Introduction to Probability and Statistics for Engineers and Scientists

584 Chapter 14*:Life Testing

whereas the corresponding probability for a smoker is, by the same reasoning,

P{A-year-old smoker reaches ageB}=exp

{

−

∫B

A

λs(t)dt

}

=exp

{

− 2

∫B

A

λn(t)dt

}

=

[

exp

{

−

∫B

A

λn(t)dt

}] 2

In other words, of two individuals of the same age, one of whom is a smoker and the
other a nonsmoker, the probability that the smoker survives to any given age is thesquare
(not one-half) of the corresponding probability for a nonsmoker. For instance, ifλn(t)=
1/20, 50≤t≤60, then the probability that a 50-year-old nonsmoker reaches age 60 is
e−1/2=.607, whereas the corresponding probability for a smoker ise−^1 =.368. ■

REMARK ON TERMINOLOGY

We will say thatXhas failure rate functionλ(t) when more precisely we mean that the
distribution function ofXhas failure rate functionλ(t).

14.3 The Exponential Distribution in Life Testing

14.3.1 Simultaneous Testing — Stopping at therth Failure

Suppose that we are testing items whose life distribution is exponential with unknown
meanθ. We putnindependent items simultaneously on test and stop the experiment
when there have been a total ofr,r≤n, failures. The problem is to then use the observed
data to estimate the meanθ.
The observed data will be the following:

Data: x 1 ≤x 2 ≤ ··· ≤xr, i 1 ,i 2 ,...,ir (14.3.1)

with the interpretation that thejth item to fail was itemijand it failed at timexj. Thus, if
we letXi,i=1,...,ndenote the lifetime of componenti, then the data will be as given
in Equation 14.3.1 if

Xi 1 =x 1 ,Xi 2 =x 2 ,...,Xir=xr

othern−rof theXjare all greater thanxr

Now the probability density ofXijis

fXij(xj)=

1

θ

e−xj/θ, j=1,...,r