Homework Exercise 6

(2a).
CL = (0.01 + 0.02 + 0.02 + 0.00 + 0.01 + 0.03 + 0.02 + 0.01 + 0.00 + 0.04 + 0.03 + 0.02) / 12
= 0.21 / 12
= 0.0175

UCL = 0.0175 + 3 √[(0.0175 x 0.9825)/400 ]
= 0.0175 + 3√(0.0172 / 400)
= 0.0372


LCL = 0.0175 - 3 √[(0.0175 x 0.9825)/400 ]
= 0.0175 - 3√(0.01719375 / 400)
= 0.0175 - 0.0393
= -0.0022, this is negative so LCL is 0.

(2b).
If there are 6 letters that are defective, 6/400 = 0.015. This is within the control limits. The process is still in control.

(6a).
X bar: CL = 30, UCL = 32.415, LCL = 27.585
R bar: CL = 5, UCL = 10.02, LCL = 0

(6b).
The sample mean = (38 + 35 + 27 + 30 + 33 + 32) / 6 = 195 /6 = 32.5
The sample range = 38 - 27 = 11
The mean of 32.5 is above the upper control limit of 32.415. The range of 11 is above the upper control limit of 10.02.
The process is not in control. we stop the process and look for an assignable cause.

(12a).
Cp=1.67 and Cpk=0.56

(12b).
To improve the process Cpk, we shift the mean closer to the center of the specification range or reduce the standard deviation.