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kvasek [131]
4 years ago
8

C-Spec, Inc., is attempting to determine whether an existing machine is capable of milling an engine part that has a key specifi

cation of {eq}4 \pm 0.003 {/eq} inches. After a trial run on this machine, C-Spec has determined that the machine has a sample mean of 4.001 inches with a standard deviation of 0.002 inch.
a. Calculate the {eq}C_{pk} {/eq} for this machine
b. Should C-Spec use this machine to produce this part? Why?
Mathematics
1 answer:
Mama L [17]4 years ago
3 0

Answer:

Step-by-step explanation:

Given that :

sample mean = 4.001 inches

sample standard deviation = 0.002 inch

a.

C_{pk} = min [\dfrac{(USL- \bar X)}{(3*std \ dev)} \ ; \  \dfrac{ (\bar X- LSL)}{(3*std \ dev)}]

specification = 4 \pm 0.003

Upper specification limit USL = 4 + 0.003  = 4.003

Lower specification limit LSL =  4 - 0.003 = 3.997

\dfrac{ (\bar X- LSL)}{(3*std  \ dev)} =  \dfrac{ (4.001-3.997)}{(3*0.002)}  = 0.667

\dfrac{(USL- \bar X)}{(3*std \ dev)} =\dfrac{(4.003-4.001)}{(3*0.002)}= 0.333

Thus ;

C_{pk} =min (0.333 , 0.667)  = 0.333

C_{pk}  is a measure of closeness to one's target and the consistency around the average performance.

b) No, C - spec should not use this machine to produce this part because C_{pk} <  1.33 which typical means that the part is not fully capable of hitting the target specification on a consistent basis .

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