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Problem 7: (8 points] The length of time to failure in hundred of hours) for a transistor is a random variable Y with c.d.f.

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(in hundred howy variable y with c.df of transistor is а random a 7) The length of time to failure given by Fly) = 2 - - expl have to a Go show forove valid pdf, we fly) is ffly) dy 2 오 4 Now, y you (fly, dy z dye ye dy yzo het 27 y 2 ) dz = z) dz z dTo find the 30th 6 bercentile of Y K be the 30th percentile of y het then, 2 2 0.30 > 0.30 P[Y<k y=k >> fly) dy yao yok >> dy-ke o =) 0.7 log 10.7). > ak? 2 2) 0.1549 K? K z-ke 0.1549 0.3936 2 z Thuse The 30th bencentile the data is of 0.39.36z Ely] fy.fly) dy 420 2 ly gerdy Note: we have already yo find in previous exerdse 2 Sayerydy z let) lyžeykdy y zo 20 y-o yThus, 20 0 05 Ery] = (- yeye fetdy to Feydy feydy O t 0 ਹੈ0 2 ਹੈ Now, o ) •u du z ਕ ( using result) and f e-42 dee 2 ਕੇ &Elyly? yun fy²flysdy yo y zoo Jy² ay eye dy yo yooo N Tydery aylly 2 Y2o het 22 y² ) dz z dydy 20_as y se 2000 as y so 2 I zzzo |-२९-२ Zz0 Z lim (-1e2 - e-2) -२९- - e-- 2500 024 2 2) - (०.०० - CO.९० - ९०) lim-z 70 lim -L 2-50 -1.etz e-३ 2 -(0-1) 2 1probability that the transistor operates 000 hours is a The Sobability for at least P[*>200] fly) dy z doo z het Z= y ² z dzP (100 54 (200) y=200 z f fly, dy yzdDO 2 42,00 (dye-y²dy Je too 22(200) 2 e 7 de 721100) 2 22600) het Zay² dz z dydy 2 X-> (

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