Question

Let X1, X2, and X3 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 μ1, μ2, and μ3 and variances σ12, σ22, and σ32, respectively. (Round your answers to four decimal places.)

LetX1 X2 and X3 represent the times necessary to perform three success ve epa r tasks at a certain service facility. Suppose they are independent normal rv's with expected values μι μ2 and μ3 and variances 01 ,02 respectively. Round your a and 03 nswers to four decimal places.) (a) If μ,-μ,-M9-90 and σ12-v22-v32-24, calculate rt, 302) and A222 s 1,5 302). PrT, s 302) - Pi2225 TO s 302)- (b) using the μ.s and σ''s given in part (a), calculate both P(82 X) and P(88 s Xs g2). (c) Using the μ.s and ơ,'s given in part (a), calculate Pt-16 SX1-0.SX2-0.5X3 8). -163 X1-0.5X2-0.5X3 8)- Interpret the quantity P(-16x10.52- 0.5X3 8). The quantity represents the probability that the difference between and the sum ofX2 and X, is between -16 and 8 The quantity represents the probability that the difference between X1 and the average or x2 and X3 is between-16 and 8 The quantity represents the probability that the difference between X3 and the average of Xi and X2 is between-16 and 8 The quantity represents the probability thatX1, Xz, and X3 are al between-16 and 8. The quantity represents the probability that the difference between X3 and the sum of X, and X2 is between -16 and 8 (d) lf 씨-30, μ,-40, μ3-50' σ' 2-12, σフ2-10, and ơ32-14, calculate PO. + Х, + Х, 128) and also P(X1 + X,2 2x3)

Answer 1

Solation Given that Ca) Hev the distoibution of ll be nosmal distoibution Gith padometeas ase μ,-Mg-Mg.qo 10+90+90 Stondood deviation 구2 8 , 1185 3 To heoe me z- scose of T, 302 1s To 30 2-2 。 8、4853 32 8.11853 乙: 3, 12 乙 :3、子8 The sevuited peobabilitys r(AS 302)-r(2-348):0.9999 Cs Scanned with CamScanne. The z-scose poo To -222 is

To 222-2 8 1853 -18 e-'1853 -5.6568 z -5.66 The reqjuised poobability is : P(253.78)-P(z 二0.9999一。 0.9999 -566) b) The sampling distsibution oF somple meon wili noomal distsibuti on with mea n Mcan 90+90+90 270 3 90 stondaod deviation 3 st Scanned with 2, 8284

The z-scose of元-82 is 82-90 2.828 4 8 2 8284 2-2.8285 50 Pf8% ): P(27/-2, 8285) The冂 68-90 2 8284 2 2 8284 The z-score FOD χ:92 is E- H 92-90 2 8284 2. 2 828y cse Scanned with CamSca n ner. '

So, e have 760 2 -0.2398 = 0.5204 .. . P(62-x): σ.qq77 Let So the distoibution of y will be noomal distsibutlon with mea n stondaod de viation Gy、| 24 +0.25 X 24+0.25%24 The z ScorC foo -16 iS -16-0 [6 6 Scanned withz -2 6667 CamScanner

9 6 2: 1.3333 So, ce hav c - P2 666 z 413333) 0.9088 O.o038 O 9050 option e is correck i e .'. The quorthity repscsents the paobobility that the dipFcoence is betacen- ond 8 and the Sum of , afrd let I aith mean So distoibution of y aill be noomal distaibution 9 120 stondord deviation 2 Scanned, with CamScanner

The z-scode Foo y- /28 is /28 120 23333 So = P(291, 3333) ,9088 so the distoibution of y will be roómal distoibution 30 + 넌0-2-50 :-30 Scanned with CamScaThenr thestoriobsd deviation

2 2 tt223 8 8318 The 2 Score of y o y-ん o- C30) 8 8318 3 ' 3970 . So Py7o) P(z73.3970) O 0003 PCye125) = 0,9088 Pt2t 3 s 128) 0 9088 CS Scanned with CamScanner