Question

The lifetime of an electronic component, L, is known to have a variance of 72 (hours2) (a) Using Chebyshev's Inequality, find a lower bound on the probability that the lifetimes are within 20 hours of the mean (b) Suppose it is found that the lifetimes actually follow an exponential distribution Determine the eract probability that the lifetimes are within 20 hours of the mean

Answer 1

V ar iance af ele ctronie componcut V(L) 2 () Inequality chery snevrs we if nandom wauiarle t'S a vau'an ce 62 tey and neith meem no pesitive fer PS uIko 1- &2 var y) PK-41C31- nouery K 6 C inequality var (L) (20)2 Se ahoue S1- 0 82 presaki Lity hond Lo wer tue that the Lifetime of nu'puln L 20 mean 0.82

epon ential di ut its he parameter exb (A) ek AO L>D 0tuerucie rsven, V IL) 1 t But ou'st with narelncto, expouential vL 2 A 2 AS O 118

1 01122 PTOb. that ufetime neituin Ym O of mean CAI) P (-4) L20 (oc 7 (r-1)) P( L L Stor 22.49 dL ->L 784g Fe - 0-12) 0.4656 -7 0.g66 2
here ,from part a we see that lower bound is 0.82 that means probability that lifetime are within 20 hrs of mean is greater than or equal to 0.82 and from part b we see that exact probability is 0.966 which is greater than the lower bound calculated from chebyshev's inequality.