##### I need help with this problem asap

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Jul 20th, 2015

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$\bar{\bar{x}}$ is the average of averages.

$\bar{\bar{x}}=\frac{1}{k}\sum\bar{x}_i$

where k is the number of groups

and $\bar{x}_i$ is the average/mean of the i th group.

$\\ \bar{\bar{x}}=\frac{156.9+153.2+155.6+157.5+156.6}{5}\\ \\ \bar{\bar{x}}=155.96\hspace{5}mm$

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Jul 20th, 2015

yes it says round that problem to two decimal places

Jul 20th, 2015

## What are the UCLx and LCLx, using 3s? Plot the data. What are the UCLR and LCLR, using 3s? Plot the data.If the true diameter mean should be 155 mm and you want this as your center ( nominal) line, what are the new UCLx andLCLx?

Jul 20th, 2015

I need help with these problems that I just post up

Jul 20th, 2015

Jul 20th, 2015

i hope so let me try it

Jul 20th, 2015

I am not sure whether these are correct. These are based on my understanding of the problem solved in

$UCL_x = \bar{\bar{x}} + A_2\bar{R}=155.96+0.308\times4.52=157.35\hspace{5}mm$

$LCL_x = \bar{\bar{x}} - A_2\bar{R}=155.96-0.308\times4.52=154.57\hspace{5}mm$

The value for $A_2$ is obtained from the table in

$\\ UCL_R = D_4\bar{R}=1.777\times4.52=8.03\\ \\ LCL_R = D_3\bar{R}=0.223\times4.52=1.01\\$

The values for $D_3\hspace{5}and\hspace{5}D_4$ are obtained from the table in

Jul 20th, 2015

thank you thats correct

Jul 20th, 2015

and this question here

## If the true diameter mean should be 155 mm and you want this as your center ( nominal) line, what are the new UCLx andLCLx?

Jul 20th, 2015

Sorry I don't know how to solve this question.

Jul 20th, 2015

its okay you did great on the other ones thank you

Jul 20th, 2015

:)

Jul 20th, 2015

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Jul 20th, 2015
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Jul 20th, 2015
Oct 23rd, 2017
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