A c chart is used to monitor surface imperfection on water heater cabinets. Each

Question

A c chart is used to monitor surface imperfection on water heater cabinets. Each cabinet is checked for nonconformities(defects) and the count entered on the c chart. Two limits are used on the chart: a control limit at +3sigma and a warning limit at +2sigma. If a point fall above the control limit or if two points in a row fall between the warning limit and the control limit, the process is stopped until the problem is corrected. The center line of the control chart is at c of 1.5. Let's represent the number of defects of each cabinet as random variable X, X can be best modeled as which one of the distribution?

(A)Find The warning limit of the c chart is equal to (keep three decimal places)?

(B)Find If the process suddenly shifts to a mean value of 4, what is the probability that the sample after the shift will have defects more than the control limit?

(C)Find if the process suddenly shifts to a mean value of 4, what is the probability of the sample fall between the warning limit and control limit?

(H)Find If the process suddenly shifts to a mean value of 4, what is the probability of having two samples in a row fall between warning limit and control limit? ( keep three decimal points.

(E)Find If the process suddenly shifts to a mean value of 4, what is the probability that the process is stopped due to violation of the aforementioned control chart rules? ( keep three decimal places)

Explanation / Answer

Variable X can be modeled with a Poisson distribution.

A.

UWL=c+raíz ( c ) x 2=3.949

LWL=c-raíz ( c ) x 2=-0.949

B.

First, lets calculate upper control limit

UCL=c+raíz ( c ) x 3=5.174

After that, let’s calculate the probability of X higher thant UCL if the new mean is 4. This can be done with Excel with this formula:

=1-POISSON.DIST(UCL,4,1)

= 0.215

C.

So, P(x>UWL) =1-POISSON.DIST(UWL,4,1) = 0.567 -> lets call this F

From previous section (B) we know P(x>UCL)=0.215 -> lets call this G

So, P (UCW<x<UCL)= F-G =0.352

H.

Because probabilities of defects are independent, just multiply:

0.351 x 0.351 = 0.124

E.

Two possibilities, 2 in a row between UCL and UWL; orr 1 upper UCL

The first probabibility was just calculated: 0.124

The latter was calculated in section B: 0.215

Finally, the sum would be the answer: 0.339

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