Question

3. There is a lawn mower used to cut a putting green. The putting green needs to be 6.53 inch long with a tolerance of +/- 1.50 inches

The lawn mower that mows the lawn has a mean of 7.70 inches with a standard deviation of 0.130 inches.

Compute the process capability ratio (Cp) and the process capability index (Cpk).

3a.) Cp = _____________ __________ (Go to 4decimal places), place answer on this line, show your work on next page. I want a number here, not a word.

3b.) Cpk = ______________________   (Go to 4 decimal places), place answer on this line, show your work on next page. I want a number here, not a word. [hint Cpk is not zero. Zero is not the answer]

3c.) Does the process meet 6 sigma quality? (yes or no).   

3d.) If not, then does the standard deviation have to be changed? In other words; is the “Cp” value not O.K? If Cp is not O.K., then your answer would be yes. If “Cp” is O.K. then answer would be no.(yes or no).                                              

3e.) If standard deviation has to be changed, what does it have to be changed to?________________

3f.) If process does not meet 6 sigma, does the mean have to be changed? In other words, is the Cpk value not O.K. If it is not O.K.; then your answer would be yes.   If Cpk is O.K. answer would be no. (yes or no).                     

3g).. If mean has to be changed, what does the mean have to be changed to in order to achieve six sigma quality.        

_________________________________ (go to four decimal places), place answer on this line, I want a number here, not a word.

Answer 1

Here USL = upper specification limit = 6.53+1.5 = 8.03

LSL = lower specification limit = 6.53-1.5 = 5.03

SD = standard deviation = 0.130

Mean = 7.7

a)

Cp = (USL – LSL)/6SD

Cp = (8.03-5.03)/6*0.130 = 3.8461

b)

Cpk = min (CpU and CpL)

CpU = (USL – mean)/3SD = (8.03-7.7)/3*0.130 = 0.8461

CpL = (Mean – LSL)/3SD = (7.7-5.03)/3*0.130 = 6.8461

Cpk = minimum of 0.8461 and 6.8461 = 0.8461

c)

No

d)

No

e)

Standard deviation does not need to be changed

f)

Yes

g)

The mean is currently at 7.7 and this is makes the CpU value to be less than 1. This is not acceptable under 6 sigma standards. Ideally the mean should be between at least below 8.03 – (3*0.130) = 7.64 or lower. In this case the CpU value will be 1 and this will make the CpK at least 1.

However, ideally the Cpk should be as high as possible and probably above 1.3. In that case the mean should be 8.03 – (3*0.130*1.33) = 7.5113 or lower in order for the process to be 6 sigma capable. However for the answer you can use 7.64