Marigold Corp. bought a machine on January 1, 2011 for $806000. The machine had an expected life of 20 years and was expected to
1 answer:
You might be interested in
Answer:
A) ≈ 3.509 grams
B) The process is not capable because the lowest Cpk which is 1.89 ≈ 1.9 is far from the ideal 1.33 capability index number ( NO )
c) 1.051 ounces
Explanation:
A) The largest standard deviation ( in grams)
This can be calculated applying the capability index formula and according to the capability index formula the value at which a process is capable is at : 1.33
hence the largest standard deviation = upper limit - lower limit / 6 * 1.33
( 354 - 326 ) / 7.98 = 3.5087 ≈ 3.509 grams
B) we first calculate the process mean and std of the box
standard deviation =
std of the six-bar box = = 2.106
process mean = 6 * ( 1.03 * 28.33) = 175.079
the mean is not centered between the upper specification and the lower specification hence we will apply the formulae used in calculating the capability index for an uncentered process
Cpk = (upper value - process mean) / (3 * std of box),
= (187 - 175.079) / ( 3 * 2.106) = 11.921 / 6.318 = 1.89
Cpk = (process mean - lower value ) / (3*std of box)
= ( 175.079 - 153 ) / (3 * 2.106) = 22.079 / 6.318 = 3.49
The process is not capable because the lowest Cpk which is 1.89 ≈ 1.9 is far from the ideal 1.33 capability index number ( N0 )
c ) lowest setting
calculate the value of the mean using capability index of 1.33
Cpk = (upper value - mean ) / 3 * std
mean = 187 - 1.33 * 3 * 2.106
= 187 - 8.40 = 178.60 grams
Cpk = ( process mean - lower value ) / 3 * std
mean = 1.33 * 3 * 2.106 + 153
= 8.40 + 153 = 161.40 grams
the lowest setting in ounces
= 178.60 / ( 6 bar * 28 .33)
= 178.60 / 169.98 = 1.051 ounces