Question

An Xത chart with three-sigma limits has parameters as follows: UCL = 104; Center Line = 100; LCL = 96; n = 5 Suppose the process quality characteristic being controlled is normally distributed with a true mean of 98 and a standard deviation of 8. What is the probability that the control chart would exhibit lack of control by at least the third point plotted?

Answer 1

Answer:

The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.

Step-by-step explanation:

We consider "lack of control by at least the third point plotted" if at least one of the three first points is over the UCL or under the LCL.

The probability of one point of being over UCL=104 is:

$z=(X-\mu)/\sigma=(104-98)/8=6/8=0.75\\\\P(X>104)=P(z>0.75)=0.227$

The probability of one point of being under LCL=96 is:

$z=(X-\mu)/\sigma=(96-98)/8=-2/8=-0.25\\\\P(X$

Then, the probability of exhibit lack of control is:

$P=P(X>104)+P(X$

The probability of having at least one point out of control in the first three points is:

$P(x\geq1)=1-P(0)\\\\P(x\geq1)=1- 0.052=0.948\\\\\\P(x=0) = \binom{11}{0} p^{0}q^{3}= 1*1*0.0515=0.0515$

The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.