NO EXCEL PLEASE. NEED it in WRITING NO EXCEL PLEASE. NEED it in WRITING 4. A nor
ID: 353052 • Letter: N
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NO EXCEL PLEASE. NEED it in WRITING
NO EXCEL PLEASE. NEED it in WRITING
4. A normally distributed quality characteristic is monitored through use of an X and R chart. These charts have the following parameters (n-3): X Chart UCL 625 R Chart UCL- 18.75 CL 620 LCL = 615 Both charts exhibit control. CL= 8.25 LCL= 0 a. What are the estimated mean () and standard deviation (6) of the process? (5 pts) b. Suppose an S chart were to be substituted for the R chart. What are the appropriate parameters of the X and S charts? (10 pts.) what are your conclusions regarding the ability of the process to produce items within these specifications? (5 pts.) 14, what is the process capability ratio and on the product were 610 c. If specificationsExplanation / Answer
(a)
The process mean is estimated through the sample mean only. Therefore them estimate for is the central line (CLx) of the x_bar chart i.e. 620.
An unbiased estimate of the process standard deviation () is calculated by dividing R_bar with d2. Here, 'd2' is the standard coefficient found from the 3-sigma factor table for X_bar/R charts; For n=3, d2 = 1.693
So, an estimate of = R_bar / d2 = CLR/d2 = 8.25/1.693 = 4.873
(b)
For constructing the S-Chart, the following parameters need to be noted from the standard table -
For n=3,
B3 = 0
B4 = 2.568
C4 = 0.8862
Estimate of process standard deviation () can also be found by the ratio "S_bar / C4"
So, S_bar / C4 = 4.873
or, S_bar = 4.873 x C4 = 4.873 x 0.8862 = 4.3185; This will be the central line of the S-Chart. So, the relevant parameters of the S-Chart are -
UCLs = B4*S_bar = 2.568 x 4.3185 = 11.09
CLs = 4.3185
LCLs = B3*S_bar = 0
(c)
USL = 610+14 = 624; LSL = 610 - 14 = 596
First of all, note that the center of the process (620) is not matching with the center of the LSL=596, USL=624 range. So, this is an off-centered process. Therefore, we will rather calculate Cpk in place of Cp to find the process capability.
Cpk = Min{(USL - )/3, ( - LSL)/3} = Min((624 - 620)/(3*4.873), (620 - 596)/(3*4.873)) = 0.2736
Since Cpk is less than 1.0, we conclude that the process is incapable of producing within the specification limit.
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