out- comes of the blood test and whether the fetus has Down syndrome. b. Assuming the sample is representative of the population, estimate the probability of a positive test for a randomly chosen pregnant woman 35 years or older. c. Given that the diagnostic test result is positive, estimate the probability that Down syndrome truly is present. " Think ltThrough‘ , K a. We’ll use the following notation for the possible outcomes of the two variables: I Down syndrome status: D ¥ Down syndrome present, DC = unaffected BIOOd test result: POS = positive, NEG = negative. ' Table 5.5 shows the four possible combinations of outcomes From the arti- cle quote, there were 54 cases of Down syndrome. This is the first row total. Of them, 48 tested positive, so 54 -— 48 = 6 tested negative. These are the counts in the first row. There were 54 Down cases out of n = 5282, so 5282 — 54 = 5228 cases were unaffected, event Dc. That’s the second row total. Now, 25% of those 5228, or 0.25 X 5228 = 1307, would have a posi- tive test. The remaining 5228 — 1307 = 3921 would have a negative test. These are the counts for the two cells in the second'row. f El The Mul‘h' [I‘Ca'h‘Ol/I Quite (P 238), /.'T_ fin ‘_ .2 W; ; ,z' ’4‘ -‘ . .’“_1/-t-..;4 (1 E la? A and, B are INDEPENDENT 8%?ng we“ PCAIB) -.-. PCA) Also PCBW = PCB). ‘ B > = m X -- .. a“) We 34 and v *‘ B me A: _ 5°“ P<B>x PCA Student: Instructor: Doug Jones Assignment: HW19—53 Date: Course: STA 2023-86728 #5,... Time: Book: Agr‘esti: Statistics: The Art and I 1 Science of Learning from Data, 2e Q For the results of a certain suwey, the estimated probability of being a binge drinker was 0.43 for males and 0.4 for females. Exp’ress each of these as a conditional probability. The probability of being a binge drinker roir’fiiéim Express this as a conditional probability. K 0A. Inorganifow QB. P(BD|M_)=0.57 @C- P(BD|M):0.43 Oil Pgns‘i’ijéoas The probabitity of being a binge drinker (BD) was 0.4 for females (M°). Express this as a conditional probability. @A- P(BD|M°) =0.4_ GB- P(M°IBD) 20.4 ()0 P(M°|BD)=0.6 GD- P(BD]M°)=0.6 Page 1 Student: Date: Time: Instructor: Doug Jones Course: STA 2023-86728 Book: Agresti: Statistics: The Art and Science of Learning from Data, 2e Assignment: HW19—5.3 Because of the increasing nuisance of spam e—mail messages, many start—up companies have emerged to develop e—mail filters. One such filter was recently advertised as being 922% accurate. This could mean one of four things. a45~u= ME.mu—"—1fiM (a) 92% of the blocked e—mail is spam, 0 www.m- W (b) 92% of the e-mail allowed through is valid, (6) 92% of spam is blocked, or (d) 92% of valid e-mail is allowed through Let S denote the event that the message is spam, and let B denote the event that the filter blocks the message. Using these events and their complements, identify each of these four possibilities as a conditional probability. (3) Express the statement "92% of the coke e—mail is s am" as a conditional probability. ' B "i B then 5 , GA. P(Scch) GB. Pals); / ll: J 00- p(s°|s°) @9- (b) Express the statement "92% of the e—mail llxed through is yalig" as a conditional probability. OA- P(BES) éa. “8%? V 17c 5f, dense. ('30 P(B°lS°) (c) Express the statement "92% ofjpam isjwlgpged” as a conditional probability. ' (DA; Mrs-«a—“'”"""'W° I " QB, 0'86) ‘I‘wfl g .91!" as. P(Blr‘w;’~r - " "' on P(SCIB°) (d) Express the statement "92% of valid e—mail is allowed through" as a conditional probability. C rem-«u :1: 3C 5 ‘J Hen 3 . OD- P(S[B) am I H M Page 2 a“ AFTEP (image- Instructor: Doug Jones Course: STA 2023-86728 Book: Agresti: Statistics: The Art and Science of Learning from Data, 2e Assignment: HW19-5.3 @ Method Used To Commute Frequency Eli A census reported the given table for how workers 16 years and over commute to work. Drove atone 97,192,412 a. What is the probability that a randomly Carpooged (f§696 146 a, 56171:“: Per”: camOOled to work@ Public transportation “’a F to “’0‘ ? Walked (3,738,611 0.- b. leen that a person carpooled to work or O h I WW” walked to work, what is the probability that t 6‘ means ’ ’ Worked at home 4,123,223 the person carpooled to work? Total 128,379,901 61/ a. P(carpooled to work or fled to work) : D w . (Round to three decimal places as needed.) ([5 695 [416 ‘i' 3 ?39 5’?>/’25’5?990 / 5'3 O.{51334?¥?5 l). P(carpooledlcarpooled or walked) 3 [j (9%] (Round to three decimal places as needed.) 15 696 I"! 6 I 5' 6,96 M'é; + 3 $386!?)S’9-80753248 33 The given table classifies auto accidents by Outcome '21 survival status (S : survived, D 2 died) and seat Belt S D Toal belt status of the individual involved in the Yes 4043' m m 04,897 b I, accident. Answer parts a-c. N0 181 690 1536 183 226 J ‘ P (L u;¢=;”"*""""”"'3 Total 586,071 583,13) a. a. Estimate the probability that the individual died (D) in the auto accident, P(D) = D (Round to three decimal places as needed.) _ LEG» 2" 4' 00 3 2 2052/56912 33> as 0.003 489 @661 [1. Estimate the probability that the individual died given that they (i) wore, (ii) did not wear a seat belt. SIG/404 893230.00! 2?43982 (3) P(DtWore seat belt) = Bio. 00! i _ (ii) P(Dldidn't wear seat belt) = (Do not round until the final answer. Then round to three decimal places as needed.) Ig3é/i852?éfi aaoa 383 089;: c. Are the events of dying and wearing a seat belt independent? t Oi). Yes, because P(Dldidn’t wear seat belt) is equal to P(D). OB. Yes, because the two events cannot happen at the same time. OC- Yes, because P(waore seat belt) is equal to P(D). Q D. No, because neither P(Dlwore seat belt) nor P(Dldidn't wear seat belt) equals P(D). Page 3 Student: Date: Time: v—u— All“ Clogs -“ Instructor: Doug Jones Course: STA 2023—86728 Book: Agresti: Statistics: The Art and Science of Leaming from Data, 2e Assignment: HWl9-5.3 One pro basketball player was known for being a good shooter. In games during 1980—1982, when he missed his first free throw, 47 out of 53 times he made the second one, and when he made his first free throw, 253 out of 285 times he made the second one. Complete parts (a) through (c). a. Form a contingency table that cross tabulates the outcome of the first free throw (made or missed) in the rows and the outcome of the second free throw (made or missed) in the columns. 1/ ‘/ 2’85 2nd free thrgwmadegnd fi'ee‘t/hrowumissedwTotai #25 a @253 [:32 D 285 32 RTE 6 am D53 D33 [3338 b. For a given pair of free throws, estimate the probability that the player made the first free throw. W lst free throw made 1st free throw missed Total (Hint: Use counts in the row margin.) 2 g 5/3 3 8 a: 0. 84 The probability is . (Round to the nearest hundredth as needed.) For a given pair of free throws, estimate the probability that the player made the second free throw. (Hint: Use count? int olumn margin.) goo/33g: s3; 0.88? 5?? 9645 0. 89 i The probability is (Round to the nearest luggdredth as needed.) c. Estimate the probability that the player made the @fiee throw, gifizn that he made the _f_i_1_'_s_t m (25:35 153/285 2: 0. 88? no 2982 The probability is (Round to the nearest hundredth as needed.) Does it seem as if his success on the second shot depends strongly on whether he made the first? C) Yes @ No 914'” ( PiSl 3" WSW), “Thereierc; "the, extents Fowl is we indefent' Page 4 Instructor: Doug Jones Assignment: HWl9-53 Date: Course: STA 2023-86728 Time: Back: Agrcsti: Statistics: The Art and ' Science of Lcaming from Data, 2e - @ regnant women were tested to see if their babies had Down syndrome (D). The results of the test are given in the table. Use the values in the table to answer parts a-c. 5&‘3 “file bQIBK Total I 1056 m —--— Blood Test POS NEG a. Given that a test result is negative OVEG , what is the probability that the fetus actually has Down ml/ 835 a: 0,690! 325 2954 P(D[NEG) 2 Die. 0 0 :3 (Do not round until the final answer. Then round tofigfir decgmabplaces as needed.) b. Given that the fetus has Down syndrome, what is the probability that the test result is negative? We) are m the "[3 world, ’ 50 use the 117-1139" 7./4g m 0,145 883 13333 PtNEGlD) = [3 (Do not round until the final answer. Then round to four decimal places as needed.) M... u n w .4“ (row, (alumru or c. Is P(DENBG) equal to P(NEG|D)? PM! " in grand finial) ab Cam . FYI QA. Yes, because PCNEGlD) deals With a much smaller pool of fetuses than P(DINEG). (13 {38. Yes, because P(NEG|D) deals with a much larger pool of fetuses than P(D{NEG). )0 No, because P EG D deals with a much smaller 001 u etuses than P D NEG . a (N I) 2 (l >C3333 00- No, because P(NEG{D) deals with a much larger pool of fetuses than P(DINEG). @ In a women's tennis tournament, the winner of the event made 74% of her first serves. When she faulted on her first serve, she made 87% of her second selves. Assuming these are typical of her serving performance, when she serves, what is the probability that she makes a double fault? P(double fault) = D (Round to four decimal places as needed.) . 033 E a a...“ o i a. u *"a— t “ Has Down syndrome) «the mlemqugzgymgflgggt “ Does not- have Down syndrome. ” " I c :0.3? fli F? Makes Fri-st; 8: Makes Second, 9., PCF)- 0'M and P(le) F‘: Mlises Fu‘fli- , P c‘F-C) u h [rm-5 es the 23—4 we are trymfi +0 fifid he“ ‘5 (S J 519:; thit she misied ' .I’ But: this 13 Much efflhPiBF "than if gouaartfifrlflwwffijfiimt 0' 26 x a e = $623.5 {mam-rt) w- . whammy/x“.- /*““‘A U", r Page 5 Student: Date: Time: We Worked Ail-er class we Instructor: Doug Jones Course: STA 2023—86728 Book: Agresti: Statistics: The Art and Science of Leaming from Data, 2e Assignment: HW19_5_3 I ...... .. A correction in the February issue of a magazine stated that the January issue that year "contained the incorrect statistic that three—fifths of voters for a presidential contender identified moral values as the most important factor in their decision. In fact, three-fifths of those identifying moral values as the most important factor of their decision were voters for the presidential contender." a. Let A be the event of identifying moral values as the most important factor in the voting decision, and let B be the event of voting for that presidential contender. The article implied that P(AI B) = 3/5. Give the notation for the actual probability that equaled 3/5 . Choose the correct answer below. Cm- P(A) OB- P(A and B) so. P(B! A) ' \\ OD. P(B) b. IfP(A) = 0.20 and P(B) : 0.51 with P(Bl A) = 35, find P(A and B) PM ml 3): POW WSW — 2.; 0,20,40,60: 0.12, P(A and B) = D (Round to the nearest hundredth as needed.) I 0. {2' c. Using (b), find P(AI B). This shows that P(AI B) is quite different from P(BI A) when P(A) is quite different from P(B). Keel!) PTA and B) e: 0&2 ) but" Pan‘hd 5) also equals P( B)$ POMS) and We»: PCAIB , 50 mas—mew FYAIES) P(AI B) = 1:] (Rdund to the nearest hundredth as needed.) 0. 24 “an to a % panama“ 0.12: 0.61 x P0543) at» Ptaaudfiim not Wile) 50 was): fi','§.~a0a23523+: I it: 0&4 Page 6 e After Class-w- Student: Instructor: Doug Jones Assignment: HW19-5.3 ' Date: Course: STA 2023-86728 Time: Bock: Agresti: Statistics: The Art and Science of Learning from Data, 2e .-- of _ . In @fiips of a coin where there is chance of flipping@let A denote {first flip is a tail}, B denote {second flip is a tail}, C dent {first two flips are tails}, and D denote {three tails on the first three flips}. Find the probabilities of A, B, C, and D, and determine which, if any, pairs of these events are independent. P(A) : D (Round to two decimal places as needed.) (.0. 56 i P(B) = D (Round to two decimal places as needed.) ‘0: 56 P(C) t D (Round to four decimal places as needed.) [6.1936 1 P(D) : D (Round to four decimal places as needed.) Which of these pairs of events are independent? 01‘) Events C and D because P(C and D) L" P(C) - P(D). 05- Events A and C because P(A and C) = I’(A) ' P(C). QC. None of the pairs of events are independent. @D- Events A and B becauseé/Eand B) = P(A) - P(i3) M T95} in But a, reallj nice Problem ii ,sgxm MM :0. media .56 Y :44 K.5(o “5131984 55w.44 -:- 0.131984 c C. t START J56X| .59 mamas == oJ’HSsis A: F.9d €1.13 is T No cola neeessangb E: Second (lift? is 'T “WWW—es No Cale hemswls a Q: TTH or T’TT, Me) = o.ssa984 + esteem skiff-1:5“ D‘. ‘TTT Pa?) =- omvsm e 0.172% 3&3 Page 7 ...
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This note was uploaded on 12/30/2011 for the course STA 2023 taught by Professor Jones during the Spring '11 term at Tallahassee Community College.

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27_20110318fri_stats_86728_p11 - 4 i i I More €513:...

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