Question

Let Xy, Xy, and X2 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 un, Uy, and uz and variances 0,2,0,2, and az, respectively. (Round your answers four decimal places.) (a) If uy = 4y = 4z = 50 and 0,2 = 02? = 032 = 21, calculate P(T, S 171) and P(108 ST. S 171). PIT, S 171) = 9959 P(108 ST 171) = 9959 (b) Using the 's and o's given in part (a), calculate both P(4357) and P(48 ST s 52). P(435) = 9959 P(48 S X < 52) = 6826 (c) Using the u,'s and 7,'s given in part (a), calculate P(-14 SX4 -0.5X2 - 0.5X3 S 7). P(-14 SX1 -0.5X2 - 0.5X3 < 7) = Interpret the quantity P(-14 SX4 - 0.5X2 - 0.5X3 S 7). The quantity represents the probability that the difference between X, and the average of X, and X2 is between – 14 and 7. The quantity represents the probability that the difference between X, and the average of X, and X, is between -14 and 7. The quantity represents the probability that the difference between X, and the sum of X, and X, is between -14 and 7. The quantity represents the probability that the difference between X, and the sum of X, and X, is between -14 and 7. The quantity represents the probability that X, X, and X, are all between -14 and 7. (d) If u, = 30, uy = 40, uz = 50,0,2 = 14,0,? = 10, and 0,2 = 12, calculate P(X, + X, + X, 5130) and also P(X, + x, > 2x2). P(X + X, +X S130) = P(X1 + X2 → 2X3) = You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

Hide : lg 2 50 5 1 - 21 (a) Total time for all three task - x + x2 + 73 So mean - 4,54, + g = 150 Variance. 2²52 253 ² = 21x3 soda √21x3 & 563 - M to 3171) 2 $ (17) -156 ) = 0.9951 up (108< To 5.171) - 01-171 - 1 108 -150 63 ndoto 2 0.9959 3 sa 4,742 +43 - 50 ? ()? 5, +(%)ę dl) • % x90 + % * 24 +2 1 - 7 P (43 **) - 1-0(43-50) = 0.9959 P (up <=392) 0/5250) - of 48-50) : 0.5503 le) - 7,-0.5X2 -0.5*3 = x (het) . 42 4,-0.542 -0.543 ²0 nz 0,2-%rz -Yo = 10.5 Ta $105 PGM Ex{ 7) = 0 (70) - p1-19-8) = 0.9946 → x = x 3-0.51,-0.5X2 42 43-0.54,-0.5 4₂ ²0 o swt n . 10.5 Ta 10.5 P (M EX &7) 8.9846

→ xa x-x₂ - X 3 M. 41-42-U3 - -50 wa 02-02-02-91 r. & probability not possiple Xex - A-42 02.-24. sara & probability not possible + P(-1* <, $7) - 09/2001 - 0 -1.50 ) 3 3.9479 41021 P (14 <A247) = pq-14 572 59) = P(1440, E7) 3.2479 x 100 AS M₂ = Mga 3 020 2 = 52.