HW6_Solutions

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

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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 ...
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## 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).

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

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