Introduction to Probability and Statistics for Engineers and Scientists

13.2Control Charts for Average Values: TheX-Control Chart 553

SinceX=3. 067,S=.122,c(4)=.9213, the control limits are

LCL=3.067−

3(.122)

2 ×.9213

=2.868

UCL=3.067+

3(.122)

2 ×.9213

=3.266

Since all theXifall within these limits, we suppose that the process is in control with
μ=3.067 andσ=S/c(4)=.1324.
Suppose now that the values of the items produced are supposed to fall within the
specifications 3±.1. Assuming that the process remains in control and that the foregoing
are accurate estimates of the true mean and standard deviation, what proportion of the
items will meet the desired specifications?

SOLUTION To answer the foregoing, we note that whenμ=3.067 andσ=.1324,

P{2.9≤X≤3.1}=P

{

2.9−3.067

.1324

≤

X−3.067

.1324

≤

3.1−3.067

.1324

}

= (.2492)− (−1.2613)

=.5984−(1−.8964)

=.4948

Hence, 49 percent of the items produced will meet the specifications. ■

REMARKS

(a)The estimatorXis equal to the average of allnkmeasurements and is thus the obvious
estimator ofμ. However, it may not immediately be clear why the sample standard
deviation of all thenkmeasurements, namely,

S≡

√√

√√∑nk

i= 1

(Xi−X)^2

nk− 1

is not used as the initial estimator ofσ. The reason it is not is that the process may not have
been in control throughout the firstksubgroups, and thus this latter estimator could be
far away from the true value. Also, it often happens that a process goes out of control by an
occurrence that results in a change of its mean valueμwhile leaving its standard deviation
unchanged. In such a case, the subgroup sample deviations would still be estimators ofσ,
whereas the entire sample standard deviation would not. Indeed, even in the case where the
process appears to be in control throughout, the estimator ofσpresented is preferred over
the sample standard deviationS. The reason for this is that we cannot be certain that the