Introduction to Probability and Statistics for Engineers and Scientists

590 Chapter 14*:Life Testing

Thus, for instance, if in Example 14.3b the true mean life was 120 hours, then the
expectation and variance of the length of the test are approximately given by

E[X(20)]≈120 log

(

100

81

)

=25.29

Var(X(20))≈(120)^2

19

100(81)

=33.78

14.3.2 Sequential Testing

Suppose now that we have an infinite supply of items, each of whose lifetime is exponential
with an unknown meanθ, which are to be tested sequentially, in that the first item is put
on test and on its failure the second is put on test, and so on. That is, as soon as an item
fails, it is immediately replaced on life test by the next item. We suppose that at some fixed
timeTthe text ends.
The observed data will consist of the following:

Data: r,x 1 ,x 2 ,...,xr

with the interpretation that there has been a total ofrfailures with theith item on test
having functioned for a timexi. Now the foregoing will be the observed data if

Xi=xi, i=1,...,r,

∑r

i= 1

xi<T (14.3.8)

Xr+ 1 >T−

∑r

i= 1

xi

whereXiis the functional lifetime of theith item to be put in use. This follows since
in order for there to be∑ rfailures, therth failure must occur before timeT— and so
r
i= 1 Xi<T— and the functional life of the (r+1)st item must exceedT−

∑r
i= 1 Xi
(see Figure 14.1).

T Time

Xr + 1

0 r + 1

i Σ= 1X i

r

i Σ= 1X i

Time of rth failure Time of (r + 1)st failure

FIGURE 14.1 r failures by time T.