Question

Two machines are currently in use in a process at the Dennis Kira Mfg. Co. The standards for this process are LSL = .440 and USL = .444. Machine One is currently producing with mean = 441 and standard deviation .ooo4. Machine two is currently producing with mean .4415 and standard deviation .0005. Which machine has the higher capability index?

Answer 1

Machine One has a higher capability index than Machine Two.

Step-by-step explanation:

First, we need to calculate the process capability index (Cpk) for both Machine One and Machine Two. Cpk is a measure of how well a process can consistently produce output within the specified upper and lower limits. It is calculated using the formula Cpk = min((USL - mean)/3σ, (mean - LSL)/3σ), where USL is the upper specification limit, LSL is the lower specification limit, mean is the process mean, and σ is the standard deviation.

For Machine One:

Cpk = min((.444 - .441)/(3*0.0004), (.441 - .44)/(3*0.0004))

Cpk = min(3/0.0012, 0.002/0.0012)

Cpk = min(2500, 1.6667)

Cpk = 1.6667

For Machine Two:

Cpk = min((.444 - .4415)/(3*0.0005), (.4415 - .44)/(3*0.0005))

Cpk = min(1.1/0.0015, 0.0015/0.0015)

Cpk = min(733.3, 1)

Cpk = 1

Comparing the Cpk values for both machines, we can conclude that Machine One has a higher capability index than Machine Two.