This preview shows pages 1–11. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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"
03828 0,3580
0, {354 ,‘ 4 2.6 25—49
5099
>/ 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” DewMl 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 202386728 #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(BDM_)=0.57
@C P(BDM):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(BDM°) =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 202386728 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 email allowed through is valid,
(6) 92% of spam is blocked, or (d) 92% of valid email 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 202386728 Book: Agresti: Statistics: The Art and
Science of Learning from Data, 2e Assignment: HW195.3 @ 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’980753248 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 ac. 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: HWl95.3 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: HWl953 Date: Course: STA 202386728 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 ac. 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 1171139" 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(NEGD)? 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(NEGD) 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 Frist; 8: Makes Second, 9., PCF) 0'M and P(le) F‘: Mlises Fu‘ﬂi , P c‘FC) u h [rm5 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 = $623.5 {mamrt) w . whammy/x“. /*““‘A U", r Page 5 Student:
Date: Time: We Worked Ailer class we Instructor: Doug Jones Course: STA 2023—86728
Book: Agresti: Statistics: The Art and
Science of Leaming from Data, 2e Assignment: HW19_5_3 I ...... .. A correction in the February issue of a magazine stated that the January issue that year "contained the
incorrect statistic that three—ﬁfths of voters for a presidential contender identified moral values as the
most important factor in their decision. In fact, threeﬁfths of those identifying moral values as the
most important factor of their decision were voters for the presidential contender." a. Let A be the event of identifying moral values as the most important factor in the voting decision,
and let B be the event of voting for that presidential contender. The article implied that P(AI B) = 3/5. Give the notation for the actual probability that equaled 3/5 . Choose the correct
answer below. Cm P(A)
OB P(A and B) so. P(B! A) ' \\ OD. P(B) b. IfP(A) = 0.20 and P(B) : 0.51 with P(Bl A) = 35, ﬁnd P(A and B) PM ml 3): POW WSW
— 2.; 0,20,40,60: 0.12,
P(A and B) = D (Round to the nearest hundredth as needed.) I 0. {2' c. Using (b), ﬁnd P(AI B). This shows that P(AI B) is quite different from P(BI A) when P(A) is
quite different from P(B). Keel!) PTA and B) e: 0&2 ) but" Pan‘hd 5) also equals P( B)$ 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 Classw Student: Instructor: Doug Jones Assignment: HW195.3 '
Date: Course: STA 202386728
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 skiff1:5“
D‘. ‘TTT Pa?) = omvsm e 0.172% 3&3 Page 7 ...
View
Full
Document
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.
 Spring '11
 Jones
 Statistics

Click to edit the document details