Question

A process is normally distributed and in control, with known mean and variance, and the usual three-sigma limits are used on the 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

Answer 1

Answer:

Check the explanation

Step-by-step explanation:

Ans=

A: For m = 5: P(³≥1) = 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)

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Going by the question and the explanation above, we derived sample values of the mean as well as standard deviation in calculating our probability, since that is the necessary value in determining the probability of an out-of-bounds point being plotted. Furthermore, we would know that that value for the possibility would likely be a poor es²ma²on, cas²ng doubt on anycalcula²ons we made using those values