contestada

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 = 5? Repeat these calculations for the cases where m = 10, m = 20, m = 30, and m = 50.

Required:
Discuss the results that you have obtained.

Respuesta :

Answer:

The solution to the issue is outlined in the following portion of the summary.

Step-by-step explanation:

The given value is:

p = 0.0027

When m = 5,

⇒ [tex]P (x \geq 1) = 1 - P (x = 0)[/tex]

                   [tex]=1 - 5C0 (0.0027)^0 (1 - 0.0027)^5[/tex]

                   [tex]= 1 - 0.9866[/tex]

                   [tex]=0.0134[/tex]

When m = 10,

⇒ [tex]P (x \geq 1) = 1 - P (x = 0)[/tex]

                   [tex]=1 - 10C0 (0.0027)^0 (1 - 0.0027)^{10}[/tex]

                   [tex]=1 - 0.9733[/tex]

                   [tex]= 0.0267[/tex]

When m = 20,

⇒ [tex]P (x \geq 1) = 1 - P (x = 0)[/tex]

                   [tex]=1 - 20C0 (0.0027)^0 (1 - 0.0027)^{20}[/tex]

                   [tex]=1 - 0.9474[/tex]

                   [tex]=0.0526[/tex]

When m = 30,

⇒ [tex]P (x \geq 1) = 1 - P (x = 0)[/tex]

                   [tex]=1 - 30C0 (0.0027)^0 (1 - 0.0027)^{30}[/tex]

                   [tex]=1 - 0.9221[/tex]

                   [tex]=0.0779[/tex]

When m = 50,

⇒ [tex]P (x \geq 1) = 1 - P (x = 0)[/tex]

                   [tex]= 1 - 50C0 (0.0027)^0 (1 - 0.0027)^{50}[/tex]

                   [tex]= 1 - 0.8736[/tex]

                   [tex]=0.1264[/tex]

Otras preguntas