13.5Control Charts for Number of Defects 561
whenYis Poisson with mean 6n. Now
p(n)≈P{Y> 4 n+ 6√
n}=P{
Y− 6 n
√
6 n>6√
n− 2 n
√
6 n}≈P{
Z>6√
n− 2 n
√
6 n}
whereZ∼N(0, 1)= 1 −(√
6 − 2√
n
6)Because each data value will be outside the control limits with probabilityp(n), it follows
that the number of data values needed to obtain one outside the limits is a geometric
random variable with parameterp(n), and thus has mean 1/p(n). Finally, since there aren
items for each data value, it follows that the number of items produced before the process
is seen to be out of control has mean valuen/p(n):
Average number of items produced while out of control =n/(1− (√
6 −√
2 n
3 ))We plot this for variousnin Table 13.2. Since larger values ofnare better when the
process is in control (because the average number of items produced before the process is
incorrectly said to be out of control is approximatelyn/.0027), it is clear from Table 13.2
that one should combine at least 9 of the items. This would mean that each data value
(equal to the number of defects in the combined set) would have mean at least 9× 4 =36.
TABLE 13.2
n Average Number of Items
1 19.6
2 20.66
3 19.80
4 19.32
5 18.80
6 18.18
7 18.13
8 18.02
918
10 18.18
11 18.33
12 18.51EXAMPLE 13.5a The following data represent the number of defects discovered at a factory
on successive units of 10 cars each.