HW6_Solutions - NDU-E14-EE’IIZ15 13: 14 'HOMEWORK6 fl 2....

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: NDU-E14-EE’IIZ15 13: 14 'HOMEWORK6 fl 2. (a) X = the number of defective gauges out of a sample of n = 4 gauges selected from N = 7 gauges of which 1‘ = 3 are defective. X is hypergeemetrie with parameters N = 7, n = 4, and r = 3. (ill?) 1 P[X=0]=P[X=OandY=4]= /——‘\ 41-4 x..____/ 1.4 Ln Lfl \u..____../' /___\ we \_________,r" .1:- 1=[3 = The Only WILKES of X and Y for which the joint density is positive are these WhiCh SatiSfY X + Y = 4. All other values ef the joint density function are ' equal to 0. P[X=3]=P[X=3andY=l (b) x 0 1 2 3 mx) 1/35 12/35 13/35 4/35 my) 4/35 13/35 12/35 1/35 Y is hypergeemetrie Li 35 35 ' *3. (1:) n0; for example, f“, (0,1) = 0 i f; (0)'fr (1) : F'.lZ|1 F' . E2 NDU-E14-EE’IIZ15 13: 15 2 4. (a) (i) Since n is a positive integer, n (H + l) (mi: 2 2 2(n—y+1)= 2 y=1x=yn(n+1)='n(n+l)y=l n(n+l) - n n+1 There are l+2+---+n = ( ) points each with probability —r~ 2 2 n(n.+1) d: = 2 " 2.7: (b) fir (x)“;n(n+l)—n(n+l), x—l,2,3,...,n _ " 2 _2(n—y+1) _ fr (y)—§n(n+—I)e-m—, y —1,2,3,i..,n (c) X and Y are not independent since (d) When n=5 the region over which (X, Y) is defined is P{X£3 andYEZ] = ii—uz—=—2uzz:(3ny+1)=£(3+2)=l_o y=l my 5(5) 30 _y=l 30 3 4 Fias31=§%z%{il]=2 2 2 5— 1 P[Y£2]=J§e(—5t—:—f—)=v32—O(S+4)—Ig The simplest way to find the above probabilities is to count the number of i points in the region that are in the event and multiply by %. 1 1 402 40 2 'i I s. (a) jjc(4x+2y+1)dydx=cj(4xy+y2+1»)!dr ‘ o 0 D u L 2 4o _ _ _ 1 =C(4x +520}O —c(6540)—l m c—6640 4“ 1 2480 “= 2 - ddx=—-i3735 (b) P[X s 20 and Y a 11 Ill 6640 (4“ ya 1) y 6640 . 2 1 1 (:3) fr (3:): M(4x+2y+l)dy£ 6640 (8x+6), Oixfi 4O 1 -2 - ‘ = =—-n--— 0_ E2 My) J6640(4x+2y+1)dx 6640(80y-Ie 0), W NEIU-Elél-EE’IES 13: 15 F'. [33 d Prat = 4 00 3240 d =flie.506 ()[ ] J6640( H )y 6640 (c) P[X‘:~ 20] n 40;(3x+6)dx—19—29e 741 ' 206640 6640 ' l 6640 1 3 +6 -— 30 +3240 (x )6640( y ) 6) no: fr (x)'fr(y)m #2 1 6640 (4.7l:+2y+1)=flW (nay) 20. (a) negative; as temperature increases, the time it takes for a diesel engine to get ready to start should decrease, and vice versa. (13) From Exercise 8(0) the marginal density fer X is 1. . ‘&«(I)=m(8x+6), Thus, 4 U l 6640 a[(sx2 + 6x)dx = 26.426 Also frcm Exercise 8(0) the marginal density fer Y is f,, (y)=—1~(30y+3240), ()5in E[X] 6640 Thus: 1 2‘ 1 EV =__.. 2 324 = _ [1 66400( J” + Oyldy —6640(6693.3333)_1,003 402 1 E[XY] = [Ixy{6640](4x+2y+l)dycit = 26.506 0 U c6v(x. Y) E[XY] 4 E[X]E[Y] = 26.586 -- (26.426)(1 .008) = 4.0615 S d]? I ‘ t 23.(a) 1 [1+i]dyax= 1 l1 ' y — ln 2(2—1) x y 2(e—1)l x+ y] 1 a 62—1 1 e (3&1 = —- l *1 1 6176: m 1 dx Mei—1);”: x +(De n 1 +] 0—1) I e_2( 1mg( : 2(el_1)[(e —l)lnx+x] NUU—E-fil—EEES 13: 15 ‘ (b)blx“l_ llz(e_1)l§+§ldym2(e-I)“l M "V =2(fll_1);fly+xlny]|ldx:2(el_1):[[(e—l)+x(l)]dx 1 x2 E_ 1 __ 2+e2—l =2(em1)[(e_l)x+3’]l"2(e—I)[(e 1) 2 ] #36—1 ' 4 36-1 E[Y] = ‘ based on similar computations (o) E[XY]= :[32(:J:1)[%+%]dydx= 1 lJ'J-[JMx]n[vc:ix: (d) no The oontrapositive of Theorem 5.2.2 is the following: If E[XY] 3'5 E[X]E[Y], then X and Y are not independent, 4 2 — _un — fiinoe $ [38 11(36 I], X and Y are not independent. 4 ‘ 1 40‘ (a) .flMx) =£Afll=g=g 8.55::‘5105 V 240 f w ( y) 2 fx (2:) because X and Y are independent; an individual’s blood eelomm level does not depend on the Specific level of her blood cholesterol. m, 120$y£240 l (bmt(y)=5"l-("‘—J’)=Zffl=zj A 0 m4? F'.lZl4 NUU—E-fil—EEES 13: 15 (‘3) Hm : Jx'fxly(x)dx= Jx'édngj 35 . as 240 240 2 film: J‘J"fm-(J’)dy= Iy-fidyflfifl 110 120 yes 44. Since fin, fl“ and x are all fixed, #le =E[le] = E[ flu + filx ] = [in + [9,): , which is the equation for a_ straight line. no, the converse of the statement “If X and Y are perfectly correlated, then the curve of regression of Y on X is linear” is “If the curve of regression of Y on X is linear, then X and Y are perfectly correlated,” which is not equivalent to the original statement. The curve of regression of Y on X is an equation that plots the average values of Y at each value of X. 48. The inverse transformation of T is T" ' {I 2: h‘ (u’ v) = u _ v I y = ha v) = v with Jacobian J _. = I ‘1 =1 1" 0 I The joint density for (U, V) is fw- (uw) 2ft? (11,10,112 (mvnlJrllz f“, (u—v,v) Therefore, )2, (u) = If” (11 ~ v, 1:) oh: F'.l2|5 NDU-E14-2E1E15 13: 16 F'. EIE: NIIIU-IZM-EE’IIZ’IS 13: 16 F'. 8'? HMHV V“ w I a ....__._‘.._._______‘_;;|__...._... . ‘ . ..__..____-_.._.‘=_. _ ‘ ____,._.....—— NDU-E14-EE’IIZ15 13: 16 F'. IZIB _..._ 1.4): — —— — — ———_ _ —_ __ ___ _ _ __ _ _— l . ‘1 ‘ F 31 .‘l I ‘ fl 5 ‘C : - a i: :. . . _ _ _ _ _ .n. — ———— —— ————— ——— — . 1 : : \ i ___ __ _...J.-.. _ ___—___ __ _____ _ __ _ _ l * t .. ‘ ‘ : ' ‘ I - ‘ ___.» 5...... _ ___ __;_u_________ _ ‘.__ u _ p m _ ‘ _ ______ __ __ , ‘ NEIU-EM-EE’IIZS 13: 1'? F'. EI‘E' v I‘ I ii i . . .._.___ Md??? m -— gm ___ _ — -—— —- -— V ._ _- --— —— —-— ___ __ ________________E_é_: ‘ - - -- - — —— — —— — ————— _— — — I—— ___— iif x “L ‘3" if}: 2-01 Fwy... AI a: ___ _I__ ‘ __ _ 1 - __ _ _. ______ T- __ _ _ _ __:§,_uc..?- ___-..._...H. _ __ ___ _ ___ _ 5| ___ _ _ ___ _ __ ___‘ji-.- _____ ___ _ _ _ ___ 13:1? NDU-E4-EEES TDTRL P.1E ...
View Full Document

This note was uploaded on 04/30/2008 for the course STAT 4710 taught by Professor N/a during the Fall '05 term at Missouri (Mizzou).

Page1 / 10

HW6_Solutions - NDU-E14-EE’IIZ15 13: 14 'HOMEWORK6 fl 2....

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online