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A true Six Sigma process aims for a centered process with Cpk = 1.5. That is, while keeping the process mean at the center (of the specification limits), it also puts the specification limits in 4.5 standard deviations away on both sides.

One hundred insurance claim forms are inspected daily at Full Life Insurance Co. over 25 working days, and the number of forms with errors have been recorded in the worksheet Prob. 8-42 in the C08Data.xlsx (In Lecture Note section, textbook data folder) file. Use 3 as the control distance.

When the process mean is 0, the process standard deviation is 1, the sample size is 64, and the control distance is 2.56, what is the probability of detecting a mean shift of +0.5 (therefore, the new mean is 0.5, SD unchanged.) To simplify the problem, we assume the population distribution will follow a normal distribution, therefore heuristic values wonÃ¢â‚¬â„¢t be needed.

(Hint: Although this problem does not give you specific value for important process parameters, such as process mean, standard deviation, specification limits, you can still solve it by relying on the information provided above. For example, you can substitute with your own values for those parameters.)

Calculate the DPMO in this case. (Note: round the answer to the next integer.)

Please re-calculate the DPMO when the process mean is shifted to the right by 1.5 standard deviation. (Note: round the answer to the next integer.)

Now, letÃ¢â‚¬â„¢s compare the process with a centered process with Cpk = 1.00. What is the DPMO for this new process?

When a system operates on Cpk = 1.5, even the mean shifts within Ã‚Â± 1.5 standard deviation, it still generates less defects than a system with Cpk = 1.0. Please discuss what are two important factors that allow you to have such a big Cpk. (Hint: think about the equation for Cpk).

Construct a p-chart. If any points occur outside the control limits, assume that assignable causes have been determined. Then construct a revised chart. Copy and paste your Excel generate control chart as a

picture

below.

Using the same data, construct an np-chart. Copy and paste your Excel generate control chart as a

picture

below.

Does the np-chart detect something that p-chart did not? Intuitively, what is the advantage of using an np-chart?

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