1 2 End PpDn Ctrl 0 Ins Del Branna Steele INEN 395- Quality Control - Spring 201

Question

1 2 End PpDn Ctrl 0 Ins Del Branna Steele INEN 395- Quality Control - Spring 2017 Classwork 10 Chapter 8-Review-October 20, 2017 4. Exercise A process has demonstrated that, when held in control, it can maintain a st certain part has specifications of 15.00 ± 0. range. Is this process capable? Also calculate the process capability indexes CP and CPK. Assume the actual mean setting is 14.90. andard deviation of 0.15 cm. A 50 cm. Test the natural process spread (60) against the specification Solution

Explanation / Answer

Mean actual = 14.90 cm

Standard deviation = 0.15 cm

For specifications Central line or mean sepcification = 15.00 cm

Lower Specification limit (LSL)= 14.50 cm

Upper Specification Limit (USL)= 15.50 cm

Here CP = [UCL - LCL]/ 6 = (15.50 - 14.50)/ (6 * 0.15) = 1.111

Cp(Lower) = [Mean - LCL]/ 3 = [14.90 - 14.50] / (3 * 0.15) = 0.89

Cp(Upper) = [UCL - Mean]/ 3 = [15.50 - 14.90] / (3 * 0.15) = 1.33

Cpk = Min [Cp(Lower) , Cp(Upper) ] = 0.89

Now we have to test that the natural process spread is against the specification range or not.

Pr[14.50 < X < 15.50] = (Z1) - (Z2)

Z1 = (15.50 - 14.90)/0.15 = 4  

Z2 = (14.50 - 14.90)/ 0.15 = -2.67

so Total specification limit = 4 - (-2.67) = 6.67

so yes, the natural process spread against the specification limit.

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