Introduction to Probability and Statistics for Engineers and Scientists

548 Chapter 13:Quality Control

0

X

m +^3 s

n

= LCL

2681012144

Subgroup

Out of control

= UCL

m −^3 s

n

FIGURE 13.1 Control chart forX,n=size of subgroup.

SOLUTION When in control the successive diameters have meanμ =3 and standard
deviationα=.1, and so withn=4 the control limits are

LCL= 3 −

3(.1)

√

4

=2.85, UCL= 3 +

3(.1)

√

4

=3.15

Because sample number 10 falls above the upper control limit, it appears that there is
reason to suspect that the mean diameter of shafts now differs from 3. (Clearly, judging
from the results of Samples 6 through 10 it appears to have increased beyond 3.) ■

REMARKS

(a)The foregoing supposes that when the process is in control the underlying distribution
is normal. However, even if this is not the case, by the central limit theorem it follows that
the subgroup averages should have a distribution that is roughly normal and so would be
unlikely to differ from its mean by more than 3 standard deviations.
(b)It is frequently the case that we do not determine the measurable qualities of all the
items produced but only those of a randomly chosen subset of items. If this is so then it is
natural to select, as a subgroup, items that are produced at roughly the same time.

It is important to note that even when the process is in control there is a chance —
namely, .0027 — that a subgroup average will fall outside the control limit and so one
would incorrectly stop the process and hunt for the nonexistent source of trouble.
Let us now suppose that the process has just gone out of control by a change in the
mean value of an item fromμtoμ+awherea>0. How long will it take (assuming