Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Control charts for up x and R are maintained for an important quality characteri

ID: 3178223 • Letter: C

Question

Control charts for up x and R are maintained for an important quality characteristic. The sample size is n = 7; x and R are computed for each sample. After 35 samples, we have found that Sigma^35_i = 1 x_i = 7805, and Sigma^35_i = 1 R_i= 1200. (a) Set up x and R charts using the data. (b) Assuming that both charts exhibit control, estimate the process mean and standard deviation. (c) If the quality characteristic is normally distributed and if the specifications are 220 plusminus 35, can the process meet specifications? Estimate the fraction nonconforming. (d) Assuming the variance to remain constant, state where the process mean should be located to minimize the fraction nonconforming. What would be the value of the fraction nonconforming under these conditions?

Explanation / Answer

1 .

Read x’ – x bar , R’ – R bar

x’’ = x ‘ /k

= 7805 / 35 = 223 .

R ‘ = R /k = 1200/35 = 34.29 .

X ‘ chart

Central line = x’’

Ucl = x ‘’ + A2* R’

Lcl = x’’ – A2 * R’

A2 can be obtained from control chart table

= .419 for n = 7

X’’ = 223

Ucl = 237.4

Lcl = 208.59

R’ chart

Std deviation is unknown

Central line = R ‘

UCL = D4R’

LCL = D3R’ , D4 & D3 can be obtained from contro chart table corresponding to n= 7

R’=34.29

UCL = 65.84

LCL = 2.60604

X’ and R ‘ chart can be drawn easily . Each requires 3 straight ine

That is central line , UCL & LCL .

2.

Process mean x’’ = 223

Std deviation = 3z/n ; z = zigma

14.4 = 3z/7

Rearranging we get z = 12.69

3.

Specification of machine is 220 +/- 35 . here variation space provided is twice actual variation

And non conforming values are minimum . No adjustment required for machine at this moment

4.

Under tolerance region given to machine

Central line = 220

Lcl = 220-35 = 185

Ucl = 220 + 35 = 255

Tolerance of machine =

Central line = 223

Ucl = 208.59

Lcl = 237.4

So to minimize non conformities maintain present condition

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote