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 arc used on the x-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 chan. 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.

Explanation / Answer

Here we have to find the probability that at least one of the averages will plot outside the control limits.

Here random variable X follows Binomial distribution with sample size is m (i.e. 5,10,20,30,50) with probability is 0.0027.

When m=5 :

P(X >=1) = 1 - P(X=0)

= 1 - (5 C 0) * (0.0027)^0 * (1 - 0.0027)^(5-0) where C is used for combination

= 1 - 0.9866

= 0.01343

When m=10,

P(X >=1) = 1 - P(X=0)

= 1 - (10 C 0) * (0.0027)^0 * (1 - 0.0027)^(10-0) where C is used for combination

= 1 - 0.9733

= 0.0267

When m=20,

P(X >=1) = 1 - P(X=0)

= 1 - (20 C 0) * (0.0027)^0 * (1 - 0.0027)^(20-0) where C is used for combination

= 1 - 0.9474

= 0.0526

When m = 30,

P(X >=1) = 1 - P(X=0)

= 1 - (30C 0) * (0.0027)^0 * (1 - 0.0027)^(30-0) where C is used for combination

= 1 - 0.9221

= 0.0779

And for m = 50,

P(X >=1) = 1 - P(X=0)

= 1 - (50 C 0) * (0.0027)^0 * (1 - 0.0027)^(50-0) where C is used for combination

= 1 - 0.8736

= 0.1264

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