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
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