Question

Let X1, X2, and Xz represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv's with expected values 11, 12, and ilz and variances 0,²,0,?, and 03?, respectively. (Round your answers to four decimal places.) (a) If My = Hy = Hz = 70 and 0,2 = 0,2 = 032 = 15, calculate PT, s 225) and P(180 ST, s 225). P(T. 5 225) = P(180 ST. 3225) = (b) Using the 's and o;'s given in part (a), calculate both P(65 s ) and P(68 s is 72). P(65 5 ) = P(68 < X < 72) = (c) Using the ;'s and 0;'s given in part (a), calculate P(-10 5 X1 - 0.5X2 - 0.5X3 = 5). P(-10 5 X1 -0.5X2 - 0.5X3 < 5) = Interpret the quantity P(-10 3 X1 -0.5x2 - 0.5X3 = 5). The quantity represents the probability that the difference between X1 and the average of X2 and X3 is between -10 and 5. The quantity represents the probability that the difference between X3 and the sum of X1 and X2 is between -10 and 5. The quantity represents the probability that the difference between X3 and the average of X1 and X2 is between -10 and 5. The quantity represents the probability that X1, X2, and X3 are all between -10 and 5. The quantity represents the probability that the difference between X1 and the sum of X2 and X3 is between -10 and 5. (d) If My = 40, H2 = 50, 4z = 60, 0,2 = 12,022 = 14, and 032 = 10, calculate P(x1 + x2 + x3 = 162) and also P(Xq + x2 2 273). P(x1 + x2 + x3 = 162) = P(X1 + X2 2X3) =

Answer 1

To heaprel qu. the quantity P(-10244-0-549-6-57945) The quantity represents the probability that the difference between xif the average of xz f x₂ 13 between 10 5. Here, el, = 40, 112=50, el3 = 60 6,2-12, 622=14, 63 = 10 Find - P(x1+x2+ X32162) O T=x+ 2+X8 ED = E(XD + E(X2)+E(X) - 40 +50+60 ECT) 150 VerrcT = Vor(x) + Vorr (72) + Vorexg 12+14+10 Vert) = 36 1.xi's enell) PCI-ECT) Z $ 162-150 =P(Z < 12 ) I Vero 5 36 = P(242) 0.97725 P(x1+x2 + x3< 162)=0.97725

Date (10) P(x-127, 2х2) 2 Р( 2X3-X - (30) Let Ti=5 X2+1 E(22-2-х). 2E (х3) - E(x) - ecx) - 2х6о -500 = \2 с) со Е (273-2-x) = 0 Ет) 90 + Very (93-х2-х) - 24ver tx - ero (х2) 4 exa(x) +о (.. Хt's cуле ) 24x10) + (+12 O4-26 Verr (T) = 66 Now Р (296 - X2-21 So) = = Р Т -Ест) с - 80 Vor(TD у 66 - Р(75 <- 2 • Gj 27 ) - Р(177. Gey 27) 2 - P( 49, 69 ) = 1 — О.g989 P2 - 2-х Ков д-Ооо 11 lie. = (P(x+12 72 19 ) = 0. Осо1 |

2.25-210 180-210 O P(180(T03 225) = P(To<225) - P(To <180) P(180 STO<225) = P(To-Ecto) 2310) Ivorrcto) J45 TO-ECTO) Tren (To) 145 P(180 <To<225) = P(2<2.26) -P(2<-4.47) 0.98745 PC 180 KTO 2225) = 0.98744 0.00001 (6) Find P(653X ) 4 P(68<x<72) x=zxi ECX = 70 & Vor(x) = Trenr(x1) Herrex2) +Ves(x3) 9 Vorr(x) = 5 OP(6537) = P(X7165) P(652X) = 1-P(X<65) P(653) - pl X-FC®) < 65-70 juar (8) √5 P(653x) = 1- P(Z < 5/5 P(653x) = 1-p( 262.236) P(65<X) 1-P (722-24) P(652X) 0.98745 ::P(655x) = 0: 01255 = P( 68-70 4 O P(6857472) X-E(X) 72-TO I TOG *P(68,5 572) =P-0.8944 <3 < 0-8944) P(68 < X <72) = P (240-8944) -P (2<-0.8946

P(684 2272) = P(220.8944) - P(Z <-0.8949 P(68<2 272) :|P(68<ZK72) 0.81327 - 0. 18673 0.62654 © P(-10 < x,-0.5X2-0.5x355) LetT=xi-0.5x2-0.5X8 E(x-0.5x2-0.543) - ECXI)-0-5 ECX2)-0,56(Xg) = 70-0.5x70 -0.5X7o. E(X1-0.5X2-0.5Xg) = 0 E(T) = 0 + ver (X1-0.5X2 - 0.5*8)= Vom (x1) +0.25 Vcro (X2) +0.25 Von (XB) +0 (xirs che ndependent = 15X0.25x15*0.25x15 Verr(x1-0.5X2-0.5*8) - 22.5 Vert = 22.5 Now het PC-10 < x1-0.5X2-0.5x355)=P(-10<T35) PROIt 12215 P(-10-04 T-ECT) 5-0 J2215 Van (T) 122.5 P( -2.1082 <23 1.054) P(221.054) -P(22-20 1082) = 8(221-0547 - P(22-2011) 0.85314 0.01543 P(-105X1-0-5X2-0.5X3, 35 ) = 0·88571 BE

Question Given- Let x,,x21x8 represent the times necesscroy to perferom 3 successive repair rusk at a ceattan survice facility X, ,X2, X8 crore independent normales condom romables el, = 112=Ug = 70 + 4² - 6 2 - 6 = 15 to find - P(To<225) To == ; 12+X3 ESTOT - IECALITATE Tot70+TO Let To = x1+x2+xz ECTO = E(x1+x2+Xg) - Tot TOHTO ECO) -210 + Vor CTO) = Ver (x1+X 24Xg) = Var(x) + Vero (x0) + Vero(xg). (xi's crore indepent) Vorr (To) = 45 Now PTO <225) P To- ECTO) 225-210 Toy 545 P(TOS 225) =P 23 2.286 : P(TO< 225) = P(Z 22-24) P(T03275 = 0.98745