Question: ( 4 Marks = 1 + 1 + 1 + 1 ) 3 . The mean and standard deviation of the above 2 1 observations

(4 Marks=1+1+1+1)
3.
The mean and standard deviation of the above 21 observations are 2.7757 ml and 0.0553 ml, respectively.
a A histogram of the above 21 observations is displayed below. Is the process capable? Why? [note that there is only one specification limit.]
The process is capable as the Mean of the total is higher than 2.5ml
b. Calculate the . Is the process capable? Why?
LSL =2.5, there is no USL
mean =2.7757
Standard devation =0.0553
Cpk =(2.7757-2.5)/(3X0.0553)=1.66
A process having Cpk greater than 1.33 is capable. So this process is capable
c. Using the A2 value (in factors table), determine the upper and lower limits for sample mean (X-bar) control chart. Is the process in control?
d. Using the D3 and D4 values, determine the upper and lower limits for sample range (R) control chart.
Is the process in control?
N A2 D3 D4
21.8803.27
31.0202.57
40.7302.28
50.5802.28
60.4802.00
70.420.081.92
80.370.141.86
90.340.181.82
100.310.221.78
110.290.261.74

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