A process is normally distributed and in control, with known mean and variance,

Question

A process is normally distributed and in control, with known mean and variance, and the usual three-sigma limits are used on the x^bar control chart, so that the probability of a single point plotting outside the control limits when the process is in control is 0.0027. Suppose that this chart is being used in phase I and the averages from a set of m samples or subgroups from this process are plotted on this chart. What is the probability that at least one of the averages will plot outside the control limits when m = 5? Repeat these calculations for the cases where m = 10, m = 20, m = 30, and m = 50. Discuss the results that you have obtained. Reconsider the situation in Question 17. Suppose that the process mean and variance were unknown and had to be estimated from the data available from the m subgroups. What complications would this introduce in the calculations that you performed in Question 17?

Explanation / Answer

17)

Ans=

A: For m = 5: P(³1) = 1 – P(³=0) = 1 – 0.9973^5 = 0.0134

M = 10: 1 – 0.9973^10 = 0.0267

M = 20: 1 – 0.9973^20 = 0.0526

M = 30: 1 – 0.9973^30 = 0.0779

M = 50: 1 – 0.9973^50 = 0.126

18)

Ans=

We would have to derive sample values of mean and standard devia²on to calculate our probability,as that is the value necessary to determine the likelihood of an out-of-bounds point being plo±ed. Andwe would know that that value for probability would likely be a poor es²ma²on, cas²ng doubt on anycalcula²ons we made using those values

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