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2009 final

# 2009 final - _—_— EESEKVE Gaff D‘A L‘ld Q...

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Unformatted text preview: _—__—. EESEKVE Gaff: D‘A L‘ld. Q Il.‘.-.I:— ‘53 ‘4‘ M Pet H 9—30 ' NAME: .................... Date: January 17, 2009 Time: 9:30—11:30 E / / Instructor: Dilek Giivenc ./ MATHZMHHNALE M: f/ / HWPORTANT //‘ ,f’ 1 Check that there are 5 questions; in y ur b oklet. ff 2 Do NOT use your mobile p‘ho as a culatorfliurn it off during the exam. 3 Show all your work. Correci‘results withoerﬁcient explanation and correct notation might not get full credit. . . // 4 Write your name on each pay . , 1. A computer store has purchased 3 computers of a certain type at 500 dollars a piece. It will sell them for 1000 dollars a piece. The manufacturer has agreed to purchase any computers still unsold after a specified period at 200 dollars a piece. Let X be the number of computers sold, and suppose that P(X=O) = 0.1, P(X=1) = 0.2, P(X=2) = 0.3 and P(X=3) = 0.4. Find the expected proﬁt. (jg; cm Vﬂfrﬁ/G/ £0)??? U AL! /J 900—200 : 20c? . Xx}; 9?" com/yaks Sﬁ/d. 1:35: /?,0/,'{:./&’_ 509x_/3_)<)300 gear X=0, /,2,3 I .. p _ 900 450 m "00 73.01 0.2. 0.3 (24 “\ _9coo {0.0 -— [@05’029 —/— 700 {0.3)+ /§oo(o,4) :700 Ola/ax:- RESERVE NAME: .......................... 2. 36 numbers are rounded to the nearest integer and then summed. Leth denote the roundoff error for the jth number (that is , the difference between the jth number and its rounded value. X} ‘s are independent and eachX} is uniformly distributed over the interval ( -0.5, 0.5) . Determine the approximate probability that the sum of the rounded numbers will equal to the rounded sum of unrounded numbers. ‘ , , _i_ ”:36 amp-To Moe/(23,), ,2 szLaiA 51mm numée/f ,Qf: {4% ”weal—cal I’lvmétcr 2X1: :(é'ﬁtg-L— 2,5...27; yré'vnxg' zyﬂéaﬂg' / ,, ,_ _.'. /9(/§ij<o‘g)wo WM /_:x‘f/40,5 6w act/(:5) 2% 3:, C _ 9 #9.; £43— 0 m- C}; f‘v/ (M35 4 Z A 3‘ > Mil I U”?- We? WM: —; /’( 02342402‘0 FK0'29)—H0'23> RESERVE NAME : .......................... 3. Two people agree to meet at a speciﬁed place at noon. Assume that arrival times of the two people are independent random variables, each having a triangular distribution on the interval (-5, 5), (see below for the expression of triangular density), Where time is measured in minutes relative to noon. a) Determine the probability that the ﬁrst person to arrive will wait more than 5 minutes. (17 points). b) What is the covariance between the two arrival times? (3 points) Note that if random variable X has triangular distribution on (-5,5) then its probability density function is : f(x)=%[ ——f~i] —5<x<5 3 X,’ Gil/f!l/Wl/ airing :20! 1'13; [aurora \ZI' // g; H Z—AJ [Pm/1,4 .. '= xiv»? MM 7/ I>Sl~ PM v>§)+/3(9’*7’4 S} 2f“ 5) 5.2.4., [Efﬁshmphsﬂaawg gafﬂ")/€V“§§ #5:” 25 s 0’5 1 o’ 2 5}. 155;)0/ S ’5 2—5.. K:___ r x : :3— f new ea m> an .25 o S 3 § C fa; :25 2 {.53_§i)/:2 (555-5)155L 5 1 M; 30 20° 0 230 ”’ ““0 a ’25 so) (L) Lu mid fir/f 3(203—I675)(3_2M> g i 1’25 (W 60 la? 5’0 m :_;=5_/_ :0,0é73 u“ — 0 (\$0 /2 r: F: 5) Ska, X Ma; ‘f/ «2/; {kqgfkwimé Cov (X»Y\:O NAME: .................. 4. Let :cl,x2,...xn be an observed random sample of random variable X which is . . 2 normal w1th mean P and variance 6 . a) Find Maximum Likelihood Estimators (MLE) of CT2 and p. . (9 points) b) Show that the estimator in part a) is a biased estimator for 02. (8 points) c) Modify the estimator in part a) so that it is an unbiased estimator for oz. ‘1 ‘t (313mg) ’1 / — {ﬁzﬂf , 1, __ ‘ 355.2. [Fl Ohm; "T __ r -. , "- _ E in / a “‘1' k}(H/r1): inZ—xkﬂ’r)"lt=, \iFl‘vri-t? n 1 _q‘14mﬁ-&ﬁﬁ—“4}j — fl 2' 91:?" agﬁglgsm—z) __ O u} 2:05: ‘TﬁQL—O 2) 2%} —/w = 0 ﬂ— ‘— _~_, pita. as M , _. , __ V "“‘ f I _,. LX ____ :3N,Léﬁeif{u/€«F—— n, X a f . r ‘1 a/” (M303- ._ ____, __ /) ZiX,-:fiw_:0 ,M-a-“E‘:1 A” , w 2 05:1. :2 ((751): be 2 _m~l+:o¢—m =’- .— P. __ '1. 'f:_4-_ fof‘rxﬂ 5) 24414.53: 5 0" 7 L! ‘— #2,) 1 ._-9. A1 I '_./?X x ,2 .2 1 b) 0“: 737 Kt ’3) F1) 0“=E[XI)—-M /* AZ 5(V' )5 I‘=I @néiuei ‘“ ' k;} ,i hFQEPHE NAME: .................... 5. The program was run six times with randomly chosen data sets, and sample mean of execution times was evaluated as 230 ms and the sample standard deviation as 14 ms. State your assumption and estimate the a) true mean execution time with a 98% conﬁdence interval, (5 points) 13) standard deviation of execution times with a 95% conﬁdence interval. (5 points) c) It is claimed that additional changes in the program increased the execution time 8 ms on the average, with a standard deviation of 2 ms. If this new version of the program and the previous one were run 36 times with randomly chosen data sets and the mean increase in execution times was observed as 9 ms, can we conclude that mean increase is more than 8 ms. Test at 5% level of signiﬁcance. Calculate P value of the test. (10 points) , Li 5): 3365 a3L4u=aasJ dee/“J'S aw 5g 0/0 of /o/ H eff: er“ E+2.36§—§—~ ) 2301:3361 VET " \ffs’ (.2/(7i7—77 249 ‘23) [5): {‘2 0.9375 A) X04225 ) as 5‘: [AA/‘9." £7- 95 / c' K 14V:f 0/“ ( 2.?4 ) 34136) (/4 75-53;; > 033 exam/Alan A“: ‘fr-fims 0‘22”“ ”=36 . t, a) Xg’ , Marta“. ’71 Ham” 2 - 3,3 ... 6 =3 .' {>Z ”I” __——9‘ 2 54'ncﬁ— p Mam: P(2>3) : x, Fra): /.- 0,998; = 6.00/3 4 0,0: W5, / fact-IL #0 3 co’qciuaie, {9%de 74’0“!— rrL-tmi lhc/éqm x'xz LXélou-ZJ/t aim-ac}; {fi’d’mﬁﬁf “Lift-am. 9’ a ____‘_.—40 RESERVE ...
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2009 final - _—_— EESEKVE Gaff D‘A L‘ld Q...

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