Question

Problem 3: [14 points) During an 8-hour shift, the proportion of time Y that a sheet- metal stamping machine is being serviced for repairs has the following distribution: f(y) | 2(1 – y), if 0<y<1, elsewhere. 0, (a) Find the probability that the machine repair-time is between 25 minutes and 40 minutes during the 8-hour shift. [3] (b) Derive the c.d.f of Y and use it to find the 25th percentile of Y. [5] (c) Find E(Y) and E(Y2) (2,2] (d) The cost C, measured in hundreds of dollars, of the time lost to repairs is given by C = 10 + 204 +4Y2. Find the expected cost. [2]

Answer 1

201-4) 0.67 042 2 0.2275 is کے OLYLI Anner during o-hour shift, the proportion of time y has density flyr f id olya elsewhere Cal The prob that the machine repair-time is b/w 25 minutes = 0.42 hrs) and 40 minutes* 0.67 hours) in P(0.422 Y<0,67 )= | 201-9) dy 20-410y = 2 |-0-you [ - (1-0,67)2 + (1-0,42) 2,421° ] = 0. (6) The calf of y ny Fly)= 201t)elt = 2 |«3+218=264-432X32] Fly) = 1-(1-3)? let y be the 25th percentile of y. Then Plycy)= 0.25 a FlY2= 1-(1-4)2=0,25 | (1-1) = 0.75 9 I-4 = 0. 86 60 = 9= 0. 340 (c) Ely) as s'y flyddy = 2/Y (1-4) dy = 2 [ lo yely-fo'y?dy] = 2(* - } ]- Ely2) = so'ya flyidy = 2 60'4401-41dy I lo'y?dy - plysdy I. -21.7941-199!]=2 [ +-4)= = (d) The cost is C= 104 20 y + 4y2 The expected cost it ECC) = E(10+20y + 4y2) = 10 +20 x + 4x4 - - = 2 = E(10) + 20 ELY)+4E(72) = 17.33