# mid2 - Mm E sournous Q-l A continuous random variable(r.v X...

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Unformatted text preview: Mm E . sournous Q-l. A continuous random variable (r.v.) X has the probability density function (p.duf.) given by 1~zx if 0";J‘<l ﬂan): Zli if 15113 0 elsewhere a) (Iﬂpts) Find me cumulative distribution function F(a:) of X . b) (5pts) Sketch the graph of the cumulative distribution function F‘(x). c) (5pts) Using the cumulative distribution function F(:z), calculate P(% < X < 2). 9g “i For 05"“ (lilrlét-ﬂ/Jc-l—T} ‘zl‘ nil- - 2, ‘ 2 0 3-,": r ~.i 1i Cyf liﬁé} Fall 2 + 4 Ii ")0 O 7140 \ V l (\ 1 ,, i L l 7,51,. 0£7€4l i t, i an i / l’m: ‘ l + 11¢“ iéxcg Ii / Z 4. “T if, Migt l X23 0 ., I 1 I‘d/4“ 1, -343 ) l){l 4X<1)~l:lzl—l:{z)~'l 1%? l2. s" " "r 3’ Q-2. A fair coin is tossed 4 times. Let X : The number of heads obtained in first 3 tosses Y : The number of heads obtained in last 2 tosses The table below shows the joint distribution (discrete) of X and Y X 0 1 2 3 0 J. A A 0 Y i6 16_. 1 Tia c ii: % 2 0 d e 75 ﬂay) a) (ﬁpts) Determine the probabilitis 0,0! and e to complete the table. Expiain briefly how did you obtain thae values. (DO NOT try to prove the other figures. They are correct l). b) {617\$} Are X and Y independent r.v.' s ? Give reasons. c) {612(5) Find the variance V(X). 4» FIN” C '- 3 Fossib'i‘lp‘gs «é (X=11‘l=‘ TI (TIT. HrTNﬂ-hTrT, whim-m) (33/1,; a“; ‘ g‘mi‘WWJJ‘bWJ'E of (Xumzsz-ﬁgv chi/)6 Y:2 pm- 3 I 7. PDJJl'Bo’lii'JtJ “g 6:117:73 : (Tl ,LLI, 14,14) ifT/T/N/ l4) .7/ “46 N0¥~'m¢‘4’xp-6n isn’t J £0, Mimi/ice 9/3,?) ; g yywuﬁisi’ 1’01)? 2? 1:3 C)LA‘43 2(3 3’ 3 1““ ;)+ (E):_; Ext) : ‘9 “'b' 3+5“? +37% ) :qf’ﬁ, VK/K):E(><L}—HL 62-21. An electronic devlce has two components. If one of them fails, the device continues operating. If two of them fails then the device stops, Let X : The time that the device stops Y : The time that one of the components fails (Obviously X > Y). It is given that the joint p.d.f. ofX and Y is # 22‘11’ if D<y<\$<ac ﬂr‘y) F { 0 elsewhere 3) {31225) Sketch in (xy)-plane the region where f(\$, y) % 0‘. In constructing the integrals in the following items ALWAYS CONSIDER THIS REGION. b) {ﬁpts} Find the marginal p.d.f. My) of the time Y to the failure of one of the components. c) {5pt5} Find E(Y), d) (5pts) Find the conditional density f(z|2) of X given Y : 2. also indicate its domain (without the domain no density is a probability density). e) (51m) Find the conditional expectation E(X|2). Tho—\L‘Aéeé neﬁfm is the éwain 0‘: £4415}; he. Elnlﬁjgéo +her~c ,K—‘I-t do} _ 5d _ e d" 1 e {—Zexi ‘ Elva—+2541 “K: Oi‘herwise pd _7_o1_ ‘90 < '2 — 6 +1. ‘11 [a Z we ta] ‘jzo O -15 X? Z_ I Zero Qincoukel'ﬂ (>6 ()0 00 7. “a; 1 ,x '1 : egg: Ax =e L_xe ] 4’86 JXJ 1 7C=L L ' Q—‘4i Random variables X'end'ysarhave tontinuousjcint‘pdi given by, " - ' . r i - r ‘ k(6—z—y) if 0<z<2,2<y<4 Hz) "{ 0 elsewhere a) (71215) Find the constant k. b) (6pm) Calculate P(1 < X < 2,2 < Y < 4), c) (10pm) Calculate P(Y < 2X), (Hint: Consider the region in the domain of f(:c, y) where the inequality 3/ < 2:: Zsatisféejl). 4 L 2 39’» 2‘70 L3 ,1 720 L 4 zl<Sq(—7ag+lo)é‘j= l<(-51—ll03)‘9l-L=k(~ \6+4o, (.4 +20)? =gl<=l —‘ l< =3}. 94 - (Jug): G’xdé ) 94,042) 4.4i544 I, Zero ClJeuukel‘f 8 4 7' 4 1.. l3) F(i <x<21 zaoﬁsi S S(6-x-3)éuéy :L Sggqgngé’ll 43 a 9:; x2.) 92 L 7:, 4 4 2‘ [—Zfé )—1—(6_)+-—l A ,l 2 -i g7- a 4 E2 .3 g 113 ? éﬁ—lﬁ—rzlég—Q (~:+:\4>‘lj=2. :—§[—3+l\$-(-Q+‘5)] :2 ‘3 u '2— 2x. 6) l”(Y<2L><l:iib S [S (éﬁ—53él11éx 4 Ls= 1- X2. (:3; xal =2- l l x 1. l at 1 7. 3 3% 3 (~4xz+l4rx~|o')éx = Jé(—ll_:_+7xL—lox3\ :l -31 2L 4 -i w J? l: + La+gﬂi+lo]'§-(||_—§—)-z§ Q-s. a) If V(X) E a} z 3, V(Y) E (7,2, = 4, Co'u(X.Y) E UXy : 73, compute i) (51%.?) V(4X 7 71/ + 6) ii) (517“) camx + Y, X 7 3y) 1)) i) {5pts} Show that if c is a constant and X a mi. then Cou(X,c) = 0. ii) (spits) Using this result (the result in b)-i)) find C’m)(U, V) where U : (1X + b, V = CY +d; a,b‘c.d being constants. (NOTE: Part a) and part b) are two different questions. They are not related). . 1 Q. s. a) 1) V(4x—7y+é L—W 4x IN): 4 v m may—1) Cw ow Mal} VM =\c (3)+ (—§a)(«3)+43 (4)=4g +16 3 H36 3 4,2 11) Cw (ZX-fo/X ~3Y): 2V(x)~6&w(><n/B + Com/mm -3 V(Y) 72C3\—§(—3)_3(43= 5 Calm/W b) 1) CW 09¢) : E(cx)_ acqvaoa: C Em) —cE(><)-=o ...
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