A company making tires for bikes is concerned about the exact width of their cyclocross tires. The company has a lower specification limit of 22.8 mm and an upper specification limit of 23.2 mm. The standard deviation is 0.15 mm and the mean is 23 mm. (Round your answer to 4 decimal places.) What is the process capability index for the process

For this process, the capability index is 0.4444

Step-by-step explanation:

Given the data in the question;

Lower specification limit = 22.8 mm

Upper specification limit = 23.2 mm

The standard deviation σ = 0.15 mm

Mean μ = 23

capability index Cpk = ?

Now, Using the formula;

Cpk = min{ [(USL - μ)/3σ], [(μ - LSL)/3σ]

where USL is the Upper specification limit, LSL is the Lower specification limit, σ is the standard deviation, μ is the mean

so we substitute

Cpk = min{ [(23.2 - 23 )/(3×0.15)], [(23 - 22.8)/(3×0.15)] }

Cpk = min{ [0.2 / 0.45 ], [0.2 / 0.45] }

Cpk = min{ [ 0.4444 ], [0.4444] }

Therefore, for this process, the capability index is 0.4444

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step-by-step explanation: