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# finalprob - FiNAL SOLU'T l ON .S (2-1. A) Consider a...

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Unformatted text preview: FiNAL SOLU'T l ON .S (2-1. A) Consider a standard normal random variable Z. Use the table on the opposite page to calculate the probabilities i) (4pm) P(Z>1.72) ii) (.{pts} P(—1.60 < Z < —0.35) iii) (4pts) P(lZ| > 0.90) B) (ﬁpts) The weekly gasoline demand at a filling station is a random variable normally distributed with mean 3000 lt/week and standard deviation 200lt . At the beginning of a certain week the storage tank of the filling station is empty. How much gasoline should the station store so that it can meet the coming week‘s demand with a probability of at least 0.95 7 a) L) PCZ>L72)=O.§000-+4He(l.72)=0‘§000_0.45¥3 30.042? it) Pageézz-o.35):tabidiaoywieiogg)r94452-03” :6-3094 Lil) P(l2l>o\o)o)=l_r’(iziéo.go);l_ “4360429).90) :l~2alnlale[o.aoy=i_2(03i53):1.. o‘églgzoségz lo ) Solve lo {rm-m 44! eqHa‘Hm P(Xéb):0 GE P(Zé b‘3000)10.6‘; 200 W 6" 1“? 42l-é4‘; 7—) ﬁble(q) :: 04% §-0.§>0i4 5" 5.3000 7’00 2' ) b:3000+200ﬂ.64§) 1’ gnaw 3315 Lt 0(- My” gkpu‘L 10g S+mﬁeéi Q-2. The joint probability density function of continuous random variables X and Y is given by 4 . Hay) :{ ——F2(1+\$2)(1+y2) If an > 0, y > 0 0 elsewhere 3) (6pts) Find the joint cumulative distribution function F(z,y) . (WARNING : It should be a function ofa: and 1;. DO NOT ATTEMPT to calculate fox fox j(a:,y,)dzdy, it is equal to 1 since given f(z,y) is a density function and constant function F(:z,y):1 can not be a cumulative distribution function). h) {6pts} Calculate P(0 < X < 1,0 < Y < ME) 0) {613133) Evaluate E(X). 4) o 1:40 «‘6440 (minhcmylo) Firm—- 9 w a " cl . i g ’42. :1 AL 5—: : Aqrc‘i'ﬁn'xaﬂ‘iuﬂj 111 H. 51- ’++_1- 111. ’- (+4} |+51 ﬂ} tzo \$110 “a 5’0 9‘7/0; \$20 l>) P(w4x4l,o<Y<5):F(4.J§i (Sum 94am?“vale I i “(Chin J; :i \E~ %:%‘ :/:6) 11\$ 111 4 1/7. The Same rerull: is Dial-nineé Llj Hm I F (inhibit {nieﬂmi g3 A) A}? 111/ F’,/' J1. a (9 Hrs IM— 00 M A i \ 1m 4 \ rL ) - ._, an: :_-_, ,, C) 3(Ki‘i S “TEL/“L -Tti In” “(9 “1 |+L 7’ n1 (Ii—x)(l+‘3 ) 3:0 7‘ ‘11"? 9.. l W. 0 (1):? ——- F‘ 7 3 fL Hut? (“\S’Lwhere o M w Q-3. a) (8pts) The number of marriage licenses issued in a certain city during the month of June may be looked upon as a random variable with the mean u = 124 and standard deviation (7 : 7.5. (we do not know whether the distribution is normal or not, it can be any distribution). Use Chebyshev‘ s inequality to find a lower bound for the probability that between 64 and 184 marriage licenses will be issued during the month of June. (A lower bound: is a number between 0 and 1 such that the required probability is equal to or greater than this number). b) Let X has binomial r.v. b(a:;n,0). As it was discussed in the class, X can be written as X = Y1+Y2+- ~+Ym where each Y.- (2‘ = 1,2,” ' ,n) is a Bernouilli r.v. (Le. Y,- : b(1, 0)), indicating whether there is a success or not on the ith trial (Y; :1 if there is a success , Y5 = 0 if there is a failure on the ith trial), Let Y1+Y2+ ---- ~+Yn Tl. i.e. )7 is the proportion of successes in 71 trials. i) (3m) Find ED”) and V07) in terms ofn and 6. ii) {5pts} Use Chebyshev' s inequality to find a lower bound for PU}? — 6) 5 e), (6 > 0 is any ﬁxed positive number). iii) (4gb) Using part ii) show that for any fixed 6 > 0, PG)7 — ﬁll S e) a 1 as n -> 00. (This is called the LAW OF LARGE NUMBERS). 4) MM <>< 4184) 39(64 -m< X-19—4<I24_)24)1p{_é0<x_l244 6°) : u G‘ﬁil Pﬂ X 124l4603 ; i— (6031 = 06343 (Since 1U. =I24 4m! “5mg aging; Rev ‘0) l‘) BC:0";L[9(W)+E(‘é)+-'+E(Yn\]r (none 1 l" NO: )= [V(\ﬁ)+V[\{1)+ ,..+ Wynn :LChQ/Hgﬂ: @[l~9) h“ n- _i, )1 CL) Pﬂ‘?relsg‘)> i- ML) ﬂ, Oreo £1 n5; [AS nae: éiyal’ ){m :1 hr) he “£1 \,,, 7V; (9 Thu; um P(i\?_oiée)=\ n40; Q-4. Customers arrive at an entertainment center at the average rate of 3 customers in 10 minutes. a) (7pts) What is the probability that in half an hour at least 10 but less that 13 customers will arrive '1 (You may leave your answer as a numerical expression without computing its final value). 13) (611”) If X is the number of customers coming in 1 hour, the profit that the entertainment center makes from them is a random variable Y thought as Y = 4X2 + 3X, (in Y.T.L.). Find E(Y), the expected proﬁt. c) (Epta) What is the probability that the ﬁrst customer will arrive in less than 5 minutes .7 dx =2 Guatemala /m;v\. iv 4) T330} kWh-1.30:6 I lo 1’ Aymvﬂls X ‘m 30 mirach {olipr 170.3500: P()(=,ci= a 5- iﬂw’l I it ,9 l'L °" P(10é X03): 9. ,9_l.b+ 32 «pi—w) lol_ 41! 12.! in) T260) (hzg-héozls ) r4351: Alla? i0 E(Y):E (4x"+ 3x324 5(x7’)+35{x) =4(5"“+ ‘3’) +3rL = 4(l3+(12)1)+’3(18):13éy +54 21422. v.11- ‘9 Let “'1 be “6 Hm MIL esp-m W- “ a '7‘ ‘~€7— #11769 Ai‘i'emﬂri’ivf’ Semi-3.0%; £53 =§AH€45+ one arrival. in [TO/’31; . “a 1,? wlittxl—i-z 1‘5 11—3— Z)C L lo 2- 7. X: xi V '5 _. _ (afa— /2. J: m =l~e Z Q-5. The probability that a patient dies from a certain blood disease is 0.20. If 25 people are known to have contracted this disease, what is the probability that at most 2 of them will die .7 Calculate this probability in three different ways 2 a) (51%) Find the exact value of this probability. b) (Spts) Find the Poisson approximation of this probability. c) {61m} Find the normal approximation with continuity correction of this probability. d) (Zpts) Which ofthe approximations in b) and c) was better ? What is the resaon for that ? 4) X : b(x;zs,o.zp) 7.5,»: 1 7‘ 7C PM“) “ Z (2: ifo-zoiiwigo) =(QS) (0.2) 5475) UJJUEJM 110 O l +(liimtroaf = 0‘ 00% +7—5fo.L) (0.004?) +20o/o-04Vooos’a) 3 2 0003:871- oro 235+o.0703 g 0.099! is) a=h9115{0.1)=‘3 ) P051): 515,5— L! P(X41l:eds(l+g+§):0\l14é 2. c) P(os><b‘,héz) g’P(—o.€<>< <25) =P{*05~§ 42416—9 \47 Haw—95> w MW» 1 F 2 PLHS < 224‘” i: mung), we (1,1,5) 10. 4910‘ 0.3944: 0.10%!) A) c) l5 55441-96 Since (-9 i5 ngi- Small (no-Josh Pailjo-m 4PVWY‘IM4HM is hol- §L(l'l:51lal€ xbh all/4%., lilo/Q M955 ma ‘nﬂ__i9): 20 >5 5w ﬁne mi awpvaiwmix'w. i4 be‘i’iém. Ila Crmr‘lJ (M7045 ' 5r». eoz-_se:»wW._.ah, Q-G. a) (Bpts) Suppose that r.v. X has the probability density function 36—31 if I > 0 ix (2) _ { 0 elsewhere Find the probability density function of the random variable Y : ~4X + 1, by using the (cumulative) distribution function technique b) i) (3px) If a r.v. X has the moment generating function (m.g.f.) Mx(t), show that the m.g.f, of the me a+bX is Mo+bx(t) =E“‘Mx(bt). ii) {7pts} Let Z be a standard normal r.v. and let X be a r,v. normally distributed with mean u and variance a“. if X and Y are also independent . find the probability distribution of Y = 2Z+ 5X using the moment generating function technique. (Hint: Notice that (2‘92) and em“) are also independent. Also use part i)). 4) Range‘f =(—<>oJ I) vaelaWYézthW'W“ ﬂaw”) €23:qu < E.” 4 T ‘_ §l—‘})/4 ’BY 30 l 35 ix: [#8 4 )2 e:— 4 ﬁyQﬂ’Ji RAB) =i i Z 0 oil—cooker? t LY) l» i) M widen” 1=Ereatarwxj:eatgfeaoxj (c-Hax z e”fo (w e’ 4 AMLLQ/él . Jc 22H; ) trz ) (x a) MYKJ,\:E[3( XJ:E[¢ 2.5”“ as e411?) 0AA egg-X) are abs inéerenécwd WVJS‘ we) ’c('§><).,r/i ri-M ‘l MYW‘ EEQ 1E“ i 22a ex“ . (/x (221 r‘—(5H+GL(EHL Park k) “' a ’L e 7’ r5 itwggzjﬁi :6; F ’L mil-j iJ‘HAa air-NV; hpmnl “(XJS—IV) 150L444.) a Wl :L'thilouk'm uélln W EM and Vet-{ante Q§SL+4- / “C. ...
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## This note was uploaded on 03/16/2012 for the course FENS 101 taught by Professor Selçukerdem during the Fall '12 term at Sabancı University.

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finalprob - FiNAL SOLU'T l ON .S (2-1. A) Consider a...

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