Sp05_EC41_key_09

Sp05_EC41_key_09 - .3130 g 1[Q We PKss PKPJ C 3[Q 7373 P4 I...

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Unformatted text preview: .3130 g] 1 [Q We? PKss PKPJ C 3 [Q 7373/ P4. I Eli? {3C3 -03.... 6033‘ JPK‘S 3. 33m __ .. A70 .. SKA MP IMd‘Pfif 140/10 P1 7‘ _ ((123) Mm: PC: _I 133" 7%,) [7.73.3 .353. . M3 5.‘ 7 __ SUP/1M? drgldaw'r‘wmflv “.13: 63043 P P @412... __ ___. «- PMPPM PPWJ pg _ i _ (Lem +1.:fiP_ an 022M! __ (no __ ______ _ Pie 0L1" __ -- 210 ______ _ __ :22 1%:de mm:f_’“ “2 _ __L._.61_m _. PEP/9P _P (Fl; . Pg» _PKEL RCFJ--. (2* ___:_.9_6 _ _ _ _ _ _.___ fir-1:33.31 {D_ _ Jié IFS: .P(E3 2 75 _:__.___:(f~:,) "'22: '_ ”2 '/ ___ 31}, :0 5‘ _ A) + NB Pam-:3 ;;‘;;fi_;“ 05 t 10? 0": ’:___ t __2_ _ U WU @1_:/____Qst€_é@_’_ {a __ _.__. ___... ___. ___, __u—Ir: ___ ._ r“ . . _ ._. . ._.—- eriéfl aflPf/MJa/w r9:- flée‘ _ h : _ My» Bfgnw _Ner'nuJ—.,_s:z>__4 W 40 _ _ I . "It“$Q..[;[email protected]flljpéf.._ wdwdw __ ‘ ‘ " _ w IPCRn u): s 35 __ pram __(éé'flfl'iégfl __ __ n: iii w J18“ P69 I)? _n ‘1 /=‘-._3£7.. 1 3037?: a; "(21957“ 11' M? X .2 (1361:} 23) 2 (g) __(lQJZ‘ (5J6 * «454.115 K' 4/ 3 :1 __ — --..._ _ Iii ) é/B (iréjfirZCJL-L(7{)Q; __, __ __ __.,__ , 1 U4 ”Image w/L 3+; __(‘ULrt’u/m -___0!_U€W1M_ _CW @11- a particular firm 10% ofexecutives are “empty suits” (have no knowledge ofthc product and are generally unhelpful, but specialize in handshaking, backslapping, and golfing). Let X = the number of empty suits in a random sample M3. 3) What are the possible values ofX and their respective probabilities? X Probabilitv 0 “g _ — (3‘) (115111 , .72a ‘f 1;: l 1? L L l l (“M (‘31 1’3 1 077 aflf'flalli ‘017 3 .ool (214190?“ .001 h) Draw 21 Probability Histogram oflthis random variable X 1:) What is the standard deviation ofthis random variable X? .{—\ hrj): Kfllffi >=r=nP/-A‘)\lm=- @ ormal Approximation to the Binomial (section 512). Consider the firm 1n question 1. Again assume 10% of 1ts cxecuthes are erupt}- suits, but now let X= the n11111bc1 ofcmpty suits in. random samples of 100. Mt l’l‘ If): 900/) l ~ /0 G“- 11 1 i H ‘ ' 3 a) Use the Normal approximation to the Binomial to find the probability that X 15 greater or GKetpliall-tloC g and lgthan or equal to ‘ 5 (v1 ithout using the continuity correction — u ithout adjusting/or the fact that Binomial is dis rete, while normal 13 :mitinuous), P(S<X<IS). K‘ .5) p 5 ~ )0 I 8 ~ 1% l3{s“éxéls) <2<fit§1 3 " 13 % < 2 < 53 = [3 (241.57} 13(2< ~L67) 1) Now use the Normal approximation to the Binomial to find the probability that X is greater or equal to 4.5 and less than or quul to l5. 5 (using the eontimStV; co1rection) P(415 <X 15 5). 1;;10<Z<’5:”’0 P{l83 <Z<l§>b p{z<12g) Pf? 4231] Fleet/303% ) Use the normal approximation to find the probability that X- — 10, (with continuity correction - this is necessary to find an stimate of the probability of one value) 1 e ,P(9. 5<)(- 10 5). 1': '5—1’042< “13—“ .~ 73(17.<2<17 Q: ’3) 5'0 1p(2<17)—p[2l<—1/L_\@ 171:0 ‘- .5674" HRH bl : ‘4525— 01/7; f} 1) “QB below shows number of‘ziuto insurance customers in each category. Rezili7c total number of customers = 2000; and n-ivers file a claim after an accident that requires repairs. Filed Claim Did Not File Claim Female: 100 800 4? 190 Male: 290“ __ __9_90_ __ __~ ______1_1l{_£_) 3m) 17 W 20120 300 )Randomly SClt c1 3 customer. The unconditional probability they filed a claim — / (x E0 ”800 U "‘ )Thc conditional probabilitv P(Filed Claim [ Female) L03 1 f _ i D ' ' 900 .lH ‘ l tl/D l Are gender and likelihood offlling 21 c laim independent) for these customer’s'. AM)- Pfllarwll-‘S‘ .lll *6 p (Clo/MA: IRandomly select a customer. The unconditional probability the» me female '— 9 Ob : ill 3 ‘ 2/ gj’b 10 ’35 Theconditionalproba‘bilityP(Fe1nalelFiled Claim): 195 K 1 23 ; 33 90 ll“ 3 ' , Supp iletes are tested for use ot'bztnned hormones. The test may be either nosrtive or not positive (negative), and the tlete may be an actual user or non-user. Assume the test correctly identifies a user 99% of the time robabili ositive given 3199); gives false positives 5% of the time (probabilitV positive given non- user' —— .05); and that 10% ofthe athletes actually 2 users (unconditional probability 0 r0 . = 10) What' IS probabili that an athlete who tests positive is actualh a user? I In:- 1 be done using Bayes’ Rule on pan .1 +l LU pfl/ll - U 1 1L) - M MHU ) E( M) (126‘) mlutitw) ”’qu ml 6 i i 697? 1( U\ 4.qu Wt 1M Pfl/3; JD __ : ‘\ fathom f’it’llfto l l’ (0?\((l0\ -‘leU// ossible values for event B? @vem A occurs with probability 0.3. If A and B ar.then what is the ran Zita") M41910 ‘ fl PQU-é‘i wat- Whig) age/9) lhe true percent return 0' lemmon stock of two companies is below. X=Percent return A__' Probability Y=Percent return B_Co Probability 3% .5 1% .3 4% .3 4% .3 % .2 7% .4 1) Find the mean ofthe random 7v7{iable X @ variance ' 6 i and standard deviation____ ['2’ : humidifier” C-Slé 371 4138* 3324-163-75 “3* *7 “”1 335" 1) Find the mean of the random variable Y @ variance_ 6; 2. i and standard deviation_ Vbl 2| :6. 26 (3363 (Mil {£167} = 4.; 946—le 4.3 6-11.52 t.H(7—4,a)l= 30277 ;) Assume the returns are independent so the true linear correlation between X and Y, pxy = 0. 3; (Z i b _ 6| 2 | _____ What is the average return (in %) r a portfolio of S 0:. A_Co and 50% B_Co? 1 371+ ,s‘ 1/133 I.) Censider the percent‘return ofthe portfolio in c above. Find the variance< i ‘ 706‘ 2nd standard deviation- G73"- ,slttf m1 of i 2 [r] OXi/gifl/ $56” i‘ .2§(6.Z)] ‘ 115251.).9523— ': {(703— )a - - large population, 59/11 ofthe people have the flu. If two people are chosen at random, what is the probability that they 0th have the flu? (write your answer to at he; ‘ ' 131 places) 029%,- ) Suppose a factor) has 400 workers and 5% ofthem (20) have the flu. If each person can be picked only once what is the obabilib. that two people chosen at random both have the flu? (write vour answci to at least 6 decimal places) 340- {iii 3&0 ‘ ico2%\ {HP/5 HUD (Wt («:qu @Fllp a fair coin and consider the numb vertical axis, and number offlips on - - n Nonnal or Binomial 01' @Flip a fair coin. Consider the number . 1138 required for the first he ' e probability ot'gening the first head on he 66' toss (i.e., 5 tails in a row, followed by a head)? (1556 - '0 [562.5— ' 5 required for the first head. If We graph probability of first head on the ’hat type of distribution is this? or Exponential probab' ' A, ' "u r V Consider tossing a coin. Draw the probability that the fiflead is on the Xth toss to the right (up to X = 5) Z . a What is the probability that the first Head is on the second toss? 8: 5’2 -, 2 (f ‘ number of toss 0 get the first “head” 1 V ' 3 < < this sum = —. $918 Geometric series can be written; 3.0%; + If 0 a: 1, l ~1- 1 _ a . . . 7 = 9 ‘ What is the sum of the geometric series if a =.2. l‘ “”3 .1 . "l r ' ‘7 _ /7""~ _L l 2 m: 12 ...
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