6.48. An Xbar chart with three-sigma limits has parameters as follows: UCL = 104

Question

6.48. An Xbar chart with three-sigma limits has parameters as follows:

UCL = 104

Center line = 100

LCL = 96

n= 5

Suppose the process quality characteristic being controlled is normally distributed with a true mean of 98 and a standard deviation of 8. What is the probability that the control chart would exhibit lack of control by at least the third point plotted?

I know the answer and how to find Beta, but i'm confused:

Beta = chart doesn't signal; 1-Beta = chart does signal. So, the answer is 1 - (beta)^3. I was thinking that I could do (1-beta)^3 can you explain to me why I cannot do it the second way? Thank you.

Explanation / Answer

This is a result from binomial modeling which follows the following derivation -

Prob(at leat one of the next 3 points will fall outside the control limits)
= 1 - Prob(none of the next 3 points will fall outside the control limits)
= 1 - BINOM.DIST(0,3,p,1)
= 1 - 3C0 x (p)0 x (1 - p)3-0
= 1 - 1 x 1 x (1 - p)3
= 1 - (1 - p)3

Where, p = P(the sample falls outside of the control chart range) = 1 - P(conclude in control when out of control)

In other words, p = 1 - beta

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