fall 2006 STAT 230 Quiz

fall 2006 STAT 230 Quiz - STAT 230 Quiz#5 3:40 — 4:10 pm...

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Unformatted text preview: STAT 230 Quiz #5 Nov 23, 2006 3:40 — 4:10 pm Name: ___—__ S LD: UWUserid:__Mark / 20 Circle your instructor and time slot: 1. Metzler(1:30) 2. Drekic(11:30) 3. Springer(10:30) 4. Dtekic(12:30) 5. Ramirez(8:30) 1‘ The daily internet time (measured in hours) used by a local web design company is a. random variable X having probability density function (p.d.£) given by __ 423(10—2) {02‘ng g 10 My) ~ { 0 elsewhexe. [2] (a) Show that. c = 1/5000. We "4145+ 54v: C algs(/O—X)o/X 1' / c (25)”; ?5/0”: / ' i c (25000-20000) 4:. I C= .5000 [l] (b) Calculate P(X = 5.5). FZX=55)=0 since X is a can+inuou3 ram/am variable, . [2] (c) Calculate P(3 < X S 7). F(3<Xs 7) = 51%4‘ 7X3Z/0-x)c/x = 50/00 (ZiX‘/'_ §)/37 = 1 /2005 [T_ I65807_ 1205+ 233—] 5000 __ / _ 5000 , IZ43G = 3/07 [4] (d) Calculate the mean and variance of daily internet time usage. 500 = 5400 JO"? *(/0 —x)0/X = (ZXS— :°>/,,’° / : 5090 (200000— 500%;00) : £2 IO 77 3 I)! = 5500 \J; z 7 p Z = 31,—;(g/a— —’;—)L — (27°) [ ( 5000000 __ 10000000) _ ﬂ 3 7 = .5000 7 = [000 _ 100 Z! 7 {5—2 3/746, [2] (e) Daily use of the internet costs the company \$0.50 per hour plus a \$2 overhead charge. What is the standard deviation of the daily cost to use the internet? LE1" C Perv—warn” “Hie a/a/ﬂ/ cos‘f' “(‘0 use ‘672 [rt/"Erna": Then) '- 0; = 0.50} = 0.5\/%%"— ago.” [4] (r) Deﬁne Y = + 2. Find the gang}; 00/75/ij Y: I/x?+Z O) ID]2 we. 170+; 4474+ Fm )8 E2) FY'Yy)= YSyT’ yeczlwifﬂ' = F(\/)—(+25)/) = Sy—Z) = m s 9.2%) : ECCy—Z)z) The 076 Y/Sj/lvlen 52/ ny—(Ec(y_z)z)> = ~F((y_.2)’-) ' 29—2) =55; . 9-2)‘(lo—9—2)‘)926/—2) 2 = 25’” 0475/0 “4"” 3 10M 2 Sys \//‘o'+2 [5] 2. Suppose W, X, and Y are random variables such that Var(W) = 4, Var(X) = 16, and Var(Y) = 9. Fhrthermore, assume that CovU’V, X) = 3, pw,y = —0.5, and X and Y are uncorrelated. Calculate the standard deviation of Z = 2W — 3X + Y. Von/(Z): Var(2w—SX+Y)"‘ - = AHMVOM) + 9 wranvmvhz BZX-3)va(Vl§X)+ (2x I )LMW) +(—3)0)CDVO(ZY):I =0 = /¢+ /’+4 +9 —/2(3)—/—4/Cov(w,y) = /é7—3é +4[‘72)(2)(3) = /(,7—3e——/2 = /.2/ "'oj7_=\/\Ar(z>= IZ/ =// STAT 230 Quiz #5 Nov 23, 2006 3:40 — 4:10 pm Name: S LD: UWUserid: Mark / 20 Circle your instructor and time slot: 1. Metzler(1:30) 2. Drekic(11:30) 3. Springer(10:30) 4. Drekic(12:30) 5. Ramirez(8:30) 1. The daily internet time (measured in hours) used by a local web design company is a random variable X having probability density function (p.d.f.) given by _ lcz(100 — 12) for 0 g z s 10 “1) > { 0 elsewhere. [2] (a) Show that k = 1/2500. Vl/e "1445+ ha Va [(J;IZ((/00~Xz)alx = / /<(50X‘— 419/5” = / k(5000—2500) =l k="i'“ Z 500 [1] (b) Calculate P(X =65). PZX=6. 5) = 0 since Xis a conﬁnum Vane/om Var; able. W,” [2] ((3) Calculate P(4 < X S 8). 8 P<4<Xsa>) = 24—003; Mme—wax if = 25%: (50xz— “air—Mfﬂ = Egg—0 (3200-/oz4 —ch0+ 64) H40 72 2500 ”” /25 =057é [4] (d) Calculate the mean and variance of daily internet time usage. Ea) = 2590 £/;Z(/00—x2)a’x = Lgaazeﬁ— 292/3 3 ﬂgaa‘L—ZOOOD) (3—9 1/ 5.3333 Val/(X) = 2;“, \52’0x3000-x‘)dx - (4 z Ewan— ¥)/;°— (ef 25/00 (250000 __ 500000)_ 254 7 [00 __ Z5é 9 T3— = 5%: 4.8839 ll [2] (9) Daily use of the internet costs the company \$0.25 per hour plus a. \$3 overhead charge. What is the standard deviation of the daily cost to use the internet? LEI“ C rarr252n+ "Hie. day-b1 005+ +0 use 'IL/Ia in+erne7li The") C1: 0.25X+3. "0;: 0.250;< = 0.25%? 2 «0,55 [4] (f) DeﬁneY=x/).( +3. Find the p.d.f. on. ConsiJer Y= 5+ 3. we» xs to, mama not. warm [\$5243]. Ferys Es,\/:?+3:|) * = {70074352,} = PM)? Sy-3) = Fmszrsﬁ) .. z E(C/—3)z) Thcr.al.‘p. o‘P Y [SJ/Van A); 3%‘(Ii((/_3)2)) = ‘P[()/-3)2)'Z()I—3) = 21530- . (y-3)2Doo—(y-3)‘ﬂ- 291—3) 3 721% (y-—3)3[/00-[)/’3)4J 1091 3 5y S VIE—+5, [5) 2. Suppose W, X, and Y are random variables such that Var(W) = 9, Var(X) = 4, and Var(Y) = 16. Furthermore, assume that Cov(X,Y) = 3, pw,x = —0.5, and W and Y are uncorrelated. Calculate the standard deviation of Z = 3W — X + 21’. VarC Z): mew—Xaw = ‘7 Var(W) + man‘ﬂ/M Y) +2ﬁ3>(—I>Gv(vyx)+(3)a)@64w +(—n(2)cov0(,'v)] = 21+4+ + é‘f — a Cov(Vt§X) —¢f(3) = M7 —/2 —6 (Va )(3)(z) = M? —/2 + k? = /55 a}: \/l/AV(Z) = Wu 2 /Z.45 ...
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