fe2005s_sol - FINAL EXAMINATION Statistics 241/251 July 30,...

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Unformatted text preview: FINAL EXAMINATION Statistics 241/251 July 30, 2005 Time: 2.5 Hours Student Name (please print): _______________________________________ __ Student Number : ___S_Q_I;:M:£LQ_N~; __________________ __ Signature : _________________________________________________________ __ Registered in: Stat 241 Stat 251 Notes: 0 This exam has 5 problems on 10 pages,including the cover. Please check to make sure that you have all pages. 0 There are a total of 65 marks, and the amount of marks for each question is shown beside the question. 0 Show all you work and / or state reasons justifying your answer in the space provided to receive full marks. 0 This is a closed book exam, although you are allowed a two-sided crib sheet on standard 8.5 x 11 paper, and any ‘clean’ statistical tables that you would like. a Good luck! I hope you enjoyed the course, and have a great rest of the summer! I Problem I ..1.. I ..2.. I ..3.. I ..4.. I ..5.. I TotalI lMarkS ll? MIC: M1117 I tSl [18 Marks] Problem 1: A local company produces square mirrors. The length X and the width Y of the mirrors are independent random variables and assumed to be normally distributed. The company takes a random sample of 22 mirrors, measures the lengths and widths and finds sample means of 54cm and 51cm and sample standard errors of 4.6mm and 3.9cm respectively. [4] (a) Calculate the mean and standard error for the perimeter of a mirror. Leif QQVQNW:P: axtay em: thway) = Jam amid? axo® \IarQP)=\lcv(9LX1)\/>t "iVNLM #fiumy) : HSH :7 SD‘KHHS : pact: ) What is the probability that the perimeter of a single mirror is more than 201cm? i3](b fitfwmq : Pl? v a:;~;:‘°)’ (’(Z> ~or7s)® = l » (icons) : 0.77 MO (c) What perimeter size would we expect yh of mirrors to be larger than? \ii’ (>0 i9< iii—e 33?.in wb’i’ck 30‘/- are ifirfiej #krn PthPajzoa a Mavi‘Zfi-L‘? 2“ [4] ((1) Calculate the mean and standard deviation for the area of a mirror. lel area 1‘ A 2X7 Em) = EMORY) = 97g~® ll VarLA3=xxarml = cam?) ~ arm‘s: amen/1) ~ [emcm tth) = ecfl~ cafe) €00) “- \lcm) » emL EN“) = WM) * EW) goNqLA) = (mm) tE(X)13(VN(y>rG(Y)Z\\ [gum-m] : 016171130: 5" 3090: alanine = 36.36 [4] (e) The owners of the company are concerned that the average length and width are not equal, and therefore their mirrors are not square. Test if there is evidence that the mirrors are “Qt Square “Sing a 1% significance level. “0: MLchj \l‘q' “a; )o(= U-Ol (l’S‘IJU-‘A\ ____._ __,_.._. K a .1 Z 0 oiog w 4 —\rk\w ( fawn ’- P(lhm\ 23333-—> om r we FTK l'lo 4- COAUW‘N lk‘l Q m Mn “talk 0% m wwmcuy NIH Mam al' lN \“)~ Sljwi'RCMq \4wl. a“ gram e~vc\u(5°< I [12 Marks] Problem 2: You are working with a team of 8 other computer engineers, making 9 of you in total. Together you are writing a computer program and each person is assigned to write a certain part of the computer code. Team members work simultaneously,‘ although independently without receiving any help from each other. It is the completion time for each team member’s task follows a uniform distribution between 5 and 55 minutes. [5] (a) What is the expected completion time for the computer program? \ u ——r-\-—’ :J’ \Qx’ .H-N Sor PJLFSOA k ) {lxd‘ b—q go ._ 9‘ _ 9‘ CL 3‘ 3;,l i'Uq‘S 5‘“)‘&“'§,E‘3dq Eo[u\ ‘go l0 -0» 5 g [2] (b) What is the probability that it takes more than 100 minutes to complete the program? 0, (CH-incl mark mlNMl€S® [5] (c) If we now assume that the completion time for each individual’s task follows an exponential distribution, and that on average it takes each person 50 minutes to finish their task. Calculate the Interquartile Range (IQR) for the completion time. > : (oh \1*{9,‘ 40M Far (‘S‘OAC ‘llr-QA ‘el— XL P (be «alone ‘0‘01x ‘KCNEKP(€=§\ ,j(x()= 0.038 ,FOLQ): \—6 «Mn» ‘l 2, Rm)? Cam)": 0— e )0 mt: (Famvonfl “ (FV‘MVO’AQQ Robin“? E3) S‘wouime 1—9 Rtm‘vms =0 0‘73 = (“a «mum “°‘°""‘ q! ' = . 'EH 4 quizsil'e ——17€ ‘l" “‘75 00 (a b) 140.031‘1 : '7; C1 A _ ' —-= M ‘ A .0). ‘O‘Olmz)e‘ o—v-‘D 9 o a 1“ Qunmwe 9 (“may 0.18 >> 0-18“ (l-€ .. /£h(l_W)= 017’)» ml,- .—6.0L 3°» IQR= lilfi’seq‘mg 2 75.680 [6 Marks] Problem 3: Suppose there is an electrical network that operates using the system of independent com- ponents (a1, a2, ..., a7) shown below, where the number below each ai is the probability that the component fails. 0.03 0.02 a2 a7 0.03 0.02 [3] (a) Calculate the reliability of the system. \i\— camsomspa“ L §c~K\S-§ Q N @Q\mbi\\¥\/= (’Uktuhihtfiumun‘s)nLAZdA§ ’ L ‘ [A ” NW“) NMUASQAE PUEVM) < €t(A.nM‘3PUA3~/\ms) )Plhwm :- [\ » emmmlb— «met/lama] [I — WAN/#1310 = (o.qquowzsqsyo-qu) = oxmxi® [3] (b) What is the probability the system fails given that it has made it through a1 and a2 successfully? fi‘moz ‘iLP COMQONAS m \ncNMJ-Ml, NSx/S‘liulls Twat) (HIM Clflaz) : P ((hsnfimfls-MUtbmmj : Humor-5) t (Jimmy - etmnmmwmwfi c «mammal *elmetm— el‘MPthlnsWMM/i ® (0.1}(oegxoil) r(o«oz\(o,01) .. (0‘l)(o.o§)(otl)(0‘03)(o.o)) : 0 ‘000 15qu r— v 6 [12 Marks] Problem 4: You and your engineer friends enjoy drinking on the weekends and so you have devised a drinking game. You get together and watch the TV show “The Smurfs”, and everytime you hear the word ‘smurf’ said you each take a shot. You are an academic alcoholic and so you would like to better understand the workings of your drinking game. You watch the show 35 times and find that on average the word ‘smurf’ is said 7 times per 60 minute show with a standard error of 3.3 minutes. We assume that the occurances of the word ‘srnurf’ are independent and somewhat rare. [4] (a) What is the probability of hearing the word ‘smurf’ 3 or more times Within the first 26 minutes of the show? \{ l-r \‘1 all +Omes "SA—Ml" l3 S‘CJA in Nxf' ab MPA‘ARS . wqesumfinl ‘A E0\\7w5‘ c. PmSM-M Jus’mtuhoA 'm =l‘f— : . A ‘ > $.03 XNPmsSoNDA) , WM “:7 ,A (,0 093 ,M Pow/33 c \— PW $1) a \» Pom) wow) \ P(x=2)(|) -3433 o 40% 4°? 1 C \_ e (3.07,)” e (3.03)" a (To?) = 0 at |l_ 2, ; \, OOHZZL~ ()‘ww—oanni = 09(350 [4] (b) What is the probability of going the last 18 minutes of the show without hearing the word ‘smurf’? . \ } ¢\\ h ward ‘Smle ‘llxn Vt‘ T: wcfllrrk) *‘mxe (Anl l lvl )‘l I TN E‘M’l'fl 7 Ho ’ PtTétl * \P 6 «x c «A «be, \e\ \I‘# HMS Smwrlst‘xfil \~ WWW“? H“ yxpilti‘lgwl) a“ , -(1)(g . X: r " ea ‘L m 90/”) 6 [(>7x%lo7 "mm _ =P =(‘Lla'7HL h [4] (c) How likely is it that the word ‘smurf’ is said more than 3 times in a. 60 minute Show? le“ \fl= H *{NS (var§\ \S SCJJ \r\c\ (DQNWWP 3k)“, fL-Qn 7"”?0‘550“ (7‘05wa W7, Ni => 15:7 0 QLXZL132 Pg“) ’ \“Hbmi “POPQ ~fixaz) ‘ my?) _ a 7 n t \p 37(1) _9 (73,6 0)le (7)? o! 0 l.‘ 2|. 3( a ‘ Ku’\hvl \1 ’5' TR ‘Zxfiockefifi‘o? (OMAC .3 \esr The,“ 9.0, H mg sho\ h 0V V N NJSMJ Pfqprox. *o we PMSSUNR & we c/(of ~O~S~7 RXMV \-?(x<w) = \“i’il‘flfi > t \,_ €(ZC-,_32) : \.. L}- 9(Z¢|.31)] ~ ()(ZLL31) : LP jraLe DER- 5sw (wuH‘S \? 5M “*3 We NOW“ “Pm-V. [17 Marks] Problem 5: You are responsible for hiring 1 new employee to work in your factory, and currently there are 3 qualified candidates. You would like to figure out if any of these potential employees work at a significantly different speed than the others to aid in your decision of who to hire. You decide to time them in completion of the task they Will be hired to perform. You let persons A, B and C complete the task 5, 7 and 6 times and find sample mean completion times of 32, 25 and 31 minutes with standard errors of 3.2, 1.6 and 2.5 minutes respectively. There is an overall mean completion time of 28.94 minutes. [5] (a) Test if the completion time for all of the potential employees is the same using a significance level of 1%. Make sure to outline the steps you have taken and state a conclusion. o’MA°Me=Mc \:»S- Hq"Mi‘-fM_', ,{VJFSoo-(Cfild' I 4:043] a. . '1 a: ‘ 33k > 2k: fic(?g*§)l= S(2L—1$¢1°1)1t7(1§~a.3.w) * G (SI—3.9.“) (70 (1‘1 Ma 3 ‘20:?” s Waiw K 1 2 z * ’ I : 73g] 336 ’Z was; a (NU-23 mom) (a, 00.5) g _ a 3329 M33“ we €36 o A I 1 . F“ %%3"5 90 FLOJMHJHlS') : .3: w—Q l‘e_\<C* Ho L COALKUA" “*7 do dfi‘" Sigmgtcwxlh, at ’W’ \éwl m [9] (b) Construct 95% simultaneous confidence intervals for all possible comparisons of means, and state which workers differ significantly in terms of completion time. a - r 0-05 _ 00$ ‘ C 0< ‘0»05 :7 06“ ‘ T 030 JCM’H‘x‘i: {(516%) -a‘bl , T ~ ..- ‘ S "“. Iifin m . + ' a.» ..|fl \otx L2L~‘1S\*/. (1M) gafw'lfi“ a? "7 _ 2m 3 , ) ‘ --—-P e~ V H401 M3 (221m) ’7. (104335-3321?sz L 2.6“, , ® , an *4, , ‘é r0133) MB ~Mcf (1§~3i3*/-(;L(a‘1\SS-9sei~g73 C 7 1) ® ,3, wort—2r B Rag A Sijmiicm'} )7 is Mr (ox-19V h“ km in“ wart“, A N C ‘ MA A :1 (A0 M) gotta S'Ijmfrcmilx/ [3] (c) If we assume that completion times follow a normal distribution and that the standard deviation is known to be 3.2, then what is the probability that person A takes more than 35 minutes to complete the task given that it has already taken them 33 minutes? iei‘ x: iv“ ear WJOAAS); 7033) 9003;) m r ‘- Yo " g /_,... P<X>33 iX>3$) " W \9UO3; ., ? p (O Qix>35)“?(2>‘3::<)=P(*«z>cw~t) om 0 = 3737) . ~3 - 270‘3\> 0 00033): «$23) a V 0.\72b » -“ HS"! Q 0(378; O 0 D 3 10 ...
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This note was uploaded on 03/28/2012 for the course MECH 260 taught by Professor Ross during the Spring '12 term at The University of British Columbia.

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fe2005s_sol - FINAL EXAMINATION Statistics 241/251 July 30,...

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