Question

The length of time to failure (in hundred of hours) for a transistor is a random variable Y with c.d.f. given by F(y) = ( 1 − exp{−y 2}, if y ≥ 0, 0, elsewhere. (a) Find the p.d.f f(y) of Y and show that it is indeed a valid p.d.f [2] (b) Find the 30th percentile of Y and interpret it [2]. (c) Find E(Y ) and V (Y ) [2] (d) Find the probability that the transistor operates for at least 200 hours. [1] (e) Find P(100 < Y < 200)

Answer 1

bet y length of time to failure in hendred of hours) Fylyl = - 412 e O -112 la fyly ay F419) any line) - -412 :.11) yoo Y~enp (1/2) 0.w P 16) s fixida 0.30 le PlYEP) -0.30 11 FIPI = 0.30 -P12 -) 1-e = 0.3 -P12 e 옥 10.7 = 0.7 2 log 2 log 107) : 30 percentile = -2 log (0.7) 0.3098 C) Elt) = 2 1 912 ye a) 아 = 2.12 = 2x1 =2. :. E141 = 2 94 xello) -) E(X) = 0 4x2 = 8 Ely? +1 very store uly = € 14²,- (E63² = 8-2² = 4 xrenpl 101 Vlx = 0 = 4 1 200 hour , Y 100 hours) (d, Plt 2 1 I- PlY & 2) - 4 12 е - 1- F1212 = el y=2 = 0.3678 An -212 F(2) - FU) ine -1/2 -1te u les P(100 < y < 2001 - ma 0.2386